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Astronomy_1007

The event horizon of a non-spinning, uncharged black hole can be thought of as a sphere with a radius equal to the Schwarzschild radius, $r_{S}=2 G M / c^{2}$ where $M$ is the mass of the black hole and $c$ is the speed of light. If the black hole at the centre of the Milky Way has a mass of $4.15 \times 10^{6} M_{\odot}$, what is the approximate average density within the event horizon? A: $\sim 1 \mathrm{~kg} \mathrm{~m}^{-3}$ B: $\sim 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ C: $\sim 10^{6} \mathrm{~kg} \mathrm{~m}^{-3}$ D: $\sim 10^{9} \mathrm{~kg} \mathrm{~m}^{-3}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: The event horizon of a non-spinning, uncharged black hole can be thought of as a sphere with a radius equal to the Schwarzschild radius, $r_{S}=2 G M / c^{2}$ where $M$ is the mass of the black hole and $c$ is the speed of light. If the black hole at the centre of the Milky Way has a mass of $4.15 \times 10^{6} M_{\odot}$, what is the approximate average density within the event horizon? A: $\sim 1 \mathrm{~kg} \mathrm{~m}^{-3}$ B: $\sim 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ C: $\sim 10^{6} \mathrm{~kg} \mathrm{~m}^{-3}$ D: $\sim 10^{9} \mathrm{~kg} \mathrm{~m}^{-3}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

text-only

Astronomy_951

The very first image released by the James Webb Space Telescope (JWST) was of a galaxy cluster called SMACS 0723. The image is considered to be Webb's first deep field, since a long exposure time of 12.5 hours was used to allow the light from very faint and distant galaxies to be seen. The spectrum of one such galaxy is shown in Figure 2. [figure1] Figure 2: Highly redshifted emission lines in the spectrum of a galaxy that is 13.1 billion years old, captured using the JWST's near-infrared spectrometer (NIRSpec). Credit: NASA, ESA, CSA, STScI. The spectrum shows four bright hydrogen lines, which are part of the Balmer series (some of which are normally seen in the visible). The rest frame wavelengths of the longest four lines in the series are $410 \mathrm{~nm}, 434 \mathrm{~nm}, 486 \mathrm{~nm}$ and $656 \mathrm{~nm}$ (not all of which are visible in the spectrum). Once a redshift is known, its recessional velocity can be calculated. At very high redshifts, such as these, General Relativity must be used. A conversion from redshift to recessional velocity is shown in Figure 3. [figure2] Figure 3: Conversion from redshift to recessional velocity for a linear approximation $(v=z c)$, using Special Relativity $\left(v=c \frac{(1+z)^{2}-1}{(1+z)^{2}+1}\right)$, and using General Relativity $\left(v=\dot{a}(z) \int_{0}^{z} \frac{c d z^{\prime}}{H\left(z^{\prime}\right)}\right)$. The grey area corresponds to a variety of values for cosmological parameters. The solid line corresponds to values approximately the same as the current measured cosmological parameters. Credit: Davis \& Lineweaver (2001). Taking measurements from the spectrum, estimate the redshift of the galaxy. [Hint: you should measure more than one line to ensure you correctly identify which rest frame wavelength corresponds to which line.]

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: The very first image released by the James Webb Space Telescope (JWST) was of a galaxy cluster called SMACS 0723. The image is considered to be Webb's first deep field, since a long exposure time of 12.5 hours was used to allow the light from very faint and distant galaxies to be seen. The spectrum of one such galaxy is shown in Figure 2. [figure1] Figure 2: Highly redshifted emission lines in the spectrum of a galaxy that is 13.1 billion years old, captured using the JWST's near-infrared spectrometer (NIRSpec). Credit: NASA, ESA, CSA, STScI. The spectrum shows four bright hydrogen lines, which are part of the Balmer series (some of which are normally seen in the visible). The rest frame wavelengths of the longest four lines in the series are $410 \mathrm{~nm}, 434 \mathrm{~nm}, 486 \mathrm{~nm}$ and $656 \mathrm{~nm}$ (not all of which are visible in the spectrum). Once a redshift is known, its recessional velocity can be calculated. At very high redshifts, such as these, General Relativity must be used. A conversion from redshift to recessional velocity is shown in Figure 3. [figure2] Figure 3: Conversion from redshift to recessional velocity for a linear approximation $(v=z c)$, using Special Relativity $\left(v=c \frac{(1+z)^{2}-1}{(1+z)^{2}+1}\right)$, and using General Relativity $\left(v=\dot{a}(z) \int_{0}^{z} \frac{c d z^{\prime}}{H\left(z^{\prime}\right)}\right)$. The grey area corresponds to a variety of values for cosmological parameters. The solid line corresponds to values approximately the same as the current measured cosmological parameters. Credit: Davis \& Lineweaver (2001). Taking measurements from the spectrum, estimate the redshift of the galaxy. [Hint: you should measure more than one line to ensure you correctly identify which rest frame wavelength corresponds to which line.] All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value.

NV

Astronomy

EN

multi-modal

Astronomy_1031

The Chandra X-ray Observatory celebrates 20 years of operation this year, and was 100 times more sensitive than previous X-ray telescopes when it was launched. All X-ray telescopes have either been space-borne or operate in near-space environments. This is because: A: X-rays cannot penetrate the Earth's atmosphere all the way to the ground B: on the ground there is too much interference from medical X-rays C: it is dangerous to be close to an X-ray telescope so it must be highly remote from human life D: the resolution of the telescope would be too poor for astronomical observations from the ground

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: The Chandra X-ray Observatory celebrates 20 years of operation this year, and was 100 times more sensitive than previous X-ray telescopes when it was launched. All X-ray telescopes have either been space-borne or operate in near-space environments. This is because: A: X-rays cannot penetrate the Earth's atmosphere all the way to the ground B: on the ground there is too much interference from medical X-rays C: it is dangerous to be close to an X-ray telescope so it must be highly remote from human life D: the resolution of the telescope would be too poor for astronomical observations from the ground You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

text-only

Astronomy_159

双中子星合并会产生引力波, 探测到引力波的存在, 可以证明爱因斯坦义相对论引力波预言的正确性。两颗中子星合并前, 绕二者连线上的某点转动, 将两颗中子星都看作是质量分布均匀的球体。若产生引力波的双中子星到地球的距离为 13.4 亿光年, 引力波的速度等于真空中的光速, 则下列关于双中子星合并产生引力波的描述正确的是 A: 该引力波传到地球的时间约为 13.4 年 B: 该引力波传到地球的时间约为 13.4 亿年 C: 双中子星合并过程中, 其运行周期逐渐变大 D: 双中子星合并过程中, 其运行周期逐渐变小

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 双中子星合并会产生引力波, 探测到引力波的存在, 可以证明爱因斯坦义相对论引力波预言的正确性。两颗中子星合并前, 绕二者连线上的某点转动, 将两颗中子星都看作是质量分布均匀的球体。若产生引力波的双中子星到地球的距离为 13.4 亿光年, 引力波的速度等于真空中的光速, 则下列关于双中子星合并产生引力波的描述正确的是 A: 该引力波传到地球的时间约为 13.4 年 B: 该引力波传到地球的时间约为 13.4 亿年 C: 双中子星合并过程中, 其运行周期逐渐变大 D: 双中子星合并过程中, 其运行周期逐渐变小 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_917

The three stars that make up Orion's belt are Alnitak, Alnilam and Mintaka, with apparent magnitudes, $m$, of $1.77,1.69$ and 2.23 respectively. What is the ratio of the apparent brightnesses of the two brightest stars? [figure1] A: 0.655 B: 1.076 C: 1.528 D: 1.644

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: The three stars that make up Orion's belt are Alnitak, Alnilam and Mintaka, with apparent magnitudes, $m$, of $1.77,1.69$ and 2.23 respectively. What is the ratio of the apparent brightnesses of the two brightest stars? [figure1] A: 0.655 B: 1.076 C: 1.528 D: 1.644 You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

multi-modal

Astronomy_882

Take a look at the following image: [figure1] Three Messier objects are circled in the image. Select the alternative that correctly matches each object with its type. A: 1 - Open cluster; 2 - Open cluster; 3 - Nebula. B: 1 - Open Cluster; 2 - Nebula; 3 - Galaxy. C: 1 - Galaxy; 2 - Nebula; 3 - Globular cluster. D: 1 - Open cluster; 2 - Galaxy; 3 - Globular cluster. E: 1 - Open cluster; 2 - Nebula; 3 - Open cluster.

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Take a look at the following image: [figure1] Three Messier objects are circled in the image. Select the alternative that correctly matches each object with its type. A: 1 - Open cluster; 2 - Open cluster; 3 - Nebula. B: 1 - Open Cluster; 2 - Nebula; 3 - Galaxy. C: 1 - Galaxy; 2 - Nebula; 3 - Globular cluster. D: 1 - Open cluster; 2 - Galaxy; 3 - Globular cluster. E: 1 - Open cluster; 2 - Nebula; 3 - Open cluster. You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

multi-modal

Astronomy_1140

The Event Horizon Telescope (EHT) is a project to use many widely-spaced radio telescopes as a Very Long Baseline Interferometer (VBLI) to create a virtual telescope as big as the Earth. This extraordinary size allows sufficient angular resolution to be able to image the space close to the event horizon of a super massive black hole (SMBH), and provide an opportunity to test the predictions of Einstein's theory of General Relativity (GR) in a very strong gravitational field. In April 2017 the EHT collaboration managed to co-ordinate time on all of the telescopes in the array so that they could observe the SMBH (called M87*) at the centre of the Virgo galaxy, M87, and they plan to also image the SMBH at the centre of our galaxy (called Sgr A*). [figure1] Figure 3: Left: The locations of all the telescopes used during the April 2017 observing run. The solid lines correspond to baselines used for observing M87, whilst the dashed lines were the baselines used for the calibration source. Credit: EHT Collaboration. Right: A simulated model of what the region near an SMBH could look like, modelled at much higher resolution than the EHT can achieve. The light comes from the accretion disc, but the paths of the photons are bent into a characteristic shape by the extreme gravity, leading to a 'shadow' in middle of the disc - this is what the EHT is trying to image. The left side of the image is brighter than the right side as light emitted from a substance moving towards an observer is brighter than that of one moving away. Credit: Hotaka Shiokawa. Some data about the locations of the eight telescopes in the array are given below in 3-D cartesian geocentric coordinates with $X$ pointing to the Greenwich meridian, $Y$ pointing $90^{\circ}$ away in the equatorial plane (eastern longitudes have positive $Y$ ), and positive $Z$ pointing in the direction of the North Pole. This is a left-handed coordinate system. | Facility | Location | $X(\mathrm{~m})$ | $Y(\mathrm{~m})$ | $Z(\mathrm{~m})$ | | :--- | :--- | :---: | :---: | :---: | | ALMA | Chile | 2225061.3 | -5440061.7 | -2481681.2 | | APEX | Chile | 2225039.5 | -5441197.6 | -2479303.4 | | JCMT | Hawaii, USA | -5464584.7 | -2493001.2 | 2150654.0 | | LMT | Mexico | -768715.6 | -5988507.1 | 2063354.9 | | PV | Spain | 5088967.8 | -301681.2 | 3825012.2 | | SMA | Hawaii, USA | -5464555.5 | -2492928.0 | 2150797.2 | | SMT | Arizona, USA | -1828796.2 | -5054406.8 | 3427865.2 | | SPT | Antarctica | 809.8 | -816.9 | -6359568.7 | The minimum angle, $\theta_{\min }$ (in radians) that can be resolved by a VLBI array is given by the equation $$ \theta_{\min }=\frac{\lambda_{\mathrm{obs}}}{d_{\max }}, $$ where $\lambda_{\text {obs }}$ is the observing wavelength and $d_{\max }$ is the longest straight line distance between two telescopes used (called the baseline), assumed perpendicular to the line of sight during the observation. An important length scale when discussing black holes is the gravitational radius, $r_{g}=\frac{G M}{c^{2}}$, where $G$ is the gravitational constant, $M$ is the mass of the black hole and $c$ is the speed of light. The familiar event horizon of a non-rotating black hole is called the Schwartzschild radius, $r_{S} \equiv 2 r_{g}$, however this is not what the EHT is able to observe - instead the closest it can see to a black hole is called the photon sphere, where photons orbit in the black hole in unstable circular orbits. On top of this the image of the black hole is gravitationally lensed by the black hole itself magnifying the apparent radius of the photon sphere to be between $(2 \sqrt{3+2 \sqrt{2}}) r_{g}$ and $(3 \sqrt{3}) r_{g}$, determined by spin and inclination; the latter corresponds to a perfectly spherical non-spinning black hole. The area within this lensed image will appear almost black and is the 'shadow' the EHT is looking for. [figure2] Figure 4: Four nights of data were taken for M87* during the observing window of the EHT, and whilst the diameter of the disk stayed relatively constant the location of bright spots moved, possibly indicating gas that is orbiting the black hole. Credit: EHT Collaboration. The EHT observed M87* on four separate occasions during the observing window (see Fig 4), and the team saw that some of the bright spots changed in that time, suggesting they may be associated with orbiting gas close to the black hole. The Innermost Stable Circular Orbit (ISCO) is the equivalent of the photon sphere but for particles with mass (and is also stable). The total conserved energy of a circular orbit close to a non-spinning black hole is given by $$ E=m c^{2}\left(\frac{1-\frac{2 r_{g}}{r}}{\sqrt{1-\frac{3 r_{g}}{r}}}\right) $$ and the radius of the ISCO, $r_{\mathrm{ISCO}}$, is the value of $r$ for which $E$ is minimised. We expect that most black holes are in fact spinning (since most stars are spinning) and the spin of a black hole is quantified with the spin parameter $a \equiv J / J_{\max }$ where $J$ is the angular momentum of the black hole and $J_{\max }=G M^{2} / c$ is the maximum possible angular momentum it can have. The value of $a$ varies from $-1 \leq a \leq 1$, where negative spins correspond to the black hole rotating in the opposite direction to its accretion disk, and positive spins in the same direction. If $a=1$ then $r_{\text {ISCO }}=r_{g}$, whilst if $a=-1$ then $r_{\text {ISCO }}=9 r_{g}$. The angular velocity of a particle in the ISCO is given by $$ \omega^{2}=\frac{G M}{\left(r_{\text {ISCO }}^{3 / 2}+a r_{g}^{3 / 2}\right)^{2}} $$ [figure3] Figure 5: Due to the curvature of spacetime, the real distance travelled by a particle moving from the ISCO to the photon sphere (indicated with the solid red arrow) is longer than you would get purely from subtracting the radial co-ordinates of those orbits (indicated with the dashed blue arrow), which would be valid for a flat spacetime. Relations between these distances are not to scale in this diagram. Credit: Modified from Bardeen et al. (1972). The spacetime near a black hole is curved, as described by the equations of GR. This means that the distance between two points can be substantially different to the distance you would expect if spacetime was flat. GR tells us that the proper distance travelled by a particle moving from radius $r_{1}$ to radius $r_{2}$ around a black hole of mass $M$ (with $r_{1}>r_{2}$ ) is given by $$ \Delta l=\int_{r_{2}}^{r_{1}}\left(1-\frac{2 r_{g}}{r}\right)^{-1 / 2} \mathrm{~d} r $$d. Determine risco for a non-spinning black hole. Give your answer in units of $\mathrm{r}_{\mathrm{g}}$.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is an expression. Here is some context information for this question, which might assist you in solving it: The Event Horizon Telescope (EHT) is a project to use many widely-spaced radio telescopes as a Very Long Baseline Interferometer (VBLI) to create a virtual telescope as big as the Earth. This extraordinary size allows sufficient angular resolution to be able to image the space close to the event horizon of a super massive black hole (SMBH), and provide an opportunity to test the predictions of Einstein's theory of General Relativity (GR) in a very strong gravitational field. In April 2017 the EHT collaboration managed to co-ordinate time on all of the telescopes in the array so that they could observe the SMBH (called M87*) at the centre of the Virgo galaxy, M87, and they plan to also image the SMBH at the centre of our galaxy (called Sgr A*). [figure1] Figure 3: Left: The locations of all the telescopes used during the April 2017 observing run. The solid lines correspond to baselines used for observing M87, whilst the dashed lines were the baselines used for the calibration source. Credit: EHT Collaboration. Right: A simulated model of what the region near an SMBH could look like, modelled at much higher resolution than the EHT can achieve. The light comes from the accretion disc, but the paths of the photons are bent into a characteristic shape by the extreme gravity, leading to a 'shadow' in middle of the disc - this is what the EHT is trying to image. The left side of the image is brighter than the right side as light emitted from a substance moving towards an observer is brighter than that of one moving away. Credit: Hotaka Shiokawa. Some data about the locations of the eight telescopes in the array are given below in 3-D cartesian geocentric coordinates with $X$ pointing to the Greenwich meridian, $Y$ pointing $90^{\circ}$ away in the equatorial plane (eastern longitudes have positive $Y$ ), and positive $Z$ pointing in the direction of the North Pole. This is a left-handed coordinate system. | Facility | Location | $X(\mathrm{~m})$ | $Y(\mathrm{~m})$ | $Z(\mathrm{~m})$ | | :--- | :--- | :---: | :---: | :---: | | ALMA | Chile | 2225061.3 | -5440061.7 | -2481681.2 | | APEX | Chile | 2225039.5 | -5441197.6 | -2479303.4 | | JCMT | Hawaii, USA | -5464584.7 | -2493001.2 | 2150654.0 | | LMT | Mexico | -768715.6 | -5988507.1 | 2063354.9 | | PV | Spain | 5088967.8 | -301681.2 | 3825012.2 | | SMA | Hawaii, USA | -5464555.5 | -2492928.0 | 2150797.2 | | SMT | Arizona, USA | -1828796.2 | -5054406.8 | 3427865.2 | | SPT | Antarctica | 809.8 | -816.9 | -6359568.7 | The minimum angle, $\theta_{\min }$ (in radians) that can be resolved by a VLBI array is given by the equation $$ \theta_{\min }=\frac{\lambda_{\mathrm{obs}}}{d_{\max }}, $$ where $\lambda_{\text {obs }}$ is the observing wavelength and $d_{\max }$ is the longest straight line distance between two telescopes used (called the baseline), assumed perpendicular to the line of sight during the observation. An important length scale when discussing black holes is the gravitational radius, $r_{g}=\frac{G M}{c^{2}}$, where $G$ is the gravitational constant, $M$ is the mass of the black hole and $c$ is the speed of light. The familiar event horizon of a non-rotating black hole is called the Schwartzschild radius, $r_{S} \equiv 2 r_{g}$, however this is not what the EHT is able to observe - instead the closest it can see to a black hole is called the photon sphere, where photons orbit in the black hole in unstable circular orbits. On top of this the image of the black hole is gravitationally lensed by the black hole itself magnifying the apparent radius of the photon sphere to be between $(2 \sqrt{3+2 \sqrt{2}}) r_{g}$ and $(3 \sqrt{3}) r_{g}$, determined by spin and inclination; the latter corresponds to a perfectly spherical non-spinning black hole. The area within this lensed image will appear almost black and is the 'shadow' the EHT is looking for. [figure2] Figure 4: Four nights of data were taken for M87* during the observing window of the EHT, and whilst the diameter of the disk stayed relatively constant the location of bright spots moved, possibly indicating gas that is orbiting the black hole. Credit: EHT Collaboration. The EHT observed M87* on four separate occasions during the observing window (see Fig 4), and the team saw that some of the bright spots changed in that time, suggesting they may be associated with orbiting gas close to the black hole. The Innermost Stable Circular Orbit (ISCO) is the equivalent of the photon sphere but for particles with mass (and is also stable). The total conserved energy of a circular orbit close to a non-spinning black hole is given by $$ E=m c^{2}\left(\frac{1-\frac{2 r_{g}}{r}}{\sqrt{1-\frac{3 r_{g}}{r}}}\right) $$ and the radius of the ISCO, $r_{\mathrm{ISCO}}$, is the value of $r$ for which $E$ is minimised. We expect that most black holes are in fact spinning (since most stars are spinning) and the spin of a black hole is quantified with the spin parameter $a \equiv J / J_{\max }$ where $J$ is the angular momentum of the black hole and $J_{\max }=G M^{2} / c$ is the maximum possible angular momentum it can have. The value of $a$ varies from $-1 \leq a \leq 1$, where negative spins correspond to the black hole rotating in the opposite direction to its accretion disk, and positive spins in the same direction. If $a=1$ then $r_{\text {ISCO }}=r_{g}$, whilst if $a=-1$ then $r_{\text {ISCO }}=9 r_{g}$. The angular velocity of a particle in the ISCO is given by $$ \omega^{2}=\frac{G M}{\left(r_{\text {ISCO }}^{3 / 2}+a r_{g}^{3 / 2}\right)^{2}} $$ [figure3] Figure 5: Due to the curvature of spacetime, the real distance travelled by a particle moving from the ISCO to the photon sphere (indicated with the solid red arrow) is longer than you would get purely from subtracting the radial co-ordinates of those orbits (indicated with the dashed blue arrow), which would be valid for a flat spacetime. Relations between these distances are not to scale in this diagram. Credit: Modified from Bardeen et al. (1972). The spacetime near a black hole is curved, as described by the equations of GR. This means that the distance between two points can be substantially different to the distance you would expect if spacetime was flat. GR tells us that the proper distance travelled by a particle moving from radius $r_{1}$ to radius $r_{2}$ around a black hole of mass $M$ (with $r_{1}>r_{2}$ ) is given by $$ \Delta l=\int_{r_{2}}^{r_{1}}\left(1-\frac{2 r_{g}}{r}\right)^{-1 / 2} \mathrm{~d} r $$ problem: d. Determine risco for a non-spinning black hole. Give your answer in units of $\mathrm{r}_{\mathrm{g}}$. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is an expression without equals signs, e.g. ANSWER=\frac{1}{2} g t^2

EX

Astronomy

EN

multi-modal

Astronomy_112

2019 年 4 月 10 日晚 9 时许，人类史上首张黑洞照片面世，有望证实广义相对论在极端条件下仍然成立！某同学查阅资料发现黑洞的半径 $R$ 和质量 $M$ 满足关系式 $R=\frac{2 G M}{c^{2}}$ (其中 $\mathrm{G}$ 为引力常量, 真空中的光速 $\mathrm{c}=3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ), 他借助太阳发出的光传播到地球需要大约 8 分钟和地球公转的周期 1 年, 估算太阳“浓缩”为黑洞时, 对应的半径约为 ( ) A: $3000 \mathrm{~m}$ B: $300 \mathrm{~m}$ C: $30 \mathrm{~m}$ D: $3 \mathrm{~m}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2019 年 4 月 10 日晚 9 时许，人类史上首张黑洞照片面世，有望证实广义相对论在极端条件下仍然成立！某同学查阅资料发现黑洞的半径 $R$ 和质量 $M$ 满足关系式 $R=\frac{2 G M}{c^{2}}$ (其中 $\mathrm{G}$ 为引力常量, 真空中的光速 $\mathrm{c}=3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ), 他借助太阳发出的光传播到地球需要大约 8 分钟和地球公转的周期 1 年, 估算太阳“浓缩”为黑洞时, 对应的半径约为 ( ) A: $3000 \mathrm{~m}$ B: $300 \mathrm{~m}$ C: $30 \mathrm{~m}$ D: $3 \mathrm{~m}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_277

公元 2032 年, 某位 2022 年进入大学的航天爱好者身着飞行机甲, 实现了年少时伴飞中国天宫空间站的梦想, 在伴飞天宫前, 该航天爱好者在近地圆轨道上做无动力飞行,然后先后通过两次的瞬间加力, 成功转移到天宫所在轨道, 图中 $\alpha$ 角为该航天爱好者第一次加力时, 天宫一号和航天爱好者相对地球球心张开的夹角, 已知, 地球半径为 $6400 \mathrm{~km}$, 天宫距地面高度约为 $400 \mathrm{~km}$, 为了以最短的时间到达天宫附近, $\alpha$ 角约为 [图1] A: $1^{\circ}$ B: $8^{\circ}$ C: $16^{\circ}$ D: $20^{\circ}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 公元 2032 年, 某位 2022 年进入大学的航天爱好者身着飞行机甲, 实现了年少时伴飞中国天宫空间站的梦想, 在伴飞天宫前, 该航天爱好者在近地圆轨道上做无动力飞行,然后先后通过两次的瞬间加力, 成功转移到天宫所在轨道, 图中 $\alpha$ 角为该航天爱好者第一次加力时, 天宫一号和航天爱好者相对地球球心张开的夹角, 已知, 地球半径为 $6400 \mathrm{~km}$, 天宫距地面高度约为 $400 \mathrm{~km}$, 为了以最短的时间到达天宫附近, $\alpha$ 角约为 [图1] A: $1^{\circ}$ B: $8^{\circ}$ C: $16^{\circ}$ D: $20^{\circ}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

ZH

multi-modal

Astronomy_569

哈雷彗星是人一生中唯一可以裸眼看能看见两次的彗星, 其绕日运行的周期为 $T$ 年,若测得它在近日点距太阳中心的距离是地球公转轨道半长轴的 $N$ 倍, 则由此估算出哈雷彗星在近日点时受到太阳的引力是在远日点受太阳引力的 A: $N^{2}$ B: $\left(2 T^{\frac{2}{3}}-N\right)^{2} N^{-2}$ C: $\left(2 T^{\frac{2}{3}} N^{-1}-1\right)$ 倍 D: $T^{\frac{4}{3}} N^{2}$ 倍

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 哈雷彗星是人一生中唯一可以裸眼看能看见两次的彗星, 其绕日运行的周期为 $T$ 年,若测得它在近日点距太阳中心的距离是地球公转轨道半长轴的 $N$ 倍, 则由此估算出哈雷彗星在近日点时受到太阳的引力是在远日点受太阳引力的 A: $N^{2}$ B: $\left(2 T^{\frac{2}{3}}-N\right)^{2} N^{-2}$ C: $\left(2 T^{\frac{2}{3}} N^{-1}-1\right)$ 倍 D: $T^{\frac{4}{3}} N^{2}$ 倍 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

ZH

text-only

Astronomy_241

2020 年 12 月 17 日嫦娥五号从月球采集月壤成功返回地球。嫦娥五号绕月运行的轨迹如图所示, 在近月点 $O$ 制动成功后被月球捕获进入近月点 $200 \mathrm{~km}$, 远月点 $5500 \mathrm{~km}$的粗圆轨道I, 在近月点再次制动进入椭圆轨道II, 第三次制动后进入离月面 $200 \mathrm{~km}$ 的环月圆轨道III, 已知月球半径约为 $1700 \mathrm{~km}$, 则 ( ) [图1] A: 嫦娥五号在I、II、III三个轨道上运行的周期相等 B: 嫦娥五号在I、II、III三个轨道上运行的机械能相等 C: 嫦娥五号在 $P 、 Q$ 两点的速度与它们到月心的距离成反比 D: $O$ 点的重力加速度约为月球表面的重力加速度的 0.8 倍

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2020 年 12 月 17 日嫦娥五号从月球采集月壤成功返回地球。嫦娥五号绕月运行的轨迹如图所示, 在近月点 $O$ 制动成功后被月球捕获进入近月点 $200 \mathrm{~km}$, 远月点 $5500 \mathrm{~km}$的粗圆轨道I, 在近月点再次制动进入椭圆轨道II, 第三次制动后进入离月面 $200 \mathrm{~km}$ 的环月圆轨道III, 已知月球半径约为 $1700 \mathrm{~km}$, 则 ( ) [图1] A: 嫦娥五号在I、II、III三个轨道上运行的周期相等 B: 嫦娥五号在I、II、III三个轨道上运行的机械能相等 C: 嫦娥五号在 $P 、 Q$ 两点的速度与它们到月心的距离成反比 D: $O$ 点的重力加速度约为月球表面的重力加速度的 0.8 倍 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

ZH

multi-modal

Astronomy_618

中国在西昌卫星发射中心成功发射“亚太九号”通信卫星, 该卫星运行的轨道示意图如图所示, 卫星先沿陏圆轨道 1 运行, 近地点为 $Q$, 远地点为 $P$ 。当卫星经过 $P$ 点时点火加速, 使卫星由粗圆轨道 1 转移到地球同步轨道 2 上运行, 下列说法正确的是 ( ) [图1] A: 卫星在轨道 1 和轨道 2 上运动时的机械能相等 B: 卫星在轨道 1 上运行经过 $P$ 点的速度大于经过 $Q$ 点的速度 C: 卫星在轨道 2 上时处于超重状态 D: 卫星在轨道 1 上运行经过 $P$ 点的加速度等于在轨道 2 上运行经过 $P$ 点的加速度

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 中国在西昌卫星发射中心成功发射“亚太九号”通信卫星, 该卫星运行的轨道示意图如图所示, 卫星先沿陏圆轨道 1 运行, 近地点为 $Q$, 远地点为 $P$ 。当卫星经过 $P$ 点时点火加速, 使卫星由粗圆轨道 1 转移到地球同步轨道 2 上运行, 下列说法正确的是 ( ) [图1] A: 卫星在轨道 1 和轨道 2 上运动时的机械能相等 B: 卫星在轨道 1 上运行经过 $P$ 点的速度大于经过 $Q$ 点的速度 C: 卫星在轨道 2 上时处于超重状态 D: 卫星在轨道 1 上运行经过 $P$ 点的加速度等于在轨道 2 上运行经过 $P$ 点的加速度 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

ZH

multi-modal

Astronomy_675

宇航员在地球表面一斜坡上 $P$ 点, 沿水平方向以初速度 $v_{0}$ 抛出一个小球, 测得小球经时间 $t$ 落到斜坡另一点 $Q$ 上现宇航员站在某质量分布均匀的星球表面相同的斜坡上 $P$点, 沿水平方向以相同的初速度 $v_{0}$ 抛出一个小球, 小球落在 $P Q$ 的中点. 已知该星球的半径为 $R$, 地球表面重力加速度为 $g$, 引力常量为 $G$, 球的体积公式是 $V=\frac{4}{3} \pi R^{3}$ 。求: 该星球的密度 $\rho$ [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 宇航员在地球表面一斜坡上 $P$ 点, 沿水平方向以初速度 $v_{0}$ 抛出一个小球, 测得小球经时间 $t$ 落到斜坡另一点 $Q$ 上现宇航员站在某质量分布均匀的星球表面相同的斜坡上 $P$点, 沿水平方向以相同的初速度 $v_{0}$ 抛出一个小球, 小球落在 $P Q$ 的中点. 已知该星球的半径为 $R$, 地球表面重力加速度为 $g$, 引力常量为 $G$, 球的体积公式是 $V=\frac{4}{3} \pi R^{3}$ 。求: 该星球的密度 $\rho$ [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

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Astronomy

ZH

multi-modal

Astronomy_35

如图所示, 航天器 $\mathrm{A}$ 和卫星 $\mathrm{B}$ 均在赤道面内, 它通过一根金属长绳相连, 在各自的轨道上绕地球做自西向东的匀速圆周运动, 不考虑绳系卫星与航天器之间的万有引力,忽略空气阻力, 不计金属长绳的质量, 则（） [图1] A: 正常运行时, 金属长绳中拉力为零 B: 绳系卫星 $\mathrm{B}$ 的线速度大于航天器 $\mathrm{A}$ 的线速度 C: 由于存在地磁场, 金属长绳上绳系卫星 $\mathrm{B}$ 端的电势高于航天器 $\mathrm{A}$ 端的电势 D: 若在绳系卫星 B 的轨道上存在另一颗独立卫星 C, 其角速度大于绳系卫星 $\mathrm{B}$ 的角速度

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 如图所示, 航天器 $\mathrm{A}$ 和卫星 $\mathrm{B}$ 均在赤道面内, 它通过一根金属长绳相连, 在各自的轨道上绕地球做自西向东的匀速圆周运动, 不考虑绳系卫星与航天器之间的万有引力,忽略空气阻力, 不计金属长绳的质量, 则（） [图1] A: 正常运行时, 金属长绳中拉力为零 B: 绳系卫星 $\mathrm{B}$ 的线速度大于航天器 $\mathrm{A}$ 的线速度 C: 由于存在地磁场, 金属长绳上绳系卫星 $\mathrm{B}$ 端的电势高于航天器 $\mathrm{A}$ 端的电势 D: 若在绳系卫星 B 的轨道上存在另一颗独立卫星 C, 其角速度大于绳系卫星 $\mathrm{B}$ 的角速度 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

ZH

multi-modal

Astronomy_179

2022 年 11 月 29 日, 神舟十五号飞行任务是中国空间站建造阶段的最后一棒, 也是空间站应用与发展阶段的第一棒. 已知空间站 $\mathrm{Q}$ 和同步卫星 $\mathrm{P}$ 环绕地球运行的轨道均可视为匀速圆周运动。如图所示, 已知 $\mathrm{P}, \mathrm{Q}$ 运动方向均沿逆时针方向, $P Q$ 与 $O P$连线的夹角最大值为 $\alpha$ 。求: 同步卫星 $\mathrm{P}$ 、空间站 $\mathrm{Q}$ 的角速度之比; [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 2022 年 11 月 29 日, 神舟十五号飞行任务是中国空间站建造阶段的最后一棒, 也是空间站应用与发展阶段的第一棒. 已知空间站 $\mathrm{Q}$ 和同步卫星 $\mathrm{P}$ 环绕地球运行的轨道均可视为匀速圆周运动。如图所示, 已知 $\mathrm{P}, \mathrm{Q}$ 运动方向均沿逆时针方向, $P Q$ 与 $O P$连线的夹角最大值为 $\alpha$ 。求: 同步卫星 $\mathrm{P}$ 、空间站 $\mathrm{Q}$ 的角速度之比; [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_177

如图所示, 是月亮女神、嫦娥 1 号绕月做圆周运行时某时刻的图片, 用 $R_{1} 、 R_{2} 、$ $T_{1} 、 T_{2}$ 分别表示月亮女神和嫦娥 1 号的轨道半径及周期, 用 $R$ 表示月亮的半径。此时二者的连线通过月心, 轨道半径之比为 $1: 4$ 。若不考虑月亮女神、嫦娥 1 号之间的引力,则下列说法正确的是（ ） [图1] A: 在图示轨道上, 月亮女神的速度小于嫦娥 1 号 B: 在图示轨道上, 嫦娥 1 号的加速度大小是月亮女神的 4 倍 C: 在图示轨道上, 且从图示位置开始经 $t=\frac{T_{1} T_{2}}{2 T_{2}-2 T_{1}}$ 二者第二次相距最近 D: 若月亮女神从图示轨道上加速, 可与嫦娥 1 号对接

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 如图所示, 是月亮女神、嫦娥 1 号绕月做圆周运行时某时刻的图片, 用 $R_{1} 、 R_{2} 、$ $T_{1} 、 T_{2}$ 分别表示月亮女神和嫦娥 1 号的轨道半径及周期, 用 $R$ 表示月亮的半径。此时二者的连线通过月心, 轨道半径之比为 $1: 4$ 。若不考虑月亮女神、嫦娥 1 号之间的引力,则下列说法正确的是（ ） [图1] A: 在图示轨道上, 月亮女神的速度小于嫦娥 1 号 B: 在图示轨道上, 嫦娥 1 号的加速度大小是月亮女神的 4 倍 C: 在图示轨道上, 且从图示位置开始经 $t=\frac{T_{1} T_{2}}{2 T_{2}-2 T_{1}}$ 二者第二次相距最近 D: 若月亮女神从图示轨道上加速, 可与嫦娥 1 号对接 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

ZH

multi-modal

Astronomy_214

太阳帆飞船是依靠太阳的光压来加速飞船的。太阳帆一般由反射率极高的韧性薄膜制成的, 面积大, 质量小。某太空探测飞船运行在地球公转轨道上, 绕太阳做匀速圆周运动, 某时刻飞船的太阳帆打开，展开的面积为 $S$, 飞船能控制帆面始终垂直太阳光线。已知太阳的总辐射功率为 $P_{0}$, 日地距离为 $r_{0}$, 光速为 $c$, 引力常量为 $G$, 太阳帆反射率 $100 \%$ 。下列说法正确的是（ ） A: 太阳帆受到的太阳光压力为 $F=\frac{2}{c} P_{0}$ B: 太阳帆受到的太阳光压力为 $F=\frac{S}{4 \pi c r_{0}^{2}} P_{0}$ C: 打开太阳帆后, 飞船将沿径向远离太阳 D: 打开太阳帆后, 飞船在以后运动中受到的光压力与太阳的引力之比恒定

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 太阳帆飞船是依靠太阳的光压来加速飞船的。太阳帆一般由反射率极高的韧性薄膜制成的, 面积大, 质量小。某太空探测飞船运行在地球公转轨道上, 绕太阳做匀速圆周运动, 某时刻飞船的太阳帆打开，展开的面积为 $S$, 飞船能控制帆面始终垂直太阳光线。已知太阳的总辐射功率为 $P_{0}$, 日地距离为 $r_{0}$, 光速为 $c$, 引力常量为 $G$, 太阳帆反射率 $100 \%$ 。下列说法正确的是（ ） A: 太阳帆受到的太阳光压力为 $F=\frac{2}{c} P_{0}$ B: 太阳帆受到的太阳光压力为 $F=\frac{S}{4 \pi c r_{0}^{2}} P_{0}$ C: 打开太阳帆后, 飞船将沿径向远离太阳 D: 打开太阳帆后, 飞船在以后运动中受到的光压力与太阳的引力之比恒定 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

ZH

text-only

Astronomy_827

Choose the values $(x, y, z)$ that would best complete the descriptions below for the 3 different types of twilights. i) civil twilight, when the Sun is $x^{\circ}$ below the horizon. We can start to see the brightest stars and the sea horizon can be clearly seen. At this point it becomes hard to read outdoors without artificial light. ii) nautical twilight, when the Sun is $y^{\circ}$ below the horizon. It is too dark to see the sea horizon and you can no longer make altitude measurements for navigation using the horizon as a reference. iii) astronomical twilight, when the Sun is $z^{\circ}$ below the horizon. Scattered sunlight becomes less than the average starlight and it is about the same brightness as the aurora or zodiacal light. A: Civil twilight $-5^{\circ}$, nautical twilight $-10^{\circ}$, astronomical twilight - $15^{\circ}$ B: Civil twilight $-6^{\circ}$, nautical twilight $-12^{\circ}$, astronomical twilight - $18^{\circ}$ (Answer) C: Civil twilight $-3^{\circ}$, nautical twilight $-6^{\circ}$, astronomical twilight - $9^{\circ}$ D: Civil twilight $-12^{\circ}$, nautical twilight $-6^{\circ}$, astronomical twilight $-18^{\circ}$ E: Civil twilight $-10^{\circ}$, nautical twilight $-20^{\circ}$, astronomical twilight $-30^{\circ}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Choose the values $(x, y, z)$ that would best complete the descriptions below for the 3 different types of twilights. i) civil twilight, when the Sun is $x^{\circ}$ below the horizon. We can start to see the brightest stars and the sea horizon can be clearly seen. At this point it becomes hard to read outdoors without artificial light. ii) nautical twilight, when the Sun is $y^{\circ}$ below the horizon. It is too dark to see the sea horizon and you can no longer make altitude measurements for navigation using the horizon as a reference. iii) astronomical twilight, when the Sun is $z^{\circ}$ below the horizon. Scattered sunlight becomes less than the average starlight and it is about the same brightness as the aurora or zodiacal light. A: Civil twilight $-5^{\circ}$, nautical twilight $-10^{\circ}$, astronomical twilight - $15^{\circ}$ B: Civil twilight $-6^{\circ}$, nautical twilight $-12^{\circ}$, astronomical twilight - $18^{\circ}$ (Answer) C: Civil twilight $-3^{\circ}$, nautical twilight $-6^{\circ}$, astronomical twilight - $9^{\circ}$ D: Civil twilight $-12^{\circ}$, nautical twilight $-6^{\circ}$, astronomical twilight $-18^{\circ}$ E: Civil twilight $-10^{\circ}$, nautical twilight $-20^{\circ}$, astronomical twilight $-30^{\circ}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

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Astronomy

EN

text-only

Astronomy_899

In January 2020, NASA's Transiting Exoplanet Survey Satellite (TESS) discovered an Earth-sized exoplanet, called TOI $700 \mathrm{~d}$, in its star's habitable zone. This is the range of distances a planet can orbit a star so that liquid water can exist on the surface, given sufficient atmospheric pressure. It was discovered using the transit method, where the planet passes directly between the observer and the star, causing a drop in brightness. [figure1] Figure 1: The three planets of the TOI 700 system, illustrated here, orbit a small, cool M dwarf star. TOI $700 \mathrm{~d}$ is the first Earth-size habitable-zone world discovered by TESS. Credit: NASA's Goddard Space Flight Center. TOI $700 \mathrm{~d}$ has radius $R_{P}=1.19 R_{E}$ orbiting a star with luminosity $0.0233 L_{\odot}$ at a distance of 0.163 au. Assume that the planet absorbs all the light that hits the surface, and that the orbit is circular. Find the value of $T_{P}$. Give your answer in ${ }^{\circ} \mathrm{C}$. [Hint: you may find it is colder than expected.]

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: In January 2020, NASA's Transiting Exoplanet Survey Satellite (TESS) discovered an Earth-sized exoplanet, called TOI $700 \mathrm{~d}$, in its star's habitable zone. This is the range of distances a planet can orbit a star so that liquid water can exist on the surface, given sufficient atmospheric pressure. It was discovered using the transit method, where the planet passes directly between the observer and the star, causing a drop in brightness. [figure1] Figure 1: The three planets of the TOI 700 system, illustrated here, orbit a small, cool M dwarf star. TOI $700 \mathrm{~d}$ is the first Earth-size habitable-zone world discovered by TESS. Credit: NASA's Goddard Space Flight Center. TOI $700 \mathrm{~d}$ has radius $R_{P}=1.19 R_{E}$ orbiting a star with luminosity $0.0233 L_{\odot}$ at a distance of 0.163 au. Assume that the planet absorbs all the light that hits the surface, and that the orbit is circular. Find the value of $T_{P}$. Give your answer in ${ }^{\circ} \mathrm{C}$. [Hint: you may find it is colder than expected.] All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of C, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_920

Why do hurricanes rotate anti-clockwise in the northern hemisphere and clockwise in the southern hemisphere? A: Due to the Earth rotating from East to West B: Due to the different ratios of land to water area between the two hemispheres C: Due to the Moon's orbit being inclined by $5^{\circ}$ above the ecliptic giving it more influence on the northern hemisphere D: Due to the Coriolis Effect causing paths of particles to curve as they travel over the Earth's surface

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Why do hurricanes rotate anti-clockwise in the northern hemisphere and clockwise in the southern hemisphere? A: Due to the Earth rotating from East to West B: Due to the different ratios of land to water area between the two hemispheres C: Due to the Moon's orbit being inclined by $5^{\circ}$ above the ecliptic giving it more influence on the northern hemisphere D: Due to the Coriolis Effect causing paths of particles to curve as they travel over the Earth's surface You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

text-only

Astronomy_654

当某一地外行星 (火星、木星、土星、天王星、海王星) 于绕日公转过程中运行到试卷第 48 页，共 150 页 与地球、太阳成一直线的状态, 且地球恰好位于太阳和外行星之间的这种天文现象叫“冲日”, 冲日前后是观测地外行星的好时机。如图所示是土星冲日示意图, 已知地球质量为 $M$, 半径为 $R$, 公转周期是 1 年, 公转半径为 $r$, 土星质量是地球的 95 倍, 土星半径是地球的 9.5 倍, 土星的公转半径是地球的 9.5 倍。求: $\left(\sqrt{9.5^{3}} \approx 29\right)$ 地球和太阳间的万有引力是土星和太阳间的几倍？ [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个数值。 问题: 当某一地外行星 (火星、木星、土星、天王星、海王星) 于绕日公转过程中运行到试卷第 48 页，共 150 页 与地球、太阳成一直线的状态, 且地球恰好位于太阳和外行星之间的这种天文现象叫“冲日”, 冲日前后是观测地外行星的好时机。如图所示是土星冲日示意图, 已知地球质量为 $M$, 半径为 $R$, 公转周期是 1 年, 公转半径为 $r$, 土星质量是地球的 95 倍, 土星半径是地球的 9.5 倍, 土星的公转半径是地球的 9.5 倍。求: $\left(\sqrt{9.5^{3}} \approx 29\right)$ 地球和太阳间的万有引力是土星和太阳间的几倍？ [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是数值。

NV

Astronomy

ZH

multi-modal

Astronomy_642

有人提出了一种不用火箭发射人造地球卫星的设想。其设想如下: 沿地球的一条弦挖一通道, 如图乙所示. 在通道的两个出口处 $A$ 和 $B$, 分别将质量为 $M$ 的物体和质量为 $m$ 的待发射卫星同时自由释放, 只要 $M$ 比 $m$ 足够大, 碰撞后, 质量为 $m$ 的物体, 即待发射的卫星就会从通道口 $B$ 冲出通道。(已知地球表面的重力加速度为 $g$, 地球半径为 $\left.R_{0}\right)$ [图1] 甲 [图2] 乙 如果在 $A$ 处释放一个质量很大的物体 $M$, 在 $B$ 处同时释放一个质量远小于 $M$ 的物体，同时达到 $O^{\prime}$ 处发生弹性正碰(由于大物体质量很大，可以认为碰后速度不变)，那么小物体返回从 $B$ 飞出, 为使飞出的速度达到地球的第一宇宙速度, $h$ 应为多大?

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 有人提出了一种不用火箭发射人造地球卫星的设想。其设想如下: 沿地球的一条弦挖一通道, 如图乙所示. 在通道的两个出口处 $A$ 和 $B$, 分别将质量为 $M$ 的物体和质量为 $m$ 的待发射卫星同时自由释放, 只要 $M$ 比 $m$ 足够大, 碰撞后, 质量为 $m$ 的物体, 即待发射的卫星就会从通道口 $B$ 冲出通道。(已知地球表面的重力加速度为 $g$, 地球半径为 $\left.R_{0}\right)$ [图1] 甲 [图2] 乙 如果在 $A$ 处释放一个质量很大的物体 $M$, 在 $B$ 处同时释放一个质量远小于 $M$ 的物体，同时达到 $O^{\prime}$ 处发生弹性正碰(由于大物体质量很大，可以认为碰后速度不变)，那么小物体返回从 $B$ 飞出, 为使飞出的速度达到地球的第一宇宙速度, $h$ 应为多大? 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_333

“天问一号”是中国自主设计的火星探测器, 已于 2021 年 3 月到达火星。已知火星 直径约为地球直径的 50\%, 火星质量约为地球质量的 10\%, 下列说法正确的是（） [图1] A: 火星表面的重力加速度大于 $9.8 \mathrm{~m} / \mathrm{s}^{2}$ B: “天问一号”的发射速度大于 $7.9 \mathrm{~km} / \mathrm{s}$ 小于 $11.2 \mathrm{~km} / \mathrm{s}$ C: “天问一号”在火星表面圆轨道上的绕行速度大于 $7.9 \mathrm{~km} / \mathrm{s}$ D: “天问一号”在火星表面圆轨道上的环绕周期小于 24 小时

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: “天问一号”是中国自主设计的火星探测器, 已于 2021 年 3 月到达火星。已知火星 直径约为地球直径的 50\%, 火星质量约为地球质量的 10\%, 下列说法正确的是（） [图1] A: 火星表面的重力加速度大于 $9.8 \mathrm{~m} / \mathrm{s}^{2}$ B: “天问一号”的发射速度大于 $7.9 \mathrm{~km} / \mathrm{s}$ 小于 $11.2 \mathrm{~km} / \mathrm{s}$ C: “天问一号”在火星表面圆轨道上的绕行速度大于 $7.9 \mathrm{~km} / \mathrm{s}$ D: “天问一号”在火星表面圆轨道上的环绕周期小于 24 小时 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

ZH

multi-modal

Astronomy_32

小行星带, 是位于火星和木星轨道之间的小行星的密集区域, 在太阳系中除了九颗大行星以外, 还有成千上万颗我们肉眼看不到的小天体, 它们沿着粗圆形的轨道不停地围绕太阳公转。如图所示, 在火星与木星轨道之间有一小行星带, 假设该带中的小行星只受到太阳的引力，并绕太阳做匀速圆周运动。下列说法正确的是（ ） A: 若知道某一小行星绕太阳运转的周期和轨道半径可求出太阳的质量 B: 若知道地球和某一小行星绕太阳运转的轨道半径, 可求出该小行星的质量 C: 小行星带内侧小行星的向心加速度值小于外侧小行星的向心加速度值 D: 在相同时间内, 地球和太阳连线扫过的面积比内侧小行星和太阳连线扫过的面积要小

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 小行星带, 是位于火星和木星轨道之间的小行星的密集区域, 在太阳系中除了九颗大行星以外, 还有成千上万颗我们肉眼看不到的小天体, 它们沿着粗圆形的轨道不停地围绕太阳公转。如图所示, 在火星与木星轨道之间有一小行星带, 假设该带中的小行星只受到太阳的引力，并绕太阳做匀速圆周运动。下列说法正确的是（ ） A: 若知道某一小行星绕太阳运转的周期和轨道半径可求出太阳的质量 B: 若知道地球和某一小行星绕太阳运转的轨道半径, 可求出该小行星的质量 C: 小行星带内侧小行星的向心加速度值小于外侧小行星的向心加速度值 D: 在相同时间内, 地球和太阳连线扫过的面积比内侧小行星和太阳连线扫过的面积要小 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_934

From the upper parts of the Sun's corona comes a stream of charged particles called the solar wind, meaning the Sun is slowly losing some of its mass (although the effect is negligible compared to the mass loss in nuclear fusion). The particles travel at supersonic speeds until the pressure from interstellar space causes their speed to drop to subsonic speeds instead - this transition is called the termination shock, and has been explored by the two Voyager probes as they leave the solar system. [figure1] Figure 3: Left: A demonstration of a termination shock formed with water flowing from a tap into a sink. Right: The Voyager spacecraft crossing the termination shock of the Solar System. At the termination shock boundary the particles slow down from their (supersonic) $u_{\infty}$ to a much slower (subsonic) speed. Another way of defining this boundary is that it is where the pressure of the interstellar medium, $P_{\mathrm{ISM}}$, is equal to the kinetic energy density (i.e. KE per unit volume) of the gas. If $P_{\mathrm{ISM}}$ is estimated to be $10^{-13} \mathrm{~Pa}$, use your values of $u_{\infty}$ and $\Delta M / \Delta t$ to calculate an estimate for the distance to the termination shock. Give your answer in $\mathrm{AU}$.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: From the upper parts of the Sun's corona comes a stream of charged particles called the solar wind, meaning the Sun is slowly losing some of its mass (although the effect is negligible compared to the mass loss in nuclear fusion). The particles travel at supersonic speeds until the pressure from interstellar space causes their speed to drop to subsonic speeds instead - this transition is called the termination shock, and has been explored by the two Voyager probes as they leave the solar system. [figure1] Figure 3: Left: A demonstration of a termination shock formed with water flowing from a tap into a sink. Right: The Voyager spacecraft crossing the termination shock of the Solar System. At the termination shock boundary the particles slow down from their (supersonic) $u_{\infty}$ to a much slower (subsonic) speed. Another way of defining this boundary is that it is where the pressure of the interstellar medium, $P_{\mathrm{ISM}}$, is equal to the kinetic energy density (i.e. KE per unit volume) of the gas. If $P_{\mathrm{ISM}}$ is estimated to be $10^{-13} \mathrm{~Pa}$, use your values of $u_{\infty}$ and $\Delta M / \Delta t$ to calculate an estimate for the distance to the termination shock. Give your answer in $\mathrm{AU}$. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of AU, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_815

. An astronomer observes an eclipsing binary star system from Earth, and he plots the following light curve. [figure1] Suppose that both stars have circular orbits and the distance between the stars is 14.8 AU. What is the total mass of the binary star system in terms of solar masses? A: $2.3 M_{\odot}$ B: $5.7 M_{\odot}$ C: $6.8 M_{\odot}$ D: $23 M_{\odot}$ E: $46 M_{\odot}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: . An astronomer observes an eclipsing binary star system from Earth, and he plots the following light curve. [figure1] Suppose that both stars have circular orbits and the distance between the stars is 14.8 AU. What is the total mass of the binary star system in terms of solar masses? A: $2.3 M_{\odot}$ B: $5.7 M_{\odot}$ C: $6.8 M_{\odot}$ D: $23 M_{\odot}$ E: $46 M_{\odot}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

multi-modal

Astronomy_1054

On $21^{\text {st }}$ December 2020, Jupiter and Saturn formed a true spectacle in the southwestern sky just after sunset in the UK, separated by only $0.102^{\circ}$. This is close enough that when viewed through a telescope both planets and their moons could be seen in the same field of view (see Fig 4). When two planets occupy the same piece of sky it is known as a conjunction, and when it is Jupiter and Saturn it is known as a great conjunction (so named because they are the rarest of the naked-eye planet conjunctions). The reason it happens is because Jupiter's orbital velocity is higher than Saturn's and so as time goes on Jupiter catches up with and overtakes Saturn (at least as viewed from Earth), with the moment of overtaking corresponding to the conjunction. The process of the two getting closer and closer together has been seen in sky throughout the year (see Fig 5). [figure1] Figure 4: Left: The view of the planets the day before their closest approach, as captured by the 16 " telescope at the Institute of Astronomy, Cambridge. Credit: Robin Catchpole. Right: The view from Arizona on the day of the closest approach as viewed with the Lowell Discovery Telescope. Credit: Levine / Elbert / Bosh / Lowell Observatory. [figure2] Figure 5: Left: Demonstrating how Jupiter and Saturn have been getting closer and closer together over the last few months. Separations are given in degrees and arcminutes $\left(1 / 60^{\text {th }}\right.$ of a degree). Credit: Pete Lawrence. Right: The positions of the Earth, Jupiter and Saturn that were responsible for the 2020 great conjunction. The precise timing of the apparent alignment is clearly sensitive to where Earth is in its orbit. Credit: timeanddate.com. The time between conjunctions is known as the synodic period. Although this period will change slightly from conjunction to conjunction due to different lines of perspective as viewed from Earth (see Fig 5), we can work out the average time between great conjunctions by considering both planets travelling on circular coplanar orbits and ignoring the position of the Earth. For circular coplanar orbits the centre of Jupiter's disc would pass in front of the centre of Saturn's disc every conjunction, and hence have an angular separation of $\theta=0^{\circ}$ (it is measured from the centre of each disc). In practice, the planets follow elliptical orbits that are in planes inclined at different angles to each other. Fig 6 shows how this affects the real values for over 8000 years' worth of data, which along with different synodic periods between conjunctions makes it a difficult problem to solve precisely without a computer. However, after one synodic period Saturn has moved about $2 / 3$ of the way around its orbit, and so roughly every 3 synodic periods it is in a similar part of the sky. Consequently every third great conjunction follows a reasonably regular pattern which can be fit with a sinusoidal function. [figure3] Figure 6: Top: All great conjunctions from 1800 to 2300, calculated for the real celestial mechanics of the Solar System. We can see that each great conjunction belongs to one of three different series or tracks, with Track A indicated with orange circles, Track B with green squares, and Track C with blue triangles. Bottom: The same idea but extended over a much larger date range, up to $10000 \mathrm{AD}$. It is clear the distinct series form broadly sinusoidal patterns which can be used with the average synodic period to give rough predictions for the separations of great conjunctions. The opacity of points is related to each conjunction's angular separation from the Sun (where low opacity means close to the Sun, so it is harder for any observers to see). Credit: Nick Koukoufilippas, but inspired by the work of Steffen Thorsen and Graham Jones / Sky \& Telescope.c. By empirically fitting a sinusoidal function (which is assumed to be the same for each track, just with a fixed phase difference between them) and assuming all conjunctions are separated by the average synodic period, we can give rough estimations for the separations of any given great conjunction. Note: be careful as your calculations will be very sensitive to rounding errors. iii. State which track the 'Star of Bethlehem' great conjunction is on, and hence use the relevant equation to predict its separation. How does it compare to the 2020 great conjunction?

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: On $21^{\text {st }}$ December 2020, Jupiter and Saturn formed a true spectacle in the southwestern sky just after sunset in the UK, separated by only $0.102^{\circ}$. This is close enough that when viewed through a telescope both planets and their moons could be seen in the same field of view (see Fig 4). When two planets occupy the same piece of sky it is known as a conjunction, and when it is Jupiter and Saturn it is known as a great conjunction (so named because they are the rarest of the naked-eye planet conjunctions). The reason it happens is because Jupiter's orbital velocity is higher than Saturn's and so as time goes on Jupiter catches up with and overtakes Saturn (at least as viewed from Earth), with the moment of overtaking corresponding to the conjunction. The process of the two getting closer and closer together has been seen in sky throughout the year (see Fig 5). [figure1] Figure 4: Left: The view of the planets the day before their closest approach, as captured by the 16 " telescope at the Institute of Astronomy, Cambridge. Credit: Robin Catchpole. Right: The view from Arizona on the day of the closest approach as viewed with the Lowell Discovery Telescope. Credit: Levine / Elbert / Bosh / Lowell Observatory. [figure2] Figure 5: Left: Demonstrating how Jupiter and Saturn have been getting closer and closer together over the last few months. Separations are given in degrees and arcminutes $\left(1 / 60^{\text {th }}\right.$ of a degree). Credit: Pete Lawrence. Right: The positions of the Earth, Jupiter and Saturn that were responsible for the 2020 great conjunction. The precise timing of the apparent alignment is clearly sensitive to where Earth is in its orbit. Credit: timeanddate.com. The time between conjunctions is known as the synodic period. Although this period will change slightly from conjunction to conjunction due to different lines of perspective as viewed from Earth (see Fig 5), we can work out the average time between great conjunctions by considering both planets travelling on circular coplanar orbits and ignoring the position of the Earth. For circular coplanar orbits the centre of Jupiter's disc would pass in front of the centre of Saturn's disc every conjunction, and hence have an angular separation of $\theta=0^{\circ}$ (it is measured from the centre of each disc). In practice, the planets follow elliptical orbits that are in planes inclined at different angles to each other. Fig 6 shows how this affects the real values for over 8000 years' worth of data, which along with different synodic periods between conjunctions makes it a difficult problem to solve precisely without a computer. However, after one synodic period Saturn has moved about $2 / 3$ of the way around its orbit, and so roughly every 3 synodic periods it is in a similar part of the sky. Consequently every third great conjunction follows a reasonably regular pattern which can be fit with a sinusoidal function. [figure3] Figure 6: Top: All great conjunctions from 1800 to 2300, calculated for the real celestial mechanics of the Solar System. We can see that each great conjunction belongs to one of three different series or tracks, with Track A indicated with orange circles, Track B with green squares, and Track C with blue triangles. Bottom: The same idea but extended over a much larger date range, up to $10000 \mathrm{AD}$. It is clear the distinct series form broadly sinusoidal patterns which can be used with the average synodic period to give rough predictions for the separations of great conjunctions. The opacity of points is related to each conjunction's angular separation from the Sun (where low opacity means close to the Sun, so it is harder for any observers to see). Credit: Nick Koukoufilippas, but inspired by the work of Steffen Thorsen and Graham Jones / Sky \& Telescope. problem: c. By empirically fitting a sinusoidal function (which is assumed to be the same for each track, just with a fixed phase difference between them) and assuming all conjunctions are separated by the average synodic period, we can give rough estimations for the separations of any given great conjunction. Note: be careful as your calculations will be very sensitive to rounding errors. iii. State which track the 'Star of Bethlehem' great conjunction is on, and hence use the relevant equation to predict its separation. How does it compare to the 2020 great conjunction? All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \circ, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_107

2017 年, 人类第一次直接探测到来自双中子星合并的引力波。根据科学家们复原的过程, 在两颗中子星合并前约 $100 \mathrm{~s}$ 时, 它们相距约 $400 \mathrm{~km}$, 绕二者连线上的某点每秒转动 12 圈, 将两颗中子星都看作是质量均匀分布的球体。由这些数据、万有引力常量并利用牛顿力学知识, 估算一下, 当两颗中子星相距约 $1600 \mathrm{~km}$ 时, 绕二者连线上的某点每秒转动（ ） A: 0.5 圈 B: 1.5 圈 C: 2.5 圈 D: 3.5 圈

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2017 年, 人类第一次直接探测到来自双中子星合并的引力波。根据科学家们复原的过程, 在两颗中子星合并前约 $100 \mathrm{~s}$ 时, 它们相距约 $400 \mathrm{~km}$, 绕二者连线上的某点每秒转动 12 圈, 将两颗中子星都看作是质量均匀分布的球体。由这些数据、万有引力常量并利用牛顿力学知识, 估算一下, 当两颗中子星相距约 $1600 \mathrm{~km}$ 时, 绕二者连线上的某点每秒转动（ ） A: 0.5 圈 B: 1.5 圈 C: 2.5 圈 D: 3.5 圈 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_948

Special Relativity (SR) tells us that two observers will disagree about the duration of a time interval measured by each one's clock if one is moving at speed $v$ relative to the other, a phenomenon called time dilation. General Relativity (GR) tells us that gravitational fields dilate time too. This has an impact on satellites, since they travel at high orbital speeds (slowing down their clocks relative to the surface) but due to their altitude they are in a weaker gravitational field (speeding up their clocks relative to the surface). Which effect is dominant varies with orbital radius. Global Positioning System (GPS) satellites must compensate for this effect, since the satellites rely on accurate measurements of the time between sending and receiving a radio signal. [figure1] Figure 4: A scale diagram of the positions of the orbits for the International Space Station (ISS), GPS satellites and geostationary satellites, along with their orbital periods In $\mathrm{SR}$, time dilation can be calculated with $$ t^{\prime}=\gamma t_{0} \quad \text { where } \quad \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \quad \text { so } \quad \Delta t_{\mathrm{SR}}=t_{0}-t^{\prime}=(1-\gamma) t_{0} $$ where $t_{0}$ is the time measured by the moving clock, $t^{\prime}$ is the time measured by the observer, $c$ is the speed of light and $v$ is the speed of the object. A negative $\Delta t$ indicates that the clocks are passing time slower relative to the observer, whilst a positive indicates they are passing quicker. Hence calculate the daily offset in GPS positions if relativity was not taken into account.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: Special Relativity (SR) tells us that two observers will disagree about the duration of a time interval measured by each one's clock if one is moving at speed $v$ relative to the other, a phenomenon called time dilation. General Relativity (GR) tells us that gravitational fields dilate time too. This has an impact on satellites, since they travel at high orbital speeds (slowing down their clocks relative to the surface) but due to their altitude they are in a weaker gravitational field (speeding up their clocks relative to the surface). Which effect is dominant varies with orbital radius. Global Positioning System (GPS) satellites must compensate for this effect, since the satellites rely on accurate measurements of the time between sending and receiving a radio signal. [figure1] Figure 4: A scale diagram of the positions of the orbits for the International Space Station (ISS), GPS satellites and geostationary satellites, along with their orbital periods In $\mathrm{SR}$, time dilation can be calculated with $$ t^{\prime}=\gamma t_{0} \quad \text { where } \quad \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \quad \text { so } \quad \Delta t_{\mathrm{SR}}=t_{0}-t^{\prime}=(1-\gamma) t_{0} $$ where $t_{0}$ is the time measured by the moving clock, $t^{\prime}$ is the time measured by the observer, $c$ is the speed of light and $v$ is the speed of the object. A negative $\Delta t$ indicates that the clocks are passing time slower relative to the observer, whilst a positive indicates they are passing quicker. Hence calculate the daily offset in GPS positions if relativity was not taken into account. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of m, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_1069

In order to move a spacecraft between orbits we must apply a thrust using rockets, which changes the velocity of the spacecraft by $\Delta v$. In this question we will ignore changes in the mass of the spacecraft due to the burning of fuel. For an object of mass $m$ in a circular orbit of radius $r$ around an object with mass $M$ (where $m \ll M$ ) the orbital velocity, $v_{\text {orb }}$, is given by the formula $v_{\text {orb }}=\sqrt{\frac{G M}{r}}$.e. Hence calculate the direct distance between Earth and Mars at the moment the spacecraft reaches Mars. How long would it take a radio message from the spacecraft to reach Earth?

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: In order to move a spacecraft between orbits we must apply a thrust using rockets, which changes the velocity of the spacecraft by $\Delta v$. In this question we will ignore changes in the mass of the spacecraft due to the burning of fuel. For an object of mass $m$ in a circular orbit of radius $r$ around an object with mass $M$ (where $m \ll M$ ) the orbital velocity, $v_{\text {orb }}$, is given by the formula $v_{\text {orb }}=\sqrt{\frac{G M}{r}}$. problem: e. Hence calculate the direct distance between Earth and Mars at the moment the spacecraft reaches Mars. How long would it take a radio message from the spacecraft to reach Earth? All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of s, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

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Astronomy

EN

text-only

Astronomy_455

宇宙空间存在两颗质量分布均匀的球体未知星球, 经过发射绕表面运行的卫星发现,两个星球的近地卫星周期相等, 同学们据此做出如下判断, 则正确的是 ( ) A: 这两个未知星球的体积一定相等 B: 这两个未知星球的密度一定相等 C: 这两个未知星球的质量若不等, 则表面的重力加速度一定不等 D: 这两个未知星球质量大的，则其表面的重力加速度较小

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 宇宙空间存在两颗质量分布均匀的球体未知星球, 经过发射绕表面运行的卫星发现,两个星球的近地卫星周期相等, 同学们据此做出如下判断, 则正确的是 ( ) A: 这两个未知星球的体积一定相等 B: 这两个未知星球的密度一定相等 C: 这两个未知星球的质量若不等, 则表面的重力加速度一定不等 D: 这两个未知星球质量大的，则其表面的重力加速度较小 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_1005

Main sequence stars fuse hydrogen atoms to form helium in their cores. About $90 \%$ of the stars in the Universe, including the Sun, are main sequence stars. These stars can range from about a tenth of the mass of the Sun to up to 200 times as massive. The main source of energy in main sequence stars is from nuclear fusion. The mass of one hydrogen nucleus is $m_{\mathrm{H}}=1.674 \times 10^{-27} \mathrm{~kg}$, and the mass of one helium nucleus is $m_{\mathrm{He}}=6.649 \times 10^{-27} \mathrm{~kg}$. [figure1] Figure 3: Left: The proton-proton chain reaction is one of two known sets of nuclear fusion reactions by which stars convert hydrogen to helium. It dominates in stars with masses less than or equal to that of the Sun's, and involves a net change of four hydrogen nuclei becoming one helium nucleus. Right: Only the core of a main sequence star will undergo nuclear fusion due to the higher temperature than the surrounding hydrogen shell. The Sun is composed of about $71 \%$ hydrogen, $27 \%$ helium and some heavier elements. However, only $13 \%$ of the hydrogen is available for hydrogen fusion in the core. The rest remains in layers of the Sun where the temperature is too low for fusion to occur. Use these figures and the answers to previous questions to calculate the lifetime in years of hydrogen fusion in the Sun (called its main sequence lifetime), assuming the Sun's luminosity remains constant.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: Main sequence stars fuse hydrogen atoms to form helium in their cores. About $90 \%$ of the stars in the Universe, including the Sun, are main sequence stars. These stars can range from about a tenth of the mass of the Sun to up to 200 times as massive. The main source of energy in main sequence stars is from nuclear fusion. The mass of one hydrogen nucleus is $m_{\mathrm{H}}=1.674 \times 10^{-27} \mathrm{~kg}$, and the mass of one helium nucleus is $m_{\mathrm{He}}=6.649 \times 10^{-27} \mathrm{~kg}$. [figure1] Figure 3: Left: The proton-proton chain reaction is one of two known sets of nuclear fusion reactions by which stars convert hydrogen to helium. It dominates in stars with masses less than or equal to that of the Sun's, and involves a net change of four hydrogen nuclei becoming one helium nucleus. Right: Only the core of a main sequence star will undergo nuclear fusion due to the higher temperature than the surrounding hydrogen shell. The Sun is composed of about $71 \%$ hydrogen, $27 \%$ helium and some heavier elements. However, only $13 \%$ of the hydrogen is available for hydrogen fusion in the core. The rest remains in layers of the Sun where the temperature is too low for fusion to occur. Use these figures and the answers to previous questions to calculate the lifetime in years of hydrogen fusion in the Sun (called its main sequence lifetime), assuming the Sun's luminosity remains constant. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of years, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_833

A supernova is triggered largely by neutrinos. In fact, $99 \%$ of the energy coming from the supernova is released in form of neutrinos. Over a time span of about three months, the supernova outputs visible light with power equivalent to 10 billion Suns. Assuming supernova neutrinos have mean energy of around $10 \mathrm{MeV}$, that all the power of the supernova is released during the time it is visible, and that all of the power released is released in the form of either visible light or neutrinos, estimate the number of neutrinos released. A: $10^{54}$ B: $10^{55}$ C: $10^{50}$ D: $10^{57}$ E: $10^{60}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: A supernova is triggered largely by neutrinos. In fact, $99 \%$ of the energy coming from the supernova is released in form of neutrinos. Over a time span of about three months, the supernova outputs visible light with power equivalent to 10 billion Suns. Assuming supernova neutrinos have mean energy of around $10 \mathrm{MeV}$, that all the power of the supernova is released during the time it is visible, and that all of the power released is released in the form of either visible light or neutrinos, estimate the number of neutrinos released. A: $10^{54}$ B: $10^{55}$ C: $10^{50}$ D: $10^{57}$ E: $10^{60}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

text-only

Astronomy_415

2021 年 5 月, “天问一号”探测器成功在火星软着陆, 我国成为第一个首次探测火星就实现“绕、落、巡”任务的国家。为了简化问题，可认为地球和火星在同一平面上绕太阳做匀速圆周运动, 如图 1 所示。已知地球的公转周期为 $T_{1}$, 公转轨道半径为 $r_{1}$, 火星的公转周期为 $T_{2}$, 火星质量为 $M$ 。如图 2 所示, 以火星为参考系, 质量为 $m_{1}$ 的探测器沿 1 号轨道到达 $B$ 点时速度为 $v_{1}, B$ 点到火星球心的距离为 $r_{3}$, 此时启动发动机, 在极短时间内一次性喷出部分气体, 喷气后探测器质量变为 $m_{2}$ 、速度变为与 $v_{1}$ 垂直的 $v_{2}$,然后进入以 $B$ 点为远火点的椭圆轨道 2 。已知万有引力势能公式 $E_{\mathrm{p}}=-\frac{G M m}{r}$, 其中 $M$为中心天体的质量, $m$ 为卫星的质量, $G$ 为引力常量, $r$ 为卫星到中心天体球心的距离。求 火星公转轨道半径 $r_{2}$;[图1] 图1 [图2] 图2

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 2021 年 5 月, “天问一号”探测器成功在火星软着陆, 我国成为第一个首次探测火星就实现“绕、落、巡”任务的国家。为了简化问题，可认为地球和火星在同一平面上绕太阳做匀速圆周运动, 如图 1 所示。已知地球的公转周期为 $T_{1}$, 公转轨道半径为 $r_{1}$, 火星的公转周期为 $T_{2}$, 火星质量为 $M$ 。如图 2 所示, 以火星为参考系, 质量为 $m_{1}$ 的探测器沿 1 号轨道到达 $B$ 点时速度为 $v_{1}, B$ 点到火星球心的距离为 $r_{3}$, 此时启动发动机, 在极短时间内一次性喷出部分气体, 喷气后探测器质量变为 $m_{2}$ 、速度变为与 $v_{1}$ 垂直的 $v_{2}$,然后进入以 $B$ 点为远火点的椭圆轨道 2 。已知万有引力势能公式 $E_{\mathrm{p}}=-\frac{G M m}{r}$, 其中 $M$为中心天体的质量, $m$ 为卫星的质量, $G$ 为引力常量, $r$ 为卫星到中心天体球心的距离。求 火星公转轨道半径 $r_{2}$;[图1] 图1 [图2] 图2 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_985

On the 1st January 2019 the New Horizons probe (having successfully flown by Pluto in 2015) had an encounter with the Kuiper belt object Ultima Thule, making it the most distant object ever visited by a spacecraft. At the time it was 43.4 au from the Sun. Given the apparent magnitude of the Sun from the Earth is - 26.74, what is the apparent magnitude of the Sun from Ultima Thule? A: -18.25 B: -18.35 C: -18.45 D: -18.55

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: On the 1st January 2019 the New Horizons probe (having successfully flown by Pluto in 2015) had an encounter with the Kuiper belt object Ultima Thule, making it the most distant object ever visited by a spacecraft. At the time it was 43.4 au from the Sun. Given the apparent magnitude of the Sun from the Earth is - 26.74, what is the apparent magnitude of the Sun from Ultima Thule? A: -18.25 B: -18.35 C: -18.45 D: -18.55 You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

text-only

Astronomy_601

如图甲所示, 小明在地球表面进行了物体在坚直方向做直线运动的实验, 弹簧原长时, 小球由静止释放, 在弹簧弹力与重力作用下, 测得小球的加速度 $a$ 与位移 $x$ 的关系图像如图乙所示。已知弹簧的劲度系数为 $k$, 地球的半径为 $R$, 万有引力常量为 $G$, 不考虑地球自转影响, 忽略空气阻力, 下列说法正确的是（ ) [图1] 甲 [图2] 乙 A: 小球的位移为 $x_{0}$ 时, 小球正好处于完全失重状态 B: 小球的最大速度为 $\sqrt{a_{0} x_{0}}$ C: 小球的质量为 $\frac{k x_{0}}{2 a_{0}}$ D: 地球的密度为 $\frac{3 a_{0}}{2 \pi G R}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 如图甲所示, 小明在地球表面进行了物体在坚直方向做直线运动的实验, 弹簧原长时, 小球由静止释放, 在弹簧弹力与重力作用下, 测得小球的加速度 $a$ 与位移 $x$ 的关系图像如图乙所示。已知弹簧的劲度系数为 $k$, 地球的半径为 $R$, 万有引力常量为 $G$, 不考虑地球自转影响, 忽略空气阻力, 下列说法正确的是（ ) [图1] 甲 [图2] 乙 A: 小球的位移为 $x_{0}$ 时, 小球正好处于完全失重状态 B: 小球的最大速度为 $\sqrt{a_{0} x_{0}}$ C: 小球的质量为 $\frac{k x_{0}}{2 a_{0}}$ D: 地球的密度为 $\frac{3 a_{0}}{2 \pi G R}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_1034

Why are there two tides every day? A: The Moon causes the one at night and the Sun causes the one during the day B: The Earth and Moon orbit a common centre of mass C: The Moon is tidally locked to the Earth D: Water waves can only travel around the Earth in 12 hours

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Why are there two tides every day? A: The Moon causes the one at night and the Sun causes the one during the day B: The Earth and Moon orbit a common centre of mass C: The Moon is tidally locked to the Earth D: Water waves can only travel around the Earth in 12 hours You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

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Astronomy

EN

text-only

Astronomy_315

如图所示, 横截面积为 $A$ 、质量为 $m$ 的柱状飞行器沿半径为 $R$ 的圆形轨道在高空绕地球做无动力运行。将地球看作质量为 $M$ 的均匀球体。万有引力常量为 $G$ 。 在飞行器运行轨道附近范围内有密度为 $\rho$ (恒量) 的稀薄空气。稀薄空气可看成是由彼此没有相互作用的均匀小颗粒组成，所有小颗粒原来都静止。假设每个小颗粒与飞行器碰撞后具有与飞行器相同的速度, 且碰撞时间很短。频繁碰撞会对飞行器产生持续阻力, 飞行器的轨道高度会逐渐降低。观察发现飞行器绕地球运行很多圈之后, 其轨道高度下降了 $\Delta H$ 。由于 $\Delta H \ll R$, 可将飞行器绕地球运动的每一圈运动均视为匀速圆周运动。已知当飞行器到地球球心距离为 $r$ 时, 飞行器与地球组成的系统具有的引力势能 $E_{\mathrm{p}}=-\frac{G M m}{r}$ 。请根据上述条件推导: 飞行器在半径为 $R$ 轨道上运行时, 所受空气阻力大小 $F$ 的表达式; [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 如图所示, 横截面积为 $A$ 、质量为 $m$ 的柱状飞行器沿半径为 $R$ 的圆形轨道在高空绕地球做无动力运行。将地球看作质量为 $M$ 的均匀球体。万有引力常量为 $G$ 。 在飞行器运行轨道附近范围内有密度为 $\rho$ (恒量) 的稀薄空气。稀薄空气可看成是由彼此没有相互作用的均匀小颗粒组成，所有小颗粒原来都静止。假设每个小颗粒与飞行器碰撞后具有与飞行器相同的速度, 且碰撞时间很短。频繁碰撞会对飞行器产生持续阻力, 飞行器的轨道高度会逐渐降低。观察发现飞行器绕地球运行很多圈之后, 其轨道高度下降了 $\Delta H$ 。由于 $\Delta H \ll R$, 可将飞行器绕地球运动的每一圈运动均视为匀速圆周运动。已知当飞行器到地球球心距离为 $r$ 时, 飞行器与地球组成的系统具有的引力势能 $E_{\mathrm{p}}=-\frac{G M m}{r}$ 。请根据上述条件推导: 飞行器在半径为 $R$ 轨道上运行时, 所受空气阻力大小 $F$ 的表达式; [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_678

我国首次发射的火星探测器“天问一号”自 2020 年 7 月 23 日成功发射入轨后, 2021 年 2 月 10 日成功被火星捕获，顺利进入环火轨道; 5 月 15 日, “天问一号”着陆巡视器顺利软着陆于火星表面。关于“天问一号”的运行，可以简化为如图所示的模型: “天问一号”先绕火星做半径为 $R_{1}$ 、周期为 $T$ 的匀速圆周运动, 在某一位置 $A$ 点改变速度,使其轨道变为粗圆, 椭圆轨道在 $B$ 点与火星表面相切, 设法使着陆巡视器落在火星上。若火星的半径为 $R_{2}$, 则下列说法正确的是 ( ) [图1] A: “天问一号”从圆轨道变为椭圆轨道, 机械能增加 B: “天问一号”在圆轨道和椭圆轨道上 $A$ 点的加速度相同 C: “天问一号”从椭圆轨道的 $A$ 点运动到 $B$ 点所需的时间为 $\frac{T}{2} \sqrt{\left(\frac{R_{1}+R_{2}}{2 R_{1}}\right)^{3}}$ D: “天问一号”在椭圆轨道 $B$ 点的速度等于火星的第一宇宙速度

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 我国首次发射的火星探测器“天问一号”自 2020 年 7 月 23 日成功发射入轨后, 2021 年 2 月 10 日成功被火星捕获，顺利进入环火轨道; 5 月 15 日, “天问一号”着陆巡视器顺利软着陆于火星表面。关于“天问一号”的运行，可以简化为如图所示的模型: “天问一号”先绕火星做半径为 $R_{1}$ 、周期为 $T$ 的匀速圆周运动, 在某一位置 $A$ 点改变速度,使其轨道变为粗圆, 椭圆轨道在 $B$ 点与火星表面相切, 设法使着陆巡视器落在火星上。若火星的半径为 $R_{2}$, 则下列说法正确的是 ( ) [图1] A: “天问一号”从圆轨道变为椭圆轨道, 机械能增加 B: “天问一号”在圆轨道和椭圆轨道上 $A$ 点的加速度相同 C: “天问一号”从椭圆轨道的 $A$ 点运动到 $B$ 点所需的时间为 $\frac{T}{2} \sqrt{\left(\frac{R_{1}+R_{2}}{2 R_{1}}\right)^{3}}$ D: “天问一号”在椭圆轨道 $B$ 点的速度等于火星的第一宇宙速度 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

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Astronomy

ZH

multi-modal

Astronomy_787

What is the name of the JWST component highlighted below? [figure1] A: Primary mirror B: Secondary mirror C: Optics subsystem D: Antenna

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: What is the name of the JWST component highlighted below? [figure1] A: Primary mirror B: Secondary mirror C: Optics subsystem D: Antenna You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

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Astronomy

EN

multi-modal

Astronomy_291

“神舟八号”与“天宫一号”对接前各自绕地球运动, 设“天宫一号”在半径为 $r_{I}$ 的圆轨道上运动, 周期为 $T_{1}$, “神舟八号”在半径为 $r_{2}$ 的圆轨道上运动, $r_{1}>r_{2}$, 则 ( ) A: “神舟八号”的周期 $T_{2}=T_{I} \sqrt{\frac{r_{2}{ }^{3}}{r_{1}^{3}}}$ B: “天宫一号”的运行速度大于 $7.9 \mathrm{~km} / \mathrm{s}$ C: 地球表面的重力加速度 $\mathrm{g}=\frac{4 \pi^{2} r_{1}}{T_{1}^{2}}$ D: 地球的质量 $\mathrm{M}=\frac{4 \pi^{2} r_{1}^{3}}{G T_{1}^{2}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: “神舟八号”与“天宫一号”对接前各自绕地球运动, 设“天宫一号”在半径为 $r_{I}$ 的圆轨道上运动, 周期为 $T_{1}$, “神舟八号”在半径为 $r_{2}$ 的圆轨道上运动, $r_{1}>r_{2}$, 则 ( ) A: “神舟八号”的周期 $T_{2}=T_{I} \sqrt{\frac{r_{2}{ }^{3}}{r_{1}^{3}}}$ B: “天宫一号”的运行速度大于 $7.9 \mathrm{~km} / \mathrm{s}$ C: 地球表面的重力加速度 $\mathrm{g}=\frac{4 \pi^{2} r_{1}}{T_{1}^{2}}$ D: 地球的质量 $\mathrm{M}=\frac{4 \pi^{2} r_{1}^{3}}{G T_{1}^{2}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_964

When two objects of unequal mass orbit around each other, they both orbit around a barycentre - this is the name given to the location of the centre of mass of the system. The masses of both objects, and the distance between their centres, affects the position of their barycentre. Imagine two objects, Object 1 and Object 2, with masses $m_{1}$ and $m_{2}$ respectively, and the average distance between the centre of both objects is $a$, then the average distance from the centre of Object 1 to the barycentre, $r$, is given by the formula: $$ r=a \frac{m_{2}}{m_{1}+m_{2}} $$ Some have claimed that a double planet should be distinguished from a planet and large moon when a system fulfils the criterion $r>R_{1}$. The Earth-Moon system does not currently satisfy that condition for a double planet despite the Moon being rather large relative to the Earth, but the Moon is slowly moving away from the Earth at roughly $4 \mathrm{~cm}$ per year. Assuming this rate stays constant, calculate the number of years until $r=\mathrm{R}_{\mathrm{E}}$. [Average distance between centres of Earth and Moon $=384400 \mathrm{~km}$, and the Earth has 83.1 times the mass of the Moon.]

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: When two objects of unequal mass orbit around each other, they both orbit around a barycentre - this is the name given to the location of the centre of mass of the system. The masses of both objects, and the distance between their centres, affects the position of their barycentre. Imagine two objects, Object 1 and Object 2, with masses $m_{1}$ and $m_{2}$ respectively, and the average distance between the centre of both objects is $a$, then the average distance from the centre of Object 1 to the barycentre, $r$, is given by the formula: $$ r=a \frac{m_{2}}{m_{1}+m_{2}} $$ Some have claimed that a double planet should be distinguished from a planet and large moon when a system fulfils the criterion $r>R_{1}$. The Earth-Moon system does not currently satisfy that condition for a double planet despite the Moon being rather large relative to the Earth, but the Moon is slowly moving away from the Earth at roughly $4 \mathrm{~cm}$ per year. Assuming this rate stays constant, calculate the number of years until $r=\mathrm{R}_{\mathrm{E}}$. [Average distance between centres of Earth and Moon $=384400 \mathrm{~km}$, and the Earth has 83.1 times the mass of the Moon.] All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of years, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

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Astronomy

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text-only

Astronomy_368

卫星携带一探测器在半径为 $4 R$ 的圆轨道 $I$ 上绕地球做匀速圆周运动。在 $\mathrm{A}$ 点, 卫星上的辅助动力装置短暂工作, 将探测器沿运动方向射出 (设辅助动力装置喷出的气体质量可忽略)。若探测器恰能完全脱离地球的引力范围, 即到达距地球无限远时的速度恰好为零, 而卫星沿新的椭圆轨道II运动, 如图所示, $A 、 B$ 两点分别是其椭圆轨道II的远地点和近地点（卫星通过 $A 、 B$ 两点时的线速度大小与其距地心距离的乘积相等）。地球质量为 $M$, 探测器的质量为 $m$, 卫星的质量为 $\sqrt{2} m$, 地球半径为 $R$, 引力常量为 $G$, 已知质量分别为 $m_{1} 、 m_{2}$ 的两个质点相距为 $r$ 时, 它们之间的引力势能为 $E_{p}=-\frac{G m_{1} m_{2}}{r}$, 求: 卫星运行到近地点 $B$ 时距地心的距离 $a$ 。 [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 卫星携带一探测器在半径为 $4 R$ 的圆轨道 $I$ 上绕地球做匀速圆周运动。在 $\mathrm{A}$ 点, 卫星上的辅助动力装置短暂工作, 将探测器沿运动方向射出 (设辅助动力装置喷出的气体质量可忽略)。若探测器恰能完全脱离地球的引力范围, 即到达距地球无限远时的速度恰好为零, 而卫星沿新的椭圆轨道II运动, 如图所示, $A 、 B$ 两点分别是其椭圆轨道II的远地点和近地点（卫星通过 $A 、 B$ 两点时的线速度大小与其距地心距离的乘积相等）。地球质量为 $M$, 探测器的质量为 $m$, 卫星的质量为 $\sqrt{2} m$, 地球半径为 $R$, 引力常量为 $G$, 已知质量分别为 $m_{1} 、 m_{2}$ 的两个质点相距为 $r$ 时, 它们之间的引力势能为 $E_{p}=-\frac{G m_{1} m_{2}}{r}$, 求: 卫星运行到近地点 $B$ 时距地心的距离 $a$ 。 [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

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Astronomy

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multi-modal

Astronomy_1027

Figure 4 below is a composite image which depicts a transit of the International Space Station (ISS) across the disc of the Sun. The image comprises 26 individual photographs which were taken at regular time intervals during the transit. The total duration of the transit was less than one second. In this question we will ignore any effects caused by the rotation of the Earth. [figure1] Figure 4: A composite of a selection of the frames taken with a high-speed camera of a transit of the ISS in front of the Sun, taken from Northamptonshire at 10:22 BST on $17^{\text {th }}$ June 2022. Credit: Jamie Cooper Photography The ISS maintained a mean height of $415 \mathrm{~km}$ above the surface of the Earth during the transit. What is the angle $\theta_{2}$ subtended by the ISS between the first and last photographs, as viewed from the centre of the Earth? [Assume that the Sun was at the zenith (directly overhead) during the transit.]

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: Figure 4 below is a composite image which depicts a transit of the International Space Station (ISS) across the disc of the Sun. The image comprises 26 individual photographs which were taken at regular time intervals during the transit. The total duration of the transit was less than one second. In this question we will ignore any effects caused by the rotation of the Earth. [figure1] Figure 4: A composite of a selection of the frames taken with a high-speed camera of a transit of the ISS in front of the Sun, taken from Northamptonshire at 10:22 BST on $17^{\text {th }}$ June 2022. Credit: Jamie Cooper Photography The ISS maintained a mean height of $415 \mathrm{~km}$ above the surface of the Earth during the transit. What is the angle $\theta_{2}$ subtended by the ISS between the first and last photographs, as viewed from the centre of the Earth? [Assume that the Sun was at the zenith (directly overhead) during the transit.] All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value.

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Astronomy

EN

multi-modal

Astronomy_652

一宇航员到达半径为 $R$ 、密度均匀的某星球表面, 做了如下实验: 用不可伸长的轻绳拴一质量为 $m$ 的小球, 上端固定于 $O$ 点, 如图甲所示, 在最低点给小球某一初速度,使其绕 $O$ 点的坚直面内做圆周运动, 测得绳的拉力 $F$ 大小随时间 $t$ 的变化规律如图乙所示, $F_{1}=3 F_{2}$, 设 $R 、 m$ 、引力常量 $G$ 和 $F_{1}$ 为已知量, 忽略各种阻力。下列说法正确的是 [图1] 甲 [图2] 乙 A: 该星球表面的重力加速度为 $\frac{F_{1}}{9 m}$ B: 星球的密度为 $\frac{F_{1}}{6 \pi G R m}$ C: 卫星绕该星球运行的最小周期为 $2 \pi \sqrt{\frac{R m}{F_{1}}}$ D: 该星球的第一宇宙速度为 $\sqrt{\frac{2 F_{1} R}{9 m}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 一宇航员到达半径为 $R$ 、密度均匀的某星球表面, 做了如下实验: 用不可伸长的轻绳拴一质量为 $m$ 的小球, 上端固定于 $O$ 点, 如图甲所示, 在最低点给小球某一初速度,使其绕 $O$ 点的坚直面内做圆周运动, 测得绳的拉力 $F$ 大小随时间 $t$ 的变化规律如图乙所示, $F_{1}=3 F_{2}$, 设 $R 、 m$ 、引力常量 $G$ 和 $F_{1}$ 为已知量, 忽略各种阻力。下列说法正确的是 [图1] 甲 [图2] 乙 A: 该星球表面的重力加速度为 $\frac{F_{1}}{9 m}$ B: 星球的密度为 $\frac{F_{1}}{6 \pi G R m}$ C: 卫星绕该星球运行的最小周期为 $2 \pi \sqrt{\frac{R m}{F_{1}}}$ D: 该星球的第一宇宙速度为 $\sqrt{\frac{2 F_{1} R}{9 m}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

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multi-modal

Astronomy_405

2021 年 6 月 17 日, 神舟十二号载人飞船与天和核心舱完成对接, 航天员聂海胜、刘伯明、汤洪波进入天和核心舱，标志着中国人首次进入了自己的空间站。对接过程的示意图如图所示, 天和核心舱处于半径为 $r_{3}$ 的圆轨道III; 神舟十二号飞船处于半径为 $r_{I}$的圆轨道I, 运行周期为 $T_{I}$, 通过变轨操作后, 沿粗圆轨道II运动到 $B$ 点与天和核心舱对接。则下列说法正确的是（） [图1] A: 神舟十二号飞船在轨道II上运动时经 $A 、 B$ 两点速率 $v_{A}: v_{B}=r_{1}: r_{3}$ B: 神舟十二号飞船沿轨道II运行的周期为 $T_{2}=T_{1} \sqrt{\left(\frac{r_{1}+r_{3}}{2 r_{1}}\right)^{3}}$ C: 神舟十二号飞船沿轨道I运行的周期大于天和核心舱沿轨道III运行的周期 D: 正常运行时, 神舟十二号飞船在轨道II上经过 $B$ 点的加速度大于在轨道III上经过 $B$ 点的加速度

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2021 年 6 月 17 日, 神舟十二号载人飞船与天和核心舱完成对接, 航天员聂海胜、刘伯明、汤洪波进入天和核心舱，标志着中国人首次进入了自己的空间站。对接过程的示意图如图所示, 天和核心舱处于半径为 $r_{3}$ 的圆轨道III; 神舟十二号飞船处于半径为 $r_{I}$的圆轨道I, 运行周期为 $T_{I}$, 通过变轨操作后, 沿粗圆轨道II运动到 $B$ 点与天和核心舱对接。则下列说法正确的是（） [图1] A: 神舟十二号飞船在轨道II上运动时经 $A 、 B$ 两点速率 $v_{A}: v_{B}=r_{1}: r_{3}$ B: 神舟十二号飞船沿轨道II运行的周期为 $T_{2}=T_{1} \sqrt{\left(\frac{r_{1}+r_{3}}{2 r_{1}}\right)^{3}}$ C: 神舟十二号飞船沿轨道I运行的周期大于天和核心舱沿轨道III运行的周期 D: 正常运行时, 神舟十二号飞船在轨道II上经过 $B$ 点的加速度大于在轨道III上经过 $B$ 点的加速度 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

ZH

multi-modal

Astronomy_839

In a certain day, when it is 0h UT, the sidereal time of Prime Meridian is 5h 56min 9.4s. For this day, with start and end based on UT, find the civil time of Chicago, whose longitude and time zone are respectively, $87.65004722^{\circ} \mathrm{W}$ and UT-6, when the sidereal time there is $20 \mathrm{~h}$. The difference between solar time and sidereal time SHOULD be accounted for. A: $14 \mathrm{~h} 1 \mathrm{~min} 32 \mathrm{~s}$ B: $13 \mathrm{~h} 26 \mathrm{~min} 17 \mathrm{~s}$ C: $14 \mathrm{~h} 36 \mathrm{~min} 47 \mathrm{~s}$ D: $14 \mathrm{~h} 0 \mathrm{~min} 43 \mathrm{~s}$ E: $13 \mathrm{~h} 51 \mathrm{~min} 11 \mathrm{~s}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: In a certain day, when it is 0h UT, the sidereal time of Prime Meridian is 5h 56min 9.4s. For this day, with start and end based on UT, find the civil time of Chicago, whose longitude and time zone are respectively, $87.65004722^{\circ} \mathrm{W}$ and UT-6, when the sidereal time there is $20 \mathrm{~h}$. The difference between solar time and sidereal time SHOULD be accounted for. A: $14 \mathrm{~h} 1 \mathrm{~min} 32 \mathrm{~s}$ B: $13 \mathrm{~h} 26 \mathrm{~min} 17 \mathrm{~s}$ C: $14 \mathrm{~h} 36 \mathrm{~min} 47 \mathrm{~s}$ D: $14 \mathrm{~h} 0 \mathrm{~min} 43 \mathrm{~s}$ E: $13 \mathrm{~h} 51 \mathrm{~min} 11 \mathrm{~s}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

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Astronomy

EN

text-only

Astronomy_1092

In July 1969 the mission Apollo 11 was the first to successfully allow humans to walk on the Moon. This was an incredible achievement as the engineering necessary to make it a possibility was an order of magnitude more complex than anything that had come before. The Apollo 11 spacecraft was launched atop the Saturn V rocket, which still stands as the most powerful rocket ever made. [figure1] Figure 1: Left: The launch of Apollo 11 upon the Saturn V rocket. Credit: NASA. Right: Showing the three stages of the Saturn V rocket (each detached once its fuel was expended), plus the Apollo spacecraft on top (containing three astronauts) which was delivered into a translunar orbit. At the base of the rocket is a person to scale, emphasising the enormous size of the rocket. Credit: Encyclopaedia Britannica. | Stage | Initial Mass $(\mathrm{t})$ | Final mass $(\mathrm{t})$ | $I_{\mathrm{sp}}(\mathrm{s})$ | Burn duration $(\mathrm{s})$ | | :---: | :---: | :---: | :---: | :---: | | S-IC | 2283.9 | 135.6 | 263 | 168 | | S-II | 483.7 | 39.9 | 421 | 384 | | S-IV (Burn 1) | 121.0 | - | 421 | 147 | | S-IV (Burn 2) | - | 13.2 | 421 | 347 | | Apollo Spacecraft | 49.7 | - | - | - | Table 1: Data about each stage of the rocket used to launch the Apollo 11 spacecraft into a translunar orbit. Masses are given in tonnes $(1 \mathrm{t}=1000 \mathrm{~kg}$ ) and for convenience include the interstage parts of the rocket too. The specific impulse, $I_{\mathrm{sp}}$, of the stage is given at sea level atmospheric pressure for S-IC and for a vacuum for S-II and S-IVB. The Saturn V rocket consisted of three stages (see Fig 1), since this was the only practical way to get the Apollo spacecraft up to the speed necessary to make the transfer to the Moon. When fully fueled the mass of the total rocket was immense, and lots of that fuel was necessary to simply lift the fuel of the later stages into high altitude - in total about $3000 \mathrm{t}(1$ tonne, $\mathrm{t}=1000 \mathrm{~kg}$ ) of rocket on the launchpad was required to send about $50 \mathrm{t}$ on a mission to the Moon. The first stage (called S-IC) was the heaviest, the second (called S-II) was considerably lighter, and the third stage (called S-IVB) was fired twice - the first to get the spacecraft into a circular 'parking' orbit around the Earth where various safety checks were made, whilst the second burn was to get the spacecraft on its way to the Moon. Once each rocket stage was fully spent it was detached from the rest of the rocket before the next stage ignited. Data about each stage is given in Table 1. The thrust of the rocket is given as $$ F=-I_{\mathrm{sp}} g_{0} \dot{m} $$ where the specific impulse, $I_{\mathrm{sp}}$, of each stage is a constant related to the type of fuel used and the shape of the rocket nozzle, $g_{0}$ is the gravitational field strength of the Earth at sea level (i.e. $g_{0}=9.81 \mathrm{~m} \mathrm{~s}^{-2}$ ) and $\dot{m} \equiv \mathrm{d} m / \mathrm{d} t$ is the rate of change of mass of the rocket with time. The thrust generated by the first two stages (S-IC and S-II) can be taken to be constant. However, the thrust generated by the third stage (S-IVB) varied in order to give a constant acceleration (taken to be the same throughout both burns of the rocket). By the end of the second burn the Apollo spacecraft, apart from a few short burns to give mid-course corrections, coasted all the way to the far side of the Moon where the engines were then fired again to circularise the orbit. All of the early Apollo missions were on a orbit known as a 'free-return trajectory', meaning that if there was a problem then they were already on an orbit that would take them back to Earth after passing around the Moon. The real shape of such a trajectory (in a rotating frame of reference) is like a stretched figure of 8 and is shown in the top panel of Fig 2. To calculate this precisely is non-trivial and required substantial computing power in the 1960s. However, we can have two simplified models that can be used to estimate the duration of the translunar coast, and they are shown in the bottom panel of the Fig 2. The first is a Hohmann transfer orbit (dashed line), which is a single ellipse with the Earth at one focus. In this model the gravitational effect of the Moon is ignored, so the spacecraft travels from A (the perigee) to B (the apogee). The second (solid line) takes advantage of a 'patched conics' approach by having two ellipses whose apoapsides coincide at point $\mathrm{C}$ where the gravitational force on the spacecraft is equal from both the Earth and the Moon. The first ellipse has a periapsis at A and ignores the gravitational effect of the Moon, whilst the second ellipse has a periapsis at B and ignores the gravitational effect of the Earth. If the spacecraft trajectory and lunar orbit are coplanar and the Moon is in a circular orbit around the Earth then the time to travel from $\mathrm{A}$ to $\mathrm{B}$ via $\mathrm{C}$ is double the value attained if taking into account the gravitational forces of the Earth and Moon together throughout the journey, which is a much better estimate of the time of a real translunar coast. [figure2] Figure 2: Top: The real shape of a translunar free-return trajectory, with the Earth on the left and the Moon on the right (orbiting around the Earth in an anti-clockwise direction). This diagram (and the one below) is shown in a co-ordinate system co-rotating with the Earth and is not to scale. Credit: NASA. Bottom: Two simplified ways of modelling the translunar trajectory. The simplest is a Hohmann transfer orbit (dashed line, outer ellipse), which is an ellipse that has the Earth at one focus and ignores the gravitational effect of the Moon. A better model (solid line, inner ellipses) of the Apollo trajectory is the use of two ellipses that meet at point $\mathrm{C}$ where the gravitational forces of the Earth and Moon on the spacecraft are equal. For the Apollo 11 journey, the end of the second burn of the S-IVB rocket (point A) was $334 \mathrm{~km}$ above the surface of the Earth, and the end of the translunar coast (point B) was $161 \mathrm{~km}$ above the surface of the Moon. The distance between the centres of mass of the Earth and the Moon at the end of the translunar coast was $3.94 \times 10^{8} \mathrm{~m}$. Take the radius of the Earth to be $6370 \mathrm{~km}$, the radius of the Moon to be $1740 \mathrm{~km}$, and the mass of the Moon to be $7.35 \times 10^{22} \mathrm{~kg}$.b. In reality, the effects of air resistance and the weight of the rocket are substantial. Once in the parking orbit it is travelling at $7.79 \mathrm{~km} \mathrm{~s}^{-1}$. i. What is its height above the Earth's surface (measured from sea level)? Give it in km.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: In July 1969 the mission Apollo 11 was the first to successfully allow humans to walk on the Moon. This was an incredible achievement as the engineering necessary to make it a possibility was an order of magnitude more complex than anything that had come before. The Apollo 11 spacecraft was launched atop the Saturn V rocket, which still stands as the most powerful rocket ever made. [figure1] Figure 1: Left: The launch of Apollo 11 upon the Saturn V rocket. Credit: NASA. Right: Showing the three stages of the Saturn V rocket (each detached once its fuel was expended), plus the Apollo spacecraft on top (containing three astronauts) which was delivered into a translunar orbit. At the base of the rocket is a person to scale, emphasising the enormous size of the rocket. Credit: Encyclopaedia Britannica. | Stage | Initial Mass $(\mathrm{t})$ | Final mass $(\mathrm{t})$ | $I_{\mathrm{sp}}(\mathrm{s})$ | Burn duration $(\mathrm{s})$ | | :---: | :---: | :---: | :---: | :---: | | S-IC | 2283.9 | 135.6 | 263 | 168 | | S-II | 483.7 | 39.9 | 421 | 384 | | S-IV (Burn 1) | 121.0 | - | 421 | 147 | | S-IV (Burn 2) | - | 13.2 | 421 | 347 | | Apollo Spacecraft | 49.7 | - | - | - | Table 1: Data about each stage of the rocket used to launch the Apollo 11 spacecraft into a translunar orbit. Masses are given in tonnes $(1 \mathrm{t}=1000 \mathrm{~kg}$ ) and for convenience include the interstage parts of the rocket too. The specific impulse, $I_{\mathrm{sp}}$, of the stage is given at sea level atmospheric pressure for S-IC and for a vacuum for S-II and S-IVB. The Saturn V rocket consisted of three stages (see Fig 1), since this was the only practical way to get the Apollo spacecraft up to the speed necessary to make the transfer to the Moon. When fully fueled the mass of the total rocket was immense, and lots of that fuel was necessary to simply lift the fuel of the later stages into high altitude - in total about $3000 \mathrm{t}(1$ tonne, $\mathrm{t}=1000 \mathrm{~kg}$ ) of rocket on the launchpad was required to send about $50 \mathrm{t}$ on a mission to the Moon. The first stage (called S-IC) was the heaviest, the second (called S-II) was considerably lighter, and the third stage (called S-IVB) was fired twice - the first to get the spacecraft into a circular 'parking' orbit around the Earth where various safety checks were made, whilst the second burn was to get the spacecraft on its way to the Moon. Once each rocket stage was fully spent it was detached from the rest of the rocket before the next stage ignited. Data about each stage is given in Table 1. The thrust of the rocket is given as $$ F=-I_{\mathrm{sp}} g_{0} \dot{m} $$ where the specific impulse, $I_{\mathrm{sp}}$, of each stage is a constant related to the type of fuel used and the shape of the rocket nozzle, $g_{0}$ is the gravitational field strength of the Earth at sea level (i.e. $g_{0}=9.81 \mathrm{~m} \mathrm{~s}^{-2}$ ) and $\dot{m} \equiv \mathrm{d} m / \mathrm{d} t$ is the rate of change of mass of the rocket with time. The thrust generated by the first two stages (S-IC and S-II) can be taken to be constant. However, the thrust generated by the third stage (S-IVB) varied in order to give a constant acceleration (taken to be the same throughout both burns of the rocket). By the end of the second burn the Apollo spacecraft, apart from a few short burns to give mid-course corrections, coasted all the way to the far side of the Moon where the engines were then fired again to circularise the orbit. All of the early Apollo missions were on a orbit known as a 'free-return trajectory', meaning that if there was a problem then they were already on an orbit that would take them back to Earth after passing around the Moon. The real shape of such a trajectory (in a rotating frame of reference) is like a stretched figure of 8 and is shown in the top panel of Fig 2. To calculate this precisely is non-trivial and required substantial computing power in the 1960s. However, we can have two simplified models that can be used to estimate the duration of the translunar coast, and they are shown in the bottom panel of the Fig 2. The first is a Hohmann transfer orbit (dashed line), which is a single ellipse with the Earth at one focus. In this model the gravitational effect of the Moon is ignored, so the spacecraft travels from A (the perigee) to B (the apogee). The second (solid line) takes advantage of a 'patched conics' approach by having two ellipses whose apoapsides coincide at point $\mathrm{C}$ where the gravitational force on the spacecraft is equal from both the Earth and the Moon. The first ellipse has a periapsis at A and ignores the gravitational effect of the Moon, whilst the second ellipse has a periapsis at B and ignores the gravitational effect of the Earth. If the spacecraft trajectory and lunar orbit are coplanar and the Moon is in a circular orbit around the Earth then the time to travel from $\mathrm{A}$ to $\mathrm{B}$ via $\mathrm{C}$ is double the value attained if taking into account the gravitational forces of the Earth and Moon together throughout the journey, which is a much better estimate of the time of a real translunar coast. [figure2] Figure 2: Top: The real shape of a translunar free-return trajectory, with the Earth on the left and the Moon on the right (orbiting around the Earth in an anti-clockwise direction). This diagram (and the one below) is shown in a co-ordinate system co-rotating with the Earth and is not to scale. Credit: NASA. Bottom: Two simplified ways of modelling the translunar trajectory. The simplest is a Hohmann transfer orbit (dashed line, outer ellipse), which is an ellipse that has the Earth at one focus and ignores the gravitational effect of the Moon. A better model (solid line, inner ellipses) of the Apollo trajectory is the use of two ellipses that meet at point $\mathrm{C}$ where the gravitational forces of the Earth and Moon on the spacecraft are equal. For the Apollo 11 journey, the end of the second burn of the S-IVB rocket (point A) was $334 \mathrm{~km}$ above the surface of the Earth, and the end of the translunar coast (point B) was $161 \mathrm{~km}$ above the surface of the Moon. The distance between the centres of mass of the Earth and the Moon at the end of the translunar coast was $3.94 \times 10^{8} \mathrm{~m}$. Take the radius of the Earth to be $6370 \mathrm{~km}$, the radius of the Moon to be $1740 \mathrm{~km}$, and the mass of the Moon to be $7.35 \times 10^{22} \mathrm{~kg}$. problem: b. In reality, the effects of air resistance and the weight of the rocket are substantial. Once in the parking orbit it is travelling at $7.79 \mathrm{~km} \mathrm{~s}^{-1}$. i. What is its height above the Earth's surface (measured from sea level)? Give it in km. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \mathrm{~km}, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_14

如图所示, $A$ 为地球赤道表面的物体, $B$ 为环绕地球运行的卫星, 此卫星在距离地球表面 $\frac{R}{2}$ 的高度处做匀速圆周运动, 且向心加速度的大小为 $a$, 地球的半径为 $R$,引力常量为 $G$. 则下列说法正确的是 ( ) [图1] A: 物体 $A$ 的向心加速度大于 $a$ B: 物体 $A$ 的线速度比卫星 $B$ 的线速度大 C: 地球的质量为 $\frac{R^{2} a}{G}$ D: 地球两极的重力加速度大小为 $\frac{9}{4} a$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 如图所示, $A$ 为地球赤道表面的物体, $B$ 为环绕地球运行的卫星, 此卫星在距离地球表面 $\frac{R}{2}$ 的高度处做匀速圆周运动, 且向心加速度的大小为 $a$, 地球的半径为 $R$,引力常量为 $G$. 则下列说法正确的是 ( ) [图1] A: 物体 $A$ 的向心加速度大于 $a$ B: 物体 $A$ 的线速度比卫星 $B$ 的线速度大 C: 地球的质量为 $\frac{R^{2} a}{G}$ D: 地球两极的重力加速度大小为 $\frac{9}{4} a$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_1146

Plotting the position of the Sun in the sky at the same time every day, you get an interesting figure-ofeight shape known as an analemma (see Figure 1). For observers in the Northern hemisphere, you might expect to always see the Sun due South at midday, however on some days the Sun has already passed through that bearing and on others it needs a few more minutes before it gets there. This is due to two effects: the axial tilt of the Earth, and the fact the Earth's orbit is not perfectly circular [figure1] Figure 1: The analemma above was composed from images taken every few days at noon near the village of Callanish in the Outer Hebrides in Scotland. In the foreground are the Callanish Stones and the main photo was taken on the winter solstice (when the maximum angle the Sun reaches above the horizon is the lowest of the year, so is at the bottom of the analemma). Credit: Giuseppe Petricca. The vertical co-ordinate of a point in the analemma is entirely determined by the Earth's axial tilt. This is known as the solar declination, $\delta$, and varies sinusoidally throughout the year. The horizontal coordinate of a point in the analemma is determined by a combination of the Earth's axial tilt and the eccentricity of the Earth's orbit. Both of these individually vary sinusoidally, but the superposition of the two is no longer sinusoidal. We will define $\alpha$ as the angle between due South and the Sun at local midday as seen from Oxford, where a positive value means the Sun has already passed through due South (so is on the right of the figure above) whilst a negative value means the Sun has yet to pass through due South. If $\alpha_{\text {tilt }}$ is the contribution due to the axial tilt and $\alpha_{\text {ecc }}$ is the contribution due to the Earth's orbital eccentricity, then $$ \alpha=\alpha_{\text {tilt }}+\alpha_{\text {ecc }} $$ If the angle of the axial tilt is $\varepsilon$ and the eccentricity of the Earth's orbit is $e$, and we assume that both are small enough that the sinusoidal approximation of $\delta, \alpha_{\text {tilt }}$, and $\alpha_{\text {ecc }}$ apply, then we find the following boundary conditions: - $\delta$ has a period of 1 year, an amplitude of $\varepsilon$, is maximum at the summer solstice (21 $21^{\text {st }}$ June) and minimum at the winter solstice $\left(21^{\text {st }}\right.$ December $)$ - $\alpha_{\text {tilt }}$ has a period of 0.5 years, an amplitude (in radians) of $\tan ^{2}(\varepsilon / 2)$, is zero at the solstices and the equinoxes (vernal equinox $=21^{\text {st }}$ March, autumnal equinox $=21^{\text {st }}$ September), and (using our sign convention) positive just after the vernal equinox - $\alpha_{\text {ecc }}$ has a period of 1 year, an amplitude (in radians) of $2 e$, is zero at the perihelion (4 $4^{\text {th }}$ January) and the aphelion ( $6^{\text {th }}$ July), and (using our sign convention) negative just after the perihelion Given the $n^{\text {th }}$ day of the year, a value can be calculated for $\delta$ and $\alpha$, and these are the co-ordinates for the analemma (it is drawn by these parametric equations). For the Earth, $\varepsilon=23.44^{\circ}$ and $e=0.0167$. Consider an alternative version of Earth, known as Earth 2.0. On this planet, the year is unchanged and the perihelion and aphelion are at the same time, but it has a different axial tilt, a different orbital eccentricity, and a different month for the vernal equinox (although it is still on the $21^{\text {st }}$ day of that month). The analemma as viewed from Earth 2.0 is show in Figure 2 below. [figure2] Figure 2: The analemma of the Sun at midday as seen by an observer on Earth 2.0. In this situation, $\alpha$ ranges from -26 mins 47 secs to 18 mins 56 secs. The circled letters correspond to the same (unknown) day of each month (for example $5^{\text {th }}$ Jan, $5^{\text {th }}$ Feb, $5^{\text {th }}$ March etc.). Credit: Bob Urschel.a. Although $\alpha$ is really an angle in radians (where $2 \pi$ radians $=360^{\circ}$ ), it is normally more useful to convert it into time units (essentially the time since the Sun was due South, or the time until the Sun reaches due South). Taking the mean solar day to be exactly 24 hours: i. Convert the amplitude of $\alpha_{\text {tilt }}$ and $\alpha_{\text {ecc }}$ for the Earth into minutes.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: Plotting the position of the Sun in the sky at the same time every day, you get an interesting figure-ofeight shape known as an analemma (see Figure 1). For observers in the Northern hemisphere, you might expect to always see the Sun due South at midday, however on some days the Sun has already passed through that bearing and on others it needs a few more minutes before it gets there. This is due to two effects: the axial tilt of the Earth, and the fact the Earth's orbit is not perfectly circular [figure1] Figure 1: The analemma above was composed from images taken every few days at noon near the village of Callanish in the Outer Hebrides in Scotland. In the foreground are the Callanish Stones and the main photo was taken on the winter solstice (when the maximum angle the Sun reaches above the horizon is the lowest of the year, so is at the bottom of the analemma). Credit: Giuseppe Petricca. The vertical co-ordinate of a point in the analemma is entirely determined by the Earth's axial tilt. This is known as the solar declination, $\delta$, and varies sinusoidally throughout the year. The horizontal coordinate of a point in the analemma is determined by a combination of the Earth's axial tilt and the eccentricity of the Earth's orbit. Both of these individually vary sinusoidally, but the superposition of the two is no longer sinusoidal. We will define $\alpha$ as the angle between due South and the Sun at local midday as seen from Oxford, where a positive value means the Sun has already passed through due South (so is on the right of the figure above) whilst a negative value means the Sun has yet to pass through due South. If $\alpha_{\text {tilt }}$ is the contribution due to the axial tilt and $\alpha_{\text {ecc }}$ is the contribution due to the Earth's orbital eccentricity, then $$ \alpha=\alpha_{\text {tilt }}+\alpha_{\text {ecc }} $$ If the angle of the axial tilt is $\varepsilon$ and the eccentricity of the Earth's orbit is $e$, and we assume that both are small enough that the sinusoidal approximation of $\delta, \alpha_{\text {tilt }}$, and $\alpha_{\text {ecc }}$ apply, then we find the following boundary conditions: - $\delta$ has a period of 1 year, an amplitude of $\varepsilon$, is maximum at the summer solstice (21 $21^{\text {st }}$ June) and minimum at the winter solstice $\left(21^{\text {st }}\right.$ December $)$ - $\alpha_{\text {tilt }}$ has a period of 0.5 years, an amplitude (in radians) of $\tan ^{2}(\varepsilon / 2)$, is zero at the solstices and the equinoxes (vernal equinox $=21^{\text {st }}$ March, autumnal equinox $=21^{\text {st }}$ September), and (using our sign convention) positive just after the vernal equinox - $\alpha_{\text {ecc }}$ has a period of 1 year, an amplitude (in radians) of $2 e$, is zero at the perihelion (4 $4^{\text {th }}$ January) and the aphelion ( $6^{\text {th }}$ July), and (using our sign convention) negative just after the perihelion Given the $n^{\text {th }}$ day of the year, a value can be calculated for $\delta$ and $\alpha$, and these are the co-ordinates for the analemma (it is drawn by these parametric equations). For the Earth, $\varepsilon=23.44^{\circ}$ and $e=0.0167$. Consider an alternative version of Earth, known as Earth 2.0. On this planet, the year is unchanged and the perihelion and aphelion are at the same time, but it has a different axial tilt, a different orbital eccentricity, and a different month for the vernal equinox (although it is still on the $21^{\text {st }}$ day of that month). The analemma as viewed from Earth 2.0 is show in Figure 2 below. [figure2] Figure 2: The analemma of the Sun at midday as seen by an observer on Earth 2.0. In this situation, $\alpha$ ranges from -26 mins 47 secs to 18 mins 56 secs. The circled letters correspond to the same (unknown) day of each month (for example $5^{\text {th }}$ Jan, $5^{\text {th }}$ Feb, $5^{\text {th }}$ March etc.). Credit: Bob Urschel. problem: a. Although $\alpha$ is really an angle in radians (where $2 \pi$ radians $=360^{\circ}$ ), it is normally more useful to convert it into time units (essentially the time since the Sun was due South, or the time until the Sun reaches due South). Taking the mean solar day to be exactly 24 hours: i. Convert the amplitude of $\alpha_{\text {tilt }}$ and $\alpha_{\text {ecc }}$ for the Earth into minutes. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \text { minutes }, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_348

已知地球的半径为 $R$, 地球的自转周期为 $T$, 地表的重力加速度为 $g$, 要在地球赤道上发射一颗近地的人造地球卫星, 使其轨道在赤道的正上方, 若不计空气的阻力, 那么 ( ) A: 向东发射与向西发射耗能相同, 均为 $\frac{1}{2} m g R-\frac{1}{2} m\left(\frac{2 \pi R}{T}\right)^{2}$ B: 向东发射耗能多, 比向西发射耗能多 $\frac{1}{2} m\left(\sqrt{g R}-\frac{2 \pi R}{T}\right)^{2}$ C: 向东发射与向西发射耗能相同, 均为 $\frac{1}{2} m\left(\sqrt{g R}+\frac{2 \pi R}{T}\right)^{2}$ D: 向西发射耗能为 $\frac{1}{2} m\left(\sqrt{g R}+\frac{2 \pi R}{T}\right)^{2}$, 比向东发射耗能多

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 已知地球的半径为 $R$, 地球的自转周期为 $T$, 地表的重力加速度为 $g$, 要在地球赤道上发射一颗近地的人造地球卫星, 使其轨道在赤道的正上方, 若不计空气的阻力, 那么 ( ) A: 向东发射与向西发射耗能相同, 均为 $\frac{1}{2} m g R-\frac{1}{2} m\left(\frac{2 \pi R}{T}\right)^{2}$ B: 向东发射耗能多, 比向西发射耗能多 $\frac{1}{2} m\left(\sqrt{g R}-\frac{2 \pi R}{T}\right)^{2}$ C: 向东发射与向西发射耗能相同, 均为 $\frac{1}{2} m\left(\sqrt{g R}+\frac{2 \pi R}{T}\right)^{2}$ D: 向西发射耗能为 $\frac{1}{2} m\left(\sqrt{g R}+\frac{2 \pi R}{T}\right)^{2}$, 比向东发射耗能多 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_145

我国天文学家通过 FAST, 在武仙座球状星团 $\mathrm{M}_{1} 3$ 中发现一个脉冲双星系统。如图所示, 假设在太空中有恒星 $A 、 B$ 双星系统绕点 $O$ 做顺时针匀速圆周运动, 运动周期为 $T_{1}$, 它们的轨道半径分别为 $R_{A} 、 R_{B}, R_{A}<R_{B}, C$ 为 $B$ 的卫星, 绕 $B$ 做逆时针匀速圆周运动, 周期为 $T_{2}$ 。忽略 $A$ 与 $C$ 之间的引力, $A$ 与 $B$ 之间的引力远大于 $C$ 与 $B$ 之间的引力。万有引力常量为 $G$, 求: $A 、 B 、 C$ 三星由图示位置到再次共线所用时间 $t$; [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 我国天文学家通过 FAST, 在武仙座球状星团 $\mathrm{M}_{1} 3$ 中发现一个脉冲双星系统。如图所示, 假设在太空中有恒星 $A 、 B$ 双星系统绕点 $O$ 做顺时针匀速圆周运动, 运动周期为 $T_{1}$, 它们的轨道半径分别为 $R_{A} 、 R_{B}, R_{A}<R_{B}, C$ 为 $B$ 的卫星, 绕 $B$ 做逆时针匀速圆周运动, 周期为 $T_{2}$ 。忽略 $A$ 与 $C$ 之间的引力, $A$ 与 $B$ 之间的引力远大于 $C$ 与 $B$ 之间的引力。万有引力常量为 $G$, 求: $A 、 B 、 C$ 三星由图示位置到再次共线所用时间 $t$; [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_1068

Recently a group of researchers announced that they had discovered an Earth-sized exoplanet around our nearest star, Proxima Centauri. Its closeness raises an intriguing possibility about whether or not we might be able to image it directly using telescopes. The difficulty comes from the small angular scales that need to be resolved and the extreme differences in brightness between the reflected light from the planet and the light given out by the star. [figure1] Figure 6: Artist's impression of the view from the surface of Proxima Centauri b. Credit: ESO / M. Kornmesser Data about the star and the planet are summarised below: | Proxima Centauri (star) | | Proxima Centauri b (planet) | | | :--- | :--- | :--- | :--- | | Distance | $1.295 \mathrm{pc}$ | Orbital period | 11.186 days | | Mass | $0.123 \mathrm{M}_{\odot}$ | Mass $(\mathrm{min})$ | $\approx 1.27 \mathrm{M}_{\oplus}$ | | Radius | $0.141 \mathrm{R}_{\odot}$ | Radius $(\mathrm{min})$ | $\approx 1.1 \mathrm{R}_{\oplus}$ | | Surface temperature | $3042 \mathrm{~K}$ | | | | Apparent magnitude | 11.13 | | | The following formulae may also be helpful: $$ m-\mathcal{M}=5 \log \left(\frac{d}{10}\right) \quad \mathcal{M}-\mathcal{M}_{\odot}=-2.5 \log \left(\frac{L}{\mathrm{~L}_{\odot}}\right) \quad \Delta m=2.5 \log C R $$ where $m$ is the apparent magnitude, $\mathcal{M}$ is the absolute magnitude, $d$ is the distance in parsecs, and the contrast ratio $(C R)$ is defined as the ratio of fluxes from the star and planet, $C R=\frac{f_{\text {star }}}{f_{\text {planet }}}$.b. Determine the luminosity of the star and hence calculate the flux received on the Earth (in $\mathrm{W} \mathrm{m}^{-2}$ ) from both the star and the planet. Use them to work out the contrast ratio and thus the apparent magnitude of the planet. Assume the planet reflects half of the incident light and that $\mathcal{M} \mathcal{M}_{\odot}=4.83$.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: Recently a group of researchers announced that they had discovered an Earth-sized exoplanet around our nearest star, Proxima Centauri. Its closeness raises an intriguing possibility about whether or not we might be able to image it directly using telescopes. The difficulty comes from the small angular scales that need to be resolved and the extreme differences in brightness between the reflected light from the planet and the light given out by the star. [figure1] Figure 6: Artist's impression of the view from the surface of Proxima Centauri b. Credit: ESO / M. Kornmesser Data about the star and the planet are summarised below: | Proxima Centauri (star) | | Proxima Centauri b (planet) | | | :--- | :--- | :--- | :--- | | Distance | $1.295 \mathrm{pc}$ | Orbital period | 11.186 days | | Mass | $0.123 \mathrm{M}_{\odot}$ | Mass $(\mathrm{min})$ | $\approx 1.27 \mathrm{M}_{\oplus}$ | | Radius | $0.141 \mathrm{R}_{\odot}$ | Radius $(\mathrm{min})$ | $\approx 1.1 \mathrm{R}_{\oplus}$ | | Surface temperature | $3042 \mathrm{~K}$ | | | | Apparent magnitude | 11.13 | | | The following formulae may also be helpful: $$ m-\mathcal{M}=5 \log \left(\frac{d}{10}\right) \quad \mathcal{M}-\mathcal{M}_{\odot}=-2.5 \log \left(\frac{L}{\mathrm{~L}_{\odot}}\right) \quad \Delta m=2.5 \log C R $$ where $m$ is the apparent magnitude, $\mathcal{M}$ is the absolute magnitude, $d$ is the distance in parsecs, and the contrast ratio $(C R)$ is defined as the ratio of fluxes from the star and planet, $C R=\frac{f_{\text {star }}}{f_{\text {planet }}}$. problem: b. Determine the luminosity of the star and hence calculate the flux received on the Earth (in $\mathrm{W} \mathrm{m}^{-2}$ ) from both the star and the planet. Use them to work out the contrast ratio and thus the apparent magnitude of the planet. Assume the planet reflects half of the incident light and that $\mathcal{M} \mathcal{M}_{\odot}=4.83$. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value.

NV

Astronomy

EN

multi-modal

Astronomy_562

遥远的星球。某星球完全由不可压缩的液态水组成。星球的表面重力加速度为 $g_{0}=9.8 \mathrm{~m} / \mathrm{s}^{2}$, 半径为 $R$, 且没有自转。水的密度 $\rho=10^{3} \mathrm{~kg} / \mathrm{m}^{3}$, 万有引力常数 $G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} \cdot \mathrm{kg}^{-2}$ 。本题中可能用到如下公式: 半径为 $R$ 的球，其体积为 $V=\frac{4}{3} \pi R^{3}$, 表面积为 $S=4 \pi R^{2}$ 。假定该星球没有大气层, 求这个星球中心处由水产生的压强。

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个数值。 问题: 遥远的星球。某星球完全由不可压缩的液态水组成。星球的表面重力加速度为 $g_{0}=9.8 \mathrm{~m} / \mathrm{s}^{2}$, 半径为 $R$, 且没有自转。水的密度 $\rho=10^{3} \mathrm{~kg} / \mathrm{m}^{3}$, 万有引力常数 $G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} \cdot \mathrm{kg}^{-2}$ 。本题中可能用到如下公式: 半径为 $R$ 的球，其体积为 $V=\frac{4}{3} \pi R^{3}$, 表面积为 $S=4 \pi R^{2}$ 。假定该星球没有大气层, 求这个星球中心处由水产生的压强。 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 请记住，你的答案应以Pa为单位计算，但在给出最终答案时，请不要包含单位。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是不包含任何单位的数值。

NV

Astronomy

ZH

text-only

Astronomy_865

An astronomer took the following picture while observing the night sky: [figure1] What is the latitude of the place where the astronomer took the picture? A: $70^{\circ} \mathrm{S}$ B: $20^{\circ} \mathrm{S}$ C: $2^{\circ} \mathrm{N}$ D: $20^{\circ} \mathrm{N}$ E: $70^{\circ} \mathrm{N}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: An astronomer took the following picture while observing the night sky: [figure1] What is the latitude of the place where the astronomer took the picture? A: $70^{\circ} \mathrm{S}$ B: $20^{\circ} \mathrm{S}$ C: $2^{\circ} \mathrm{N}$ D: $20^{\circ} \mathrm{N}$ E: $70^{\circ} \mathrm{N}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

multi-modal

Astronomy_635

如图为一退役卫星绕地球 $\mathrm{M}$ 运动的示意图, 卫星先绕虚线圆轨道运行, 在 $B$ 点处变轨进入椭圆轨道, $A 、 B$ 分别为椭圆轨道的近地点和远地点, $A$ 点与地球球心的距离为 $a, B$ 点与地球球心的距离为 $b$, 半短轴的长度为 $c$, 当退役卫星运动到 $A$ 点时, 再次变轨进入大气层以实现回收太空垃圾的目的。则下列说法中正确的是（引力常量为 $G$,地球质量为 $M)(\quad)$ [图1] A: 退役卫星沿粗圆轨道从 $B$ 点经 $C$ 点运动到 $A$ 点的过程中, 速率先减小后增大 B: 退役卫星在 $C$ 点的加速度大小 $a=\frac{4 G M}{(b-a)^{2}+4 c^{2}}$ C: 若要将退役卫星带回地球大气层, 微型电力推进器需提供动力使其加速运动 D: 退役卫星在 $A$ 点变轨刚要进入大气层的速度比虚线圆轨道的速度小

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 如图为一退役卫星绕地球 $\mathrm{M}$ 运动的示意图, 卫星先绕虚线圆轨道运行, 在 $B$ 点处变轨进入椭圆轨道, $A 、 B$ 分别为椭圆轨道的近地点和远地点, $A$ 点与地球球心的距离为 $a, B$ 点与地球球心的距离为 $b$, 半短轴的长度为 $c$, 当退役卫星运动到 $A$ 点时, 再次变轨进入大气层以实现回收太空垃圾的目的。则下列说法中正确的是（引力常量为 $G$,地球质量为 $M)(\quad)$ [图1] A: 退役卫星沿粗圆轨道从 $B$ 点经 $C$ 点运动到 $A$ 点的过程中, 速率先减小后增大 B: 退役卫星在 $C$ 点的加速度大小 $a=\frac{4 G M}{(b-a)^{2}+4 c^{2}}$ C: 若要将退役卫星带回地球大气层, 微型电力推进器需提供动力使其加速运动 D: 退役卫星在 $A$ 点变轨刚要进入大气层的速度比虚线圆轨道的速度小 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_594

2020 年 12 月 7 日, 嫦娥五号成功返回地球创造了我国到月球取土的伟大历史, 如图所示, 嫦娥五号取土后, 在 $P$ 处由圆形轨道I变轨到椭圆轨道II, 以便返回地球。下列说法正确的是（ ） ## 月球 [图1] A: 嫦娥五号在轨道I和II运行时均处于失重状态 B: 嫦娥五号在轨道I运行时的机械能比在轨道II运行时机械能大 C: 嫦娥五号在轨道I和II运行时与月球中心的连线在相等时间内扫过的面积相等 D: 嫦娥五号在轨道 I 和II运行至 $P$ 处时向心加速度大小相等

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 2020 年 12 月 7 日, 嫦娥五号成功返回地球创造了我国到月球取土的伟大历史, 如图所示, 嫦娥五号取土后, 在 $P$ 处由圆形轨道I变轨到椭圆轨道II, 以便返回地球。下列说法正确的是（ ） ## 月球 [图1] A: 嫦娥五号在轨道I和II运行时均处于失重状态 B: 嫦娥五号在轨道I运行时的机械能比在轨道II运行时机械能大 C: 嫦娥五号在轨道I和II运行时与月球中心的连线在相等时间内扫过的面积相等 D: 嫦娥五号在轨道 I 和II运行至 $P$ 处时向心加速度大小相等 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_86

2018 年 12 月 27 日, 我国北斗卫星导航系统开始提供全球服务, 标志着北斗系统正式迈入全球时代。覆盖全球的北斗卫星导航系统是由静止轨道卫星 (即地球同步卫星)和非静止轨道卫星共 35 颗组成的。卫星绕地球近似做匀速圆周运动。已知其中一颗地球同步卫星距离地球表面的高度为 $h$, 地球质量为 $M$, 地球半径为 $R$, 引力常量为 $G$ 。 求该同步卫星绕地球运动的线速度 $v$ 的大小; [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 2018 年 12 月 27 日, 我国北斗卫星导航系统开始提供全球服务, 标志着北斗系统正式迈入全球时代。覆盖全球的北斗卫星导航系统是由静止轨道卫星 (即地球同步卫星)和非静止轨道卫星共 35 颗组成的。卫星绕地球近似做匀速圆周运动。已知其中一颗地球同步卫星距离地球表面的高度为 $h$, 地球质量为 $M$, 地球半径为 $R$, 引力常量为 $G$ 。 求该同步卫星绕地球运动的线速度 $v$ 的大小; [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_982

The Crab Nebula is the expanding remnant of a supernova that was observed by Chinese astronomers about 1000 years ago. Although faint in the visible, it is one of the brightest objects in the sky both in the radio and in X-ray. Figure 4 shows it as observed with the Very Large Array (a radio telescope) operating at a frequency of $3.0 \mathrm{GHz}$. The whole image has an angular size of 530 by 530 arcseconds (where there are 3600 arcseconds in a degree). [figure1] Figure 4: The Crab Nebula as seen at the radio frequency of $3.0 \mathrm{GHz}$, imaged by the Very Large Array in 2017. Credit: NRAO/AUI/NSF. For long wavelengths (such as those found in the radio), Planck's Law for black-body radiation simplifies to the Rayleigh-Jeans Law, $$ B_{\nu}=\frac{2 \nu^{2} k_{B} T}{c^{2}} $$ where $B$ is the emitted intensity at frequency $\nu$ per unit frequency per steradian (a unit of solid angle, for which there are $4 \pi$ steradians in a sphere), $k_{B}$ is the Boltzmann constant, $T$ is the temperature of the source, and $c$ is the speed of light. Typically what is measured is the emitted intensity per unit frequency, $I_{\nu}$, which is related to $B_{\nu}$ as $$ I_{\nu}=B_{\nu} \Omega $$ where $\Omega$ is the solid angle in steradians subtended by the emitting source as seen by the detector. A small angular ellipse subtends approximately $\Omega=\pi a b$ steradians, where $a$ is the semi-major axis and $b$ is the semi-minor axis of the ellipse, both specified in radians. The total measured intensity in Figure 4 is $I_{\nu}=7.43 \times 10^{-24} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~Hz}^{-1}$. By taking suitable measurements from Figure 4, calculate the temperature of the nebula.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: The Crab Nebula is the expanding remnant of a supernova that was observed by Chinese astronomers about 1000 years ago. Although faint in the visible, it is one of the brightest objects in the sky both in the radio and in X-ray. Figure 4 shows it as observed with the Very Large Array (a radio telescope) operating at a frequency of $3.0 \mathrm{GHz}$. The whole image has an angular size of 530 by 530 arcseconds (where there are 3600 arcseconds in a degree). [figure1] Figure 4: The Crab Nebula as seen at the radio frequency of $3.0 \mathrm{GHz}$, imaged by the Very Large Array in 2017. Credit: NRAO/AUI/NSF. For long wavelengths (such as those found in the radio), Planck's Law for black-body radiation simplifies to the Rayleigh-Jeans Law, $$ B_{\nu}=\frac{2 \nu^{2} k_{B} T}{c^{2}} $$ where $B$ is the emitted intensity at frequency $\nu$ per unit frequency per steradian (a unit of solid angle, for which there are $4 \pi$ steradians in a sphere), $k_{B}$ is the Boltzmann constant, $T$ is the temperature of the source, and $c$ is the speed of light. Typically what is measured is the emitted intensity per unit frequency, $I_{\nu}$, which is related to $B_{\nu}$ as $$ I_{\nu}=B_{\nu} \Omega $$ where $\Omega$ is the solid angle in steradians subtended by the emitting source as seen by the detector. A small angular ellipse subtends approximately $\Omega=\pi a b$ steradians, where $a$ is the semi-major axis and $b$ is the semi-minor axis of the ellipse, both specified in radians. The total measured intensity in Figure 4 is $I_{\nu}=7.43 \times 10^{-24} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~Hz}^{-1}$. By taking suitable measurements from Figure 4, calculate the temperature of the nebula. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of K, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_324

某同学在学习中记录了一些与地球、月球有关的数据资料如表中所示, 利用这些数据来计算地球表面与月球表面之间的最近距离, 则下列表达式中正确的是（ ） | 地球半径 | $R=6.4 \times 10^{6} \mathrm{~m}$ | | :--- | :--- | | 月球半径 | $t=1.74 \times 10^{6} \mathrm{~m}$ | | 地球表面重力加速度 | $g_{0}=9.80 \mathrm{~m} / \mathrm{s}^{2}$ | | 月球表面重力加速度 | $g=1.56 \mathrm{~m} / \mathrm{s}^{2}$ | | 月球绕地球转动的线速度 | $V=1 \mathrm{~km} / \mathrm{s}$ | | 月球绕地球转动的周期 | $T=27.3$ 天 | | 光速 | $c=3.0 \times 10^{3} \mathrm{~m} / \mathrm{s}$ | | 用溦光器从地球表面上正对月球表面处向月球表面发射溦光束, 经过 $t=2.56 s$ 接 <br> 收到反射回来的溦光信号 | | A: $s=c t$ B: $s=\frac{V T}{2 \pi}-R-r$ C: $s=\frac{V^{2}}{g}-R-r$ D: $s=\sqrt[3]{\frac{g_{0} R^{2} T^{2}}{4 \pi^{2}}}-R-r$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 某同学在学习中记录了一些与地球、月球有关的数据资料如表中所示, 利用这些数据来计算地球表面与月球表面之间的最近距离, 则下列表达式中正确的是（ ） | 地球半径 | $R=6.4 \times 10^{6} \mathrm{~m}$ | | :--- | :--- | | 月球半径 | $t=1.74 \times 10^{6} \mathrm{~m}$ | | 地球表面重力加速度 | $g_{0}=9.80 \mathrm{~m} / \mathrm{s}^{2}$ | | 月球表面重力加速度 | $g=1.56 \mathrm{~m} / \mathrm{s}^{2}$ | | 月球绕地球转动的线速度 | $V=1 \mathrm{~km} / \mathrm{s}$ | | 月球绕地球转动的周期 | $T=27.3$ 天 | | 光速 | $c=3.0 \times 10^{3} \mathrm{~m} / \mathrm{s}$ | | 用溦光器从地球表面上正对月球表面处向月球表面发射溦光束, 经过 $t=2.56 s$ 接 <br> 收到反射回来的溦光信号 | | A: $s=c t$ B: $s=\frac{V T}{2 \pi}-R-r$ C: $s=\frac{V^{2}}{g}-R-r$ D: $s=\sqrt[3]{\frac{g_{0} R^{2} T^{2}}{4 \pi^{2}}}-R-r$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_329

宇宙中, 两颗靠得比较近的恒星, 只受到彼此之间的万有引力作用互相绕转, 称之为双星系统. 在浩瀚的银河系中, 多数恒星都是双星系统. 设某双星系统 $A 、 B$ 绕其连线上的 $O$ 点做匀速圆周运动, 如图所示, 若 $A O>O B$, 则 ( ) [图1] A: 星球 $A$ 的质量一定大于 $B$ 的质量 B: 星球 $A$ 的线速度一定大于 $B$ 的线速度 C: 双星间距离一定, 双星的总质量越大, 其转动周期越小 D: 双星的总质量一定, 双星之间的距离越大，其转动周期越小

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 宇宙中, 两颗靠得比较近的恒星, 只受到彼此之间的万有引力作用互相绕转, 称之为双星系统. 在浩瀚的银河系中, 多数恒星都是双星系统. 设某双星系统 $A 、 B$ 绕其连线上的 $O$ 点做匀速圆周运动, 如图所示, 若 $A O>O B$, 则 ( ) [图1] A: 星球 $A$ 的质量一定大于 $B$ 的质量 B: 星球 $A$ 的线速度一定大于 $B$ 的线速度 C: 双星间距离一定, 双星的总质量越大, 其转动周期越小 D: 双星的总质量一定, 双星之间的距离越大，其转动周期越小 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_821

Consider the binary system Kepler-16, which has the primary star Kepler-16A and the secondary star Kepler-16B. It has an orbital period $P=41.08$ days and the measured parallax is $p=$ 13.29 mas. Calculate the total mass of the stars, using the fact that their maximum angular separation measured from Earth is $\theta=2.98$ mas and they are on an edge-on orbit. A: $0.756 M_{\odot}$ B: $0.803 M_{\odot}$ C: $0.891 M_{\odot}$ D: $0.987 M_{\odot}$ E: $1.326 M_{\odot}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Consider the binary system Kepler-16, which has the primary star Kepler-16A and the secondary star Kepler-16B. It has an orbital period $P=41.08$ days and the measured parallax is $p=$ 13.29 mas. Calculate the total mass of the stars, using the fact that their maximum angular separation measured from Earth is $\theta=2.98$ mas and they are on an edge-on orbit. A: $0.756 M_{\odot}$ B: $0.803 M_{\odot}$ C: $0.891 M_{\odot}$ D: $0.987 M_{\odot}$ E: $1.326 M_{\odot}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

text-only

Astronomy_1196

The James Webb Space Telescope (JWST) is an incredibly exciting next generation telescope that was successfully launched on $25^{\text {th }}$ December 2021 . Its mirror is approximately $6.5 \mathrm{~m}$ in diameter, much larger than the $2.4 \mathrm{~m}$ mirror of the Hubble Space Telescope (HST), and so it has far greater resolution and sensitivity. Whilst HST largely imaged in the visible, JWST will do most of its work in the nearand mid-infrared (NIR and MIR respectively). This will allow it to pick up heavily redshifted light, such as that from the first generation of stars in the very first galaxies. [figure1] Figure 5: Left: A full-scale model of JWST next to some of the scientists and engineers involved in its development at the Goddard Space Flight Center. Credit: NASA / Goddard Space Flight Center / Pat Izzo. Right: The position of the second Lagrangian point, $L_{2}$, relative to the Earth. Credit: ESA. The resolution limit of a telescope is set by the amount of diffraction light rays experience as they enter the system, and is related to the diameter of a telescope, $D$, and the wavelength being observed, $\lambda$. The resolution limit of a CCD is set by the size of the pixels. Three of the imaging cameras on JWST are tabulated with some properties below: | Instrument | Wavelength range $(\mu \mathrm{m})$ | CCD plate scale (arcseconds / pixel) | | :---: | :---: | :---: | | NIRCam (short wave) | $0.6-2.3$ | 0.031 | | NIRCam (long wave) | $2.4-5.0$ | 0.065 | | MIRI | $5.6-25.5$ | 0.11 | An arcsecond is a measure of angle where $1^{\circ}=3600$ arcseconds. The familiar variation in intensity on a screen, $I_{\text {slit }}$, due to diffraction through an infinitely tall single slit is given as $$ I_{\text {slit }}=I_{0}\left(\frac{\sin (x)}{x}\right)^{2}, \text { where } \quad x=\frac{\pi D \theta}{\lambda} $$ and $I_{0}$ is the initial intensity. For a circular aperture, the formula is slightly different and is given as $$ I_{\mathrm{circ}}=I_{0}\left(\frac{2 J_{1}(x)}{x}\right)^{2} . $$ Here $J_{1}(x)$ is the Bessel function of the first kind and is calculated as $$ J_{n}(x)=\sum_{r=0}^{\infty} \frac{(-1)^{r}}{r !(n+r) !}\left(\frac{x}{2}\right)^{n+2 r} \quad \text { so } \quad J_{1}(x)=\frac{x}{2}\left(1-\frac{x^{2}}{8}+\frac{x^{4}}{192}-\ldots\right) . $$ The $x$-axis intercepts and shape of the maxima are quite different, as shown in Figure 6. The position of the first minimum of $I_{\text {slit }}$ is at $x_{\min }=\pi$ meaning that $\theta_{\min , \text { slit }}=\lambda / D$, whilst for $I_{\text {circ }}$ it is at $x_{\min }=3.8317 \ldots$ so $\theta_{\min , \mathrm{circ}} \approx 1.22 \lambda / D$. This is one way of defining the minimum angular resolution, although since the flux drops off so steeply away from the central maximum a more convenient one for use with CCDs is the angle corresponding to the full width half maximum (FWHM). [figure2] Figure 6: Left: The $I_{\text {slit }}$ (purple) and $I_{\text {circ }}$ (blue - the wider central maximum) functions, normalised so that $I_{0}=1$. You can see the shapes and $x$-intercepts are different. Right: How $x_{\min }$ and the full width half maximum (FWHM) are defined. Here it is shown for $I_{\text {circ }}$. As well as having the largest mirror of any space telescope ever launched, it is also one of the most sensitive, with its greatest sensitivity in the NIRCam F200W filter (centred on a wavelength of $1.989 \mu \mathrm{m})$ where after $10^{4}$ seconds it can detect a flux of $9.1 \mathrm{nJy}\left(1 \mathrm{Jy}=10^{-26} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~Hz}^{-1}\right.$ ) with a signal-to-noise ratio (S/N) of 10 , corresponding to an apparent magnitude of $m=29.0$. This extraordinary sensitivity can be used to pick up light from the earliest galaxies in the Universe. The scale factor, $a$, parameterises the expansion of the Universe since the Big Bang, and is related to the redshift, $z$, as $$ a=(1+z)^{-1} \quad \text { where } \quad z \equiv \frac{\lambda_{\text {obs }}-\lambda_{\mathrm{emit}}}{\lambda_{\mathrm{emit}}} $$ with $\lambda_{\text {obs }}$ the observed wavelength and $\lambda_{\text {emit }}$ the rest frame wavelength. The current rate of expansion of the Universe is given by the Hubble constant, $H_{0}$, and this is related to the current Hubble time, $t_{\mathrm{H}_{0}}$, and current Hubble distance, $D_{\mathrm{H}_{0}}$, as $$ t_{\mathrm{H}_{0}} \equiv H_{0}^{-1} \quad \text { and } \quad D_{\mathrm{H}_{0}} \equiv c t_{\mathrm{H}_{0}} \text {. } $$ Here the subscript 0 indicates the values are measured today. The Hubble constant is more appropriately known as the Hubble parameter as it is a function of time, and the evolution of $H$ as a function of $z$ is $$ E(z)=\frac{H}{H_{0}} \equiv\left[\Omega_{0, m}(1+z)^{3}+\Omega_{0, \Lambda}+\Omega_{0, r}(1+z)^{4}\right]^{1 / 2}, $$ where $\Omega$ is the normalised density parameter, and the subscript $m, r$, and $\Lambda$ indicate the contribution to $\Omega$ from matter, radiation, and dark energy, respectively. The proper age of the Universe $t(z)$ at redshift $z$ is best evaluated in terms of $a$ as $$ t=t_{\mathrm{H}_{0}} \int_{0}^{(1+z)^{-1}} \frac{a}{\left(\Omega_{0, m} a+\Omega_{0, \Lambda} a^{4}+\Omega_{0, r}\right)^{1 / 2}} \mathrm{~d} a $$ If $\Omega_{0, r}=0$ and $\Omega_{0, m}+\Omega_{0, \Lambda}=1$ (corresponding to what it known as a flat Universe), then via the standard integral $\int\left(b^{2}+x^{2}\right)^{-1 / 2} \mathrm{~d} x=\ln \left(x+\sqrt{b^{2}+x^{2}}\right)+C$ this integral can be evaluated analytically to give $$ t=t_{\mathrm{H}_{0}} \frac{2}{3 \Omega_{0, \Lambda}^{1 / 2}} \ln \left[\left(\frac{\Omega_{0, \Lambda}}{\Omega_{0, m}}\right)^{1 / 2}(1+z)^{-3 / 2}+\left(\frac{\Omega_{0, \Lambda}}{\Omega_{0, m}(1+z)^{3}}+1\right)^{1 / 2}\right] $$ Finally, the luminosity distance, $D_{L}(z)$, corresponding to the distance away that an object appears to be due to its measured flux given its intrinsic luminosity (i.e. $f \equiv L / 4 \pi D_{L}^{2}$ ) is given as $$ D_{L}=\left(1+z_{i}\right) D_{\mathrm{H}_{0}} \int_{0}^{z_{i}} \frac{1}{E(z)} \mathrm{d} z=\left(1+z_{i}\right) D_{\mathrm{H}_{0}} \int_{a_{i}}^{1} \frac{1}{\left(\Omega_{0, m} a+\Omega_{0, \Lambda} a^{4}+\Omega_{0, r}\right)^{1 / 2}} \mathrm{~d} a $$ where $z_{i}$ is the redshift of interest and $a_{i}$ is the equivalent scale factor. Even for the flat Universe case with $\Omega_{0, r}=0$ this integral cannot be be done analytically so must be evaluated numerically.a. The telescope will spend its expected 10-20 year mission in a halo orbit about the second Lagrangian point, L2 (see Figure 5). This is one of five special points in the Sun-Earth system where the gravitational forces from the two bodies provide the centripetal force required to have a (small mass) object there have an orbital period identical to the Earth. At the L2 point, this means it orbits quicker than you would expect for an object that distance from the Sun. ii. The JWST is on an orbit that will take it to within $200000 \mathrm{~km}$ of $\mathrm{L2}$, where it will then do a final large burn of the rockets to insert it into the halo orbit around L2. Assuming it is on a simple elliptical transfer orbit ignoring the influence of the Sun, and had a perigee at an altitude of $2100 \mathrm{~km}$ above the surface of the Earth, how long will it take JWST to get to the L2 orbital insertion phase of its mission? Give your answer in days.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: The James Webb Space Telescope (JWST) is an incredibly exciting next generation telescope that was successfully launched on $25^{\text {th }}$ December 2021 . Its mirror is approximately $6.5 \mathrm{~m}$ in diameter, much larger than the $2.4 \mathrm{~m}$ mirror of the Hubble Space Telescope (HST), and so it has far greater resolution and sensitivity. Whilst HST largely imaged in the visible, JWST will do most of its work in the nearand mid-infrared (NIR and MIR respectively). This will allow it to pick up heavily redshifted light, such as that from the first generation of stars in the very first galaxies. [figure1] Figure 5: Left: A full-scale model of JWST next to some of the scientists and engineers involved in its development at the Goddard Space Flight Center. Credit: NASA / Goddard Space Flight Center / Pat Izzo. Right: The position of the second Lagrangian point, $L_{2}$, relative to the Earth. Credit: ESA. The resolution limit of a telescope is set by the amount of diffraction light rays experience as they enter the system, and is related to the diameter of a telescope, $D$, and the wavelength being observed, $\lambda$. The resolution limit of a CCD is set by the size of the pixels. Three of the imaging cameras on JWST are tabulated with some properties below: | Instrument | Wavelength range $(\mu \mathrm{m})$ | CCD plate scale (arcseconds / pixel) | | :---: | :---: | :---: | | NIRCam (short wave) | $0.6-2.3$ | 0.031 | | NIRCam (long wave) | $2.4-5.0$ | 0.065 | | MIRI | $5.6-25.5$ | 0.11 | An arcsecond is a measure of angle where $1^{\circ}=3600$ arcseconds. The familiar variation in intensity on a screen, $I_{\text {slit }}$, due to diffraction through an infinitely tall single slit is given as $$ I_{\text {slit }}=I_{0}\left(\frac{\sin (x)}{x}\right)^{2}, \text { where } \quad x=\frac{\pi D \theta}{\lambda} $$ and $I_{0}$ is the initial intensity. For a circular aperture, the formula is slightly different and is given as $$ I_{\mathrm{circ}}=I_{0}\left(\frac{2 J_{1}(x)}{x}\right)^{2} . $$ Here $J_{1}(x)$ is the Bessel function of the first kind and is calculated as $$ J_{n}(x)=\sum_{r=0}^{\infty} \frac{(-1)^{r}}{r !(n+r) !}\left(\frac{x}{2}\right)^{n+2 r} \quad \text { so } \quad J_{1}(x)=\frac{x}{2}\left(1-\frac{x^{2}}{8}+\frac{x^{4}}{192}-\ldots\right) . $$ The $x$-axis intercepts and shape of the maxima are quite different, as shown in Figure 6. The position of the first minimum of $I_{\text {slit }}$ is at $x_{\min }=\pi$ meaning that $\theta_{\min , \text { slit }}=\lambda / D$, whilst for $I_{\text {circ }}$ it is at $x_{\min }=3.8317 \ldots$ so $\theta_{\min , \mathrm{circ}} \approx 1.22 \lambda / D$. This is one way of defining the minimum angular resolution, although since the flux drops off so steeply away from the central maximum a more convenient one for use with CCDs is the angle corresponding to the full width half maximum (FWHM). [figure2] Figure 6: Left: The $I_{\text {slit }}$ (purple) and $I_{\text {circ }}$ (blue - the wider central maximum) functions, normalised so that $I_{0}=1$. You can see the shapes and $x$-intercepts are different. Right: How $x_{\min }$ and the full width half maximum (FWHM) are defined. Here it is shown for $I_{\text {circ }}$. As well as having the largest mirror of any space telescope ever launched, it is also one of the most sensitive, with its greatest sensitivity in the NIRCam F200W filter (centred on a wavelength of $1.989 \mu \mathrm{m})$ where after $10^{4}$ seconds it can detect a flux of $9.1 \mathrm{nJy}\left(1 \mathrm{Jy}=10^{-26} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~Hz}^{-1}\right.$ ) with a signal-to-noise ratio (S/N) of 10 , corresponding to an apparent magnitude of $m=29.0$. This extraordinary sensitivity can be used to pick up light from the earliest galaxies in the Universe. The scale factor, $a$, parameterises the expansion of the Universe since the Big Bang, and is related to the redshift, $z$, as $$ a=(1+z)^{-1} \quad \text { where } \quad z \equiv \frac{\lambda_{\text {obs }}-\lambda_{\mathrm{emit}}}{\lambda_{\mathrm{emit}}} $$ with $\lambda_{\text {obs }}$ the observed wavelength and $\lambda_{\text {emit }}$ the rest frame wavelength. The current rate of expansion of the Universe is given by the Hubble constant, $H_{0}$, and this is related to the current Hubble time, $t_{\mathrm{H}_{0}}$, and current Hubble distance, $D_{\mathrm{H}_{0}}$, as $$ t_{\mathrm{H}_{0}} \equiv H_{0}^{-1} \quad \text { and } \quad D_{\mathrm{H}_{0}} \equiv c t_{\mathrm{H}_{0}} \text {. } $$ Here the subscript 0 indicates the values are measured today. The Hubble constant is more appropriately known as the Hubble parameter as it is a function of time, and the evolution of $H$ as a function of $z$ is $$ E(z)=\frac{H}{H_{0}} \equiv\left[\Omega_{0, m}(1+z)^{3}+\Omega_{0, \Lambda}+\Omega_{0, r}(1+z)^{4}\right]^{1 / 2}, $$ where $\Omega$ is the normalised density parameter, and the subscript $m, r$, and $\Lambda$ indicate the contribution to $\Omega$ from matter, radiation, and dark energy, respectively. The proper age of the Universe $t(z)$ at redshift $z$ is best evaluated in terms of $a$ as $$ t=t_{\mathrm{H}_{0}} \int_{0}^{(1+z)^{-1}} \frac{a}{\left(\Omega_{0, m} a+\Omega_{0, \Lambda} a^{4}+\Omega_{0, r}\right)^{1 / 2}} \mathrm{~d} a $$ If $\Omega_{0, r}=0$ and $\Omega_{0, m}+\Omega_{0, \Lambda}=1$ (corresponding to what it known as a flat Universe), then via the standard integral $\int\left(b^{2}+x^{2}\right)^{-1 / 2} \mathrm{~d} x=\ln \left(x+\sqrt{b^{2}+x^{2}}\right)+C$ this integral can be evaluated analytically to give $$ t=t_{\mathrm{H}_{0}} \frac{2}{3 \Omega_{0, \Lambda}^{1 / 2}} \ln \left[\left(\frac{\Omega_{0, \Lambda}}{\Omega_{0, m}}\right)^{1 / 2}(1+z)^{-3 / 2}+\left(\frac{\Omega_{0, \Lambda}}{\Omega_{0, m}(1+z)^{3}}+1\right)^{1 / 2}\right] $$ Finally, the luminosity distance, $D_{L}(z)$, corresponding to the distance away that an object appears to be due to its measured flux given its intrinsic luminosity (i.e. $f \equiv L / 4 \pi D_{L}^{2}$ ) is given as $$ D_{L}=\left(1+z_{i}\right) D_{\mathrm{H}_{0}} \int_{0}^{z_{i}} \frac{1}{E(z)} \mathrm{d} z=\left(1+z_{i}\right) D_{\mathrm{H}_{0}} \int_{a_{i}}^{1} \frac{1}{\left(\Omega_{0, m} a+\Omega_{0, \Lambda} a^{4}+\Omega_{0, r}\right)^{1 / 2}} \mathrm{~d} a $$ where $z_{i}$ is the redshift of interest and $a_{i}$ is the equivalent scale factor. Even for the flat Universe case with $\Omega_{0, r}=0$ this integral cannot be be done analytically so must be evaluated numerically. problem: a. The telescope will spend its expected 10-20 year mission in a halo orbit about the second Lagrangian point, L2 (see Figure 5). This is one of five special points in the Sun-Earth system where the gravitational forces from the two bodies provide the centripetal force required to have a (small mass) object there have an orbital period identical to the Earth. At the L2 point, this means it orbits quicker than you would expect for an object that distance from the Sun. ii. The JWST is on an orbit that will take it to within $200000 \mathrm{~km}$ of $\mathrm{L2}$, where it will then do a final large burn of the rockets to insert it into the halo orbit around L2. Assuming it is on a simple elliptical transfer orbit ignoring the influence of the Sun, and had a perigee at an altitude of $2100 \mathrm{~km}$ above the surface of the Earth, how long will it take JWST to get to the L2 orbital insertion phase of its mission? Give your answer in days. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \mathrm{~s}, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_337

三颗人造卫星 $\mathrm{A} 、 \mathrm{~B} 、 \mathrm{C}$ 都在赤道正上方同方向绕地球做匀速圆周运动, $\mathrm{A} 、 \mathrm{C}$ 为地球同步卫星, 某时刻 $\mathrm{A} 、 \mathrm{~B}$ 相距最近, 如图所示。已知地球自转周期为 $T_{1}, \mathrm{~B}$ 的运行周期为 $T_{2}$, 则下列说法正确的是（ ） [图1] A: C 加速可追上同一轨道上的 A B: 经过时间 $\frac{T_{1} T_{2}}{2\left(T_{1}-T_{2}\right)}, \mathrm{A} 、 \mathrm{~B}$ 相距最远 C: A、C 向心加速度大小相等, 且小于 $\mathrm{B}$ 的向心加速度 D: 在相同时间内, $\mathrm{A}$ 与地心连线扫过的面积大于 $\mathrm{B}$ 与地心连线扫过的面积

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 三颗人造卫星 $\mathrm{A} 、 \mathrm{~B} 、 \mathrm{C}$ 都在赤道正上方同方向绕地球做匀速圆周运动, $\mathrm{A} 、 \mathrm{C}$ 为地球同步卫星, 某时刻 $\mathrm{A} 、 \mathrm{~B}$ 相距最近, 如图所示。已知地球自转周期为 $T_{1}, \mathrm{~B}$ 的运行周期为 $T_{2}$, 则下列说法正确的是（ ） [图1] A: C 加速可追上同一轨道上的 A B: 经过时间 $\frac{T_{1} T_{2}}{2\left(T_{1}-T_{2}\right)}, \mathrm{A} 、 \mathrm{~B}$ 相距最远 C: A、C 向心加速度大小相等, 且小于 $\mathrm{B}$ 的向心加速度 D: 在相同时间内, $\mathrm{A}$ 与地心连线扫过的面积大于 $\mathrm{B}$ 与地心连线扫过的面积 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_258

三体问题是天体力学中的基本模型, 即探究三个质量、初始位置和初始速度都任意的可视为质点的天体，在相互之间万有引力的作用下的运动规律。三体问题同时也是一个著名的数学难题, 1772 年, 拉格朗日在“平面限制性三体问题”条件下找到了 5 个特解,它就是著名的拉格朗日点。在该点上，小天体在两个大天体的引力作用下能基本保持相对静止。如图是日地系统的 5 个拉格朗日点 $\left(L_{1} 、 L_{2} 、 L_{3} 、 L_{4} 、 L_{5}\right)$, 设想未来人类在这五个点上都建立了太空站, 若不考虑其它天体对太空站的引力, 则下列说法正确的是 [图1] A: 位于 $L_{l}$ 点的太空站处于受力平衡状态 B: 位于 $L_{2}$ 点的太空站的线速度小于地球的线速度 C: 位于 $L_{3}$ 点的太空站的向心加速度大于位于 $L_{1}$ 点的太空站的向心加速度 D: 位于 $L_{4}$ 点的太空站向心力大小一定等于位于 $L_{5}$ 点的太空站向心力大小

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 三体问题是天体力学中的基本模型, 即探究三个质量、初始位置和初始速度都任意的可视为质点的天体，在相互之间万有引力的作用下的运动规律。三体问题同时也是一个著名的数学难题, 1772 年, 拉格朗日在“平面限制性三体问题”条件下找到了 5 个特解,它就是著名的拉格朗日点。在该点上，小天体在两个大天体的引力作用下能基本保持相对静止。如图是日地系统的 5 个拉格朗日点 $\left(L_{1} 、 L_{2} 、 L_{3} 、 L_{4} 、 L_{5}\right)$, 设想未来人类在这五个点上都建立了太空站, 若不考虑其它天体对太空站的引力, 则下列说法正确的是 [图1] A: 位于 $L_{l}$ 点的太空站处于受力平衡状态 B: 位于 $L_{2}$ 点的太空站的线速度小于地球的线速度 C: 位于 $L_{3}$ 点的太空站的向心加速度大于位于 $L_{1}$ 点的太空站的向心加速度 D: 位于 $L_{4}$ 点的太空站向心力大小一定等于位于 $L_{5}$ 点的太空站向心力大小 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_801

The habitable zone of a star is defined as the one where water in the liquid state can exist in the surface of a planet. Therefore, considering that the planets are ideal black bodies with fast rotation, determine the maximum eccentricity that the orbit of a planet can have so that it can be home to life. Ignore any thermodynamic effects that might happen in the atmosphere or the sidereal space. Consider that the melting point of water is $273 \mathrm{~K}$ and the boiling point is $373 \mathrm{~K}$. A: 0.274 B: 0.302 C: 0.316 D: 0.328 E: 0.540

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: The habitable zone of a star is defined as the one where water in the liquid state can exist in the surface of a planet. Therefore, considering that the planets are ideal black bodies with fast rotation, determine the maximum eccentricity that the orbit of a planet can have so that it can be home to life. Ignore any thermodynamic effects that might happen in the atmosphere or the sidereal space. Consider that the melting point of water is $273 \mathrm{~K}$ and the boiling point is $373 \mathrm{~K}$. A: 0.274 B: 0.302 C: 0.316 D: 0.328 E: 0.540 You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

text-only

Astronomy_432

万有引力定律 $F_{\text {引 }}=G \frac{m_{1} m_{2}}{r^{2}}$ 和库仑定律 $F_{\text {电 }}=k \frac{q_{1} q_{2}}{r^{2}}$ 都满足力与距离平方成反比关系。如图所示, 计算物体从距离地球球心 $r_{1}$ 处, 远离至与地心距离 $r_{2}$ 处, 万有引力对物体做功时, 由于力的大小随距离而变化, 一般需采用微元法。也可采用从 $r_{1}$ 到 $r_{2}$ 过程的平均力, 即 $\overline{F_{\text {引 }}}=G \frac{m_{1} m_{2}}{r_{1} \cdot r_{2}}$ 计算做功。已知物体质量为 $m$, 地球质量为 $M$, 半径为 $R$, 引力常量为 $G$ 。 氢原子是最简单的原子, 电子绕原子核做匀速圆周运动与人造卫星绕地球做匀速圆周运动类似。已知电子质量为 $m$, 带电量为 $-e$, 氢原子核带电量为 $+e$, 电子绕核运动半径为 $r$, 静电力常量为 $k$, 求电子绕核运动的速度 $v_{l}$ 大小; 若要使氢原子电离 (使核外电子运动至无穷远, 逃出原子核的电场范围), 则至少额外需要提供多大的能量 $\Delta E$ 。 [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题包含多个待求解的量。 问题: 万有引力定律 $F_{\text {引 }}=G \frac{m_{1} m_{2}}{r^{2}}$ 和库仑定律 $F_{\text {电 }}=k \frac{q_{1} q_{2}}{r^{2}}$ 都满足力与距离平方成反比关系。如图所示, 计算物体从距离地球球心 $r_{1}$ 处, 远离至与地心距离 $r_{2}$ 处, 万有引力对物体做功时, 由于力的大小随距离而变化, 一般需采用微元法。也可采用从 $r_{1}$ 到 $r_{2}$ 过程的平均力, 即 $\overline{F_{\text {引 }}}=G \frac{m_{1} m_{2}}{r_{1} \cdot r_{2}}$ 计算做功。已知物体质量为 $m$, 地球质量为 $M$, 半径为 $R$, 引力常量为 $G$ 。 氢原子是最简单的原子, 电子绕原子核做匀速圆周运动与人造卫星绕地球做匀速圆周运动类似。已知电子质量为 $m$, 带电量为 $-e$, 氢原子核带电量为 $+e$, 电子绕核运动半径为 $r$, 静电力常量为 $k$, 求电子绕核运动的速度 $v_{l}$ 大小; 若要使氢原子电离 (使核外电子运动至无穷远, 逃出原子核的电场范围), 则至少额外需要提供多大的能量 $\Delta E$ 。 [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你的最终解答的量应该按以下顺序输出：[电子绕核运动的速度, 至少额外提供多大的能量] 它们的答案类型依次是[表达式, 表达式] 你需要在输出的最后用以下格式总结答案：“最终答案是\boxed{ANSWER}”，其中ANSWER应为你的最终答案序列，用英文逗号分隔，例如：5, 7, 2.5

MPV

Astronomy

ZH

multi-modal

Astronomy_1136

Which of the following planets has the longest day, defined as the period of a complete rotation about its axis? A: Venus([figure1]) B: Earth([figure2]) C: Mars([figure3]) D: Jupiter([figure4])

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Which of the following planets has the longest day, defined as the period of a complete rotation about its axis? A: Venus([figure1]) B: Earth([figure2]) C: Mars([figure3]) D: Jupiter([figure4]) You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

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Astronomy

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multi-modal

Astronomy_613

$\mathrm{A} 、 \mathrm{~B}$ 两颗卫星在同一平面内沿同一方向绕地球做匀速圆周运动, 它们之间的距离 $\triangle r$ 随时间变化的关系如图所示。已知地球的半径为 $0.8 r$, 万有引力常量为 $G$, 卫星 A 的线速度大于卫星 $\mathrm{B}$ 的线速度, 不考虑 $\mathrm{A} 、 \mathrm{~B}$ 之间的万有引力, 则下列说法正确的是 [图1] A: 卫星 A 的加速度大于卫星 $\mathrm{B}$ 的加速度 B: 卫星 A 的发射速度可能大于第二宇宙速度 C: 地球的质量为 $\frac{256 \pi^{2} r^{3}}{49 G T^{2}}$ D: 地球的第一宇宙速度为 $\frac{8 \sqrt{5} \pi r}{7 T}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: $\mathrm{A} 、 \mathrm{~B}$ 两颗卫星在同一平面内沿同一方向绕地球做匀速圆周运动, 它们之间的距离 $\triangle r$ 随时间变化的关系如图所示。已知地球的半径为 $0.8 r$, 万有引力常量为 $G$, 卫星 A 的线速度大于卫星 $\mathrm{B}$ 的线速度, 不考虑 $\mathrm{A} 、 \mathrm{~B}$ 之间的万有引力, 则下列说法正确的是 [图1] A: 卫星 A 的加速度大于卫星 $\mathrm{B}$ 的加速度 B: 卫星 A 的发射速度可能大于第二宇宙速度 C: 地球的质量为 $\frac{256 \pi^{2} r^{3}}{49 G T^{2}}$ D: 地球的第一宇宙速度为 $\frac{8 \sqrt{5} \pi r}{7 T}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_267

2017 年, 人类第一次直接探测到来自双中子星合并的引力波。根据科学家们复原的过程, 在两颗中子星合并前约 $100 \mathrm{~s}$ 时, 它们相距约 $r=400 \mathrm{~km}$, 绕二者连线上的某点每秒转动 12 圈, 将两颗中子星都看作是质量均匀分布的球体, 由这些数据、万有引力常量并利用牛顿力学知识, 可以估算出这一时刻两颗中子星 ( ) A: 各自的速率为 $2 \pi r f$ B: 各自的自转角速度为 $2 \pi f$ C: 质量之和为 $\frac{(2 \pi f)^{2} r^{3}}{G}$ D: 速率之积为 $2 \pi r f$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2017 年, 人类第一次直接探测到来自双中子星合并的引力波。根据科学家们复原的过程, 在两颗中子星合并前约 $100 \mathrm{~s}$ 时, 它们相距约 $r=400 \mathrm{~km}$, 绕二者连线上的某点每秒转动 12 圈, 将两颗中子星都看作是质量均匀分布的球体, 由这些数据、万有引力常量并利用牛顿力学知识, 可以估算出这一时刻两颗中子星 ( ) A: 各自的速率为 $2 \pi r f$ B: 各自的自转角速度为 $2 \pi f$ C: 质量之和为 $\frac{(2 \pi f)^{2} r^{3}}{G}$ D: 速率之积为 $2 \pi r f$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

ZH

text-only

Astronomy_608

在力学发展的过程中, 许多物理学家的科学发现推动了物理学的进步。对以下几位物理学家所作科学贡献的表述中, 与事实不相符的是 ( ) A: 伽利略首先建立平均速度、瞬时速度和加速度等描述运动的概念 B: 胡克提出如果行星的轨道是圆形, 太阳与行星间的引力与距离的平方成反比 C: 卡文迪许是测量地球质量的第一人 D: 伽利略根据理想斜面实验, 得出自由落体运动是匀变速直线运动

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 在力学发展的过程中, 许多物理学家的科学发现推动了物理学的进步。对以下几位物理学家所作科学贡献的表述中, 与事实不相符的是 ( ) A: 伽利略首先建立平均速度、瞬时速度和加速度等描述运动的概念 B: 胡克提出如果行星的轨道是圆形, 太阳与行星间的引力与距离的平方成反比 C: 卡文迪许是测量地球质量的第一人 D: 伽利略根据理想斜面实验, 得出自由落体运动是匀变速直线运动 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

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text-only

Astronomy_655

《流浪地球 2 》中太空电梯非常吸引观众眼球。太空电梯通过超级缆绳连接地球赤道上的固定基地与配重空间站, 它们随地球以同步静止状态一起旋转，如图所示。图中配重空间站质量为 $m$, 比同步卫星更高, 距地面高达 $9 R$ 。若地球半径为 $R$, 自转周期为 $T$, 重力加速度为 $g$, 求: 配重空间站受到缆绳的力大小为多少; [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 《流浪地球 2 》中太空电梯非常吸引观众眼球。太空电梯通过超级缆绳连接地球赤道上的固定基地与配重空间站, 它们随地球以同步静止状态一起旋转，如图所示。图中配重空间站质量为 $m$, 比同步卫星更高, 距地面高达 $9 R$ 。若地球半径为 $R$, 自转周期为 $T$, 重力加速度为 $g$, 求: 配重空间站受到缆绳的力大小为多少; [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

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Astronomy

ZH

multi-modal

Astronomy_36

2021 年 2 月 10 日, 在历经近 7 个月的太空飞行后, 我国首个火星探测器“天问一号”成功“太空刹车”, 顺利被火星捕获，进入环火星轨道。物体在万有引力场中具有的势能叫作引力势能, 若取两物体相距无穷远时的引力势能为零, 一个质量为 $m$ 的质点距质量为 $M$ 的引力源中心为 $r$ 时, 其引力势能 $E_{\mathrm{p}}=-\frac{G M m}{r}$ (式中 $G$ 为引力常量)。已知地球半径约为 $6400 \mathrm{~km}$, 地球的第一宇宙速度为 $7.9 \mathrm{~km} / \mathrm{s}$, 火星半径约为地球半径的 $\frac{1}{2}$, 火星质量约为球质量的 $\frac{1}{9}$ 。则“天问一号”刹车后相对于火星的速度不可能为 ( ) A: $7.9 \mathrm{~km} / \mathrm{s}$ B: $5.5 \mathrm{~km} / \mathrm{s}$ C: $4.0 \mathrm{~km} / \mathrm{s}$ D: $3.2 \mathrm{~km} / \mathrm{s}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 2021 年 2 月 10 日, 在历经近 7 个月的太空飞行后, 我国首个火星探测器“天问一号”成功“太空刹车”, 顺利被火星捕获，进入环火星轨道。物体在万有引力场中具有的势能叫作引力势能, 若取两物体相距无穷远时的引力势能为零, 一个质量为 $m$ 的质点距质量为 $M$ 的引力源中心为 $r$ 时, 其引力势能 $E_{\mathrm{p}}=-\frac{G M m}{r}$ (式中 $G$ 为引力常量)。已知地球半径约为 $6400 \mathrm{~km}$, 地球的第一宇宙速度为 $7.9 \mathrm{~km} / \mathrm{s}$, 火星半径约为地球半径的 $\frac{1}{2}$, 火星质量约为球质量的 $\frac{1}{9}$ 。则“天问一号”刹车后相对于火星的速度不可能为 ( ) A: $7.9 \mathrm{~km} / \mathrm{s}$ B: $5.5 \mathrm{~km} / \mathrm{s}$ C: $4.0 \mathrm{~km} / \mathrm{s}$ D: $3.2 \mathrm{~km} / \mathrm{s}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

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Astronomy

ZH

text-only

Astronomy_845

Knowing that the following image was taken at at $11: 59 \mathrm{pm}$, determine the name of which constellation was the sun passing in front of in that same day. [figure1] A: Scorpius B: Virgo C: Big Dipper D: Cancer E: Taurus

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Knowing that the following image was taken at at $11: 59 \mathrm{pm}$, determine the name of which constellation was the sun passing in front of in that same day. [figure1] A: Scorpius B: Virgo C: Big Dipper D: Cancer E: Taurus You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

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Astronomy

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multi-modal

Astronomy_804

Taking the radius of the black hole to be the Schwarzschild radius (the radius at which the escape velocity of an object would be equal to the speed of light), what is the surface area of a black hole of mass M? A: $A=\frac{16 \pi}{c^{4}} G^{2} M^{2}$ B: $A=\frac{4 \pi}{c^{4}} G^{2} M^{2}$ C: $A=\frac{4 \pi}{3 c^{4}} G^{2} M^{2}$ D: $A=16 \pi G^{2} M^{2}$ E: $A=\frac{16 \pi}{c^{2}} G^{2} M^{2}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Taking the radius of the black hole to be the Schwarzschild radius (the radius at which the escape velocity of an object would be equal to the speed of light), what is the surface area of a black hole of mass M? A: $A=\frac{16 \pi}{c^{4}} G^{2} M^{2}$ B: $A=\frac{4 \pi}{c^{4}} G^{2} M^{2}$ C: $A=\frac{4 \pi}{3 c^{4}} G^{2} M^{2}$ D: $A=16 \pi G^{2} M^{2}$ E: $A=\frac{16 \pi}{c^{2}} G^{2} M^{2}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

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Astronomy

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text-only

Astronomy_23

建造一条能通向太空的天梯, 是人们长期的梦想。当今在美国宇航局（NASA）支持下, 洛斯阿拉莫斯国家实验室的科学家已在进行这方面的研究。一种简单的设计是把天梯看作一条长度达千万层楼高的质量均匀分布的缆绳, 它由一种高强度、很轻的纳米碳管制成，由传统的太空飞船运到太空上，然后慢慢垂到地球表面。最后达到这样的状态和位置: 天梯本身呈直线状; 其上端指向太空, 下端刚与地面接触但与地面之间无相互作用; 整个天梯相对于地球静止不动。如果只考虑地球对天梯的万有引力, 试求此天梯的长度。已知地球半径 $R_{0}=6.37 \times 10^{6} \mathrm{~m}$, 地球表面处的重力加速度 $\mathrm{g}=9.80 \mathrm{~m} \cdot \mathrm{s}^{-2}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个数值。 问题: 建造一条能通向太空的天梯, 是人们长期的梦想。当今在美国宇航局（NASA）支持下, 洛斯阿拉莫斯国家实验室的科学家已在进行这方面的研究。一种简单的设计是把天梯看作一条长度达千万层楼高的质量均匀分布的缆绳, 它由一种高强度、很轻的纳米碳管制成，由传统的太空飞船运到太空上，然后慢慢垂到地球表面。最后达到这样的状态和位置: 天梯本身呈直线状; 其上端指向太空, 下端刚与地面接触但与地面之间无相互作用; 整个天梯相对于地球静止不动。如果只考虑地球对天梯的万有引力, 试求此天梯的长度。已知地球半径 $R_{0}=6.37 \times 10^{6} \mathrm{~m}$, 地球表面处的重力加速度 $\mathrm{g}=9.80 \mathrm{~m} \cdot \mathrm{s}^{-2}$ 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 请记住，你的答案应以m为单位计算，但在给出最终答案时，请不要包含单位。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是不包含任何单位的数值。

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Astronomy

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text-only

Astronomy_621

如图所示, 质量分别为 $m$ 和 $M$ 的两个星球 $A$ 和 $B$ 在引力作用下都绕 $O$ 点做匀速圆周运动, 星球 $A$ 和 $B$ 两者中心之间距离为 $L$ 。已知 $A 、 B$ 的中心和 $O$ 三点始终共线, $A$和 $B$ 分别在 $O$ 的两侧, 引力常量为 $G$ 。求: 两星球做圆周运动的周期; [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 如图所示, 质量分别为 $m$ 和 $M$ 的两个星球 $A$ 和 $B$ 在引力作用下都绕 $O$ 点做匀速圆周运动, 星球 $A$ 和 $B$ 两者中心之间距离为 $L$ 。已知 $A 、 B$ 的中心和 $O$ 三点始终共线, $A$和 $B$ 分别在 $O$ 的两侧, 引力常量为 $G$ 。求: 两星球做圆周运动的周期; [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

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Astronomy

ZH

multi-modal

Astronomy_1102

The Event Horizon Telescope (EHT) is a project to use many widely-spaced radio telescopes as a Very Long Baseline Interferometer (VBLI) to create a virtual telescope as big as the Earth. This extraordinary size allows sufficient angular resolution to be able to image the space close to the event horizon of a super massive black hole (SMBH), and provide an opportunity to test the predictions of Einstein's theory of General Relativity (GR) in a very strong gravitational field. In April 2017 the EHT collaboration managed to co-ordinate time on all of the telescopes in the array so that they could observe the SMBH (called M87*) at the centre of the Virgo galaxy, M87, and they plan to also image the SMBH at the centre of our galaxy (called Sgr A*). [figure1] Figure 3: Left: The locations of all the telescopes used during the April 2017 observing run. The solid lines correspond to baselines used for observing M87, whilst the dashed lines were the baselines used for the calibration source. Credit: EHT Collaboration. Right: A simulated model of what the region near an SMBH could look like, modelled at much higher resolution than the EHT can achieve. The light comes from the accretion disc, but the paths of the photons are bent into a characteristic shape by the extreme gravity, leading to a 'shadow' in middle of the disc - this is what the EHT is trying to image. The left side of the image is brighter than the right side as light emitted from a substance moving towards an observer is brighter than that of one moving away. Credit: Hotaka Shiokawa. Some data about the locations of the eight telescopes in the array are given below in 3-D cartesian geocentric coordinates with $X$ pointing to the Greenwich meridian, $Y$ pointing $90^{\circ}$ away in the equatorial plane (eastern longitudes have positive $Y$ ), and positive $Z$ pointing in the direction of the North Pole. This is a left-handed coordinate system. | Facility | Location | $X(\mathrm{~m})$ | $Y(\mathrm{~m})$ | $Z(\mathrm{~m})$ | | :--- | :--- | :---: | :---: | :---: | | ALMA | Chile | 2225061.3 | -5440061.7 | -2481681.2 | | APEX | Chile | 2225039.5 | -5441197.6 | -2479303.4 | | JCMT | Hawaii, USA | -5464584.7 | -2493001.2 | 2150654.0 | | LMT | Mexico | -768715.6 | -5988507.1 | 2063354.9 | | PV | Spain | 5088967.8 | -301681.2 | 3825012.2 | | SMA | Hawaii, USA | -5464555.5 | -2492928.0 | 2150797.2 | | SMT | Arizona, USA | -1828796.2 | -5054406.8 | 3427865.2 | | SPT | Antarctica | 809.8 | -816.9 | -6359568.7 | The minimum angle, $\theta_{\min }$ (in radians) that can be resolved by a VLBI array is given by the equation $$ \theta_{\min }=\frac{\lambda_{\mathrm{obs}}}{d_{\max }}, $$ where $\lambda_{\text {obs }}$ is the observing wavelength and $d_{\max }$ is the longest straight line distance between two telescopes used (called the baseline), assumed perpendicular to the line of sight during the observation. An important length scale when discussing black holes is the gravitational radius, $r_{g}=\frac{G M}{c^{2}}$, where $G$ is the gravitational constant, $M$ is the mass of the black hole and $c$ is the speed of light. The familiar event horizon of a non-rotating black hole is called the Schwartzschild radius, $r_{S} \equiv 2 r_{g}$, however this is not what the EHT is able to observe - instead the closest it can see to a black hole is called the photon sphere, where photons orbit in the black hole in unstable circular orbits. On top of this the image of the black hole is gravitationally lensed by the black hole itself magnifying the apparent radius of the photon sphere to be between $(2 \sqrt{3+2 \sqrt{2}}) r_{g}$ and $(3 \sqrt{3}) r_{g}$, determined by spin and inclination; the latter corresponds to a perfectly spherical non-spinning black hole. The area within this lensed image will appear almost black and is the 'shadow' the EHT is looking for. [figure2] Figure 4: Four nights of data were taken for M87* during the observing window of the EHT, and whilst the diameter of the disk stayed relatively constant the location of bright spots moved, possibly indicating gas that is orbiting the black hole. Credit: EHT Collaboration. The EHT observed M87* on four separate occasions during the observing window (see Fig 4), and the team saw that some of the bright spots changed in that time, suggesting they may be associated with orbiting gas close to the black hole. The Innermost Stable Circular Orbit (ISCO) is the equivalent of the photon sphere but for particles with mass (and is also stable). The total conserved energy of a circular orbit close to a non-spinning black hole is given by $$ E=m c^{2}\left(\frac{1-\frac{2 r_{g}}{r}}{\sqrt{1-\frac{3 r_{g}}{r}}}\right) $$ and the radius of the ISCO, $r_{\mathrm{ISCO}}$, is the value of $r$ for which $E$ is minimised. We expect that most black holes are in fact spinning (since most stars are spinning) and the spin of a black hole is quantified with the spin parameter $a \equiv J / J_{\max }$ where $J$ is the angular momentum of the black hole and $J_{\max }=G M^{2} / c$ is the maximum possible angular momentum it can have. The value of $a$ varies from $-1 \leq a \leq 1$, where negative spins correspond to the black hole rotating in the opposite direction to its accretion disk, and positive spins in the same direction. If $a=1$ then $r_{\text {ISCO }}=r_{g}$, whilst if $a=-1$ then $r_{\text {ISCO }}=9 r_{g}$. The angular velocity of a particle in the ISCO is given by $$ \omega^{2}=\frac{G M}{\left(r_{\text {ISCO }}^{3 / 2}+a r_{g}^{3 / 2}\right)^{2}} $$ [figure3] Figure 5: Due to the curvature of spacetime, the real distance travelled by a particle moving from the ISCO to the photon sphere (indicated with the solid red arrow) is longer than you would get purely from subtracting the radial co-ordinates of those orbits (indicated with the dashed blue arrow), which would be valid for a flat spacetime. Relations between these distances are not to scale in this diagram. Credit: Modified from Bardeen et al. (1972). The spacetime near a black hole is curved, as described by the equations of GR. This means that the distance between two points can be substantially different to the distance you would expect if spacetime was flat. GR tells us that the proper distance travelled by a particle moving from radius $r_{1}$ to radius $r_{2}$ around a black hole of mass $M$ (with $r_{1}>r_{2}$ ) is given by $$ \Delta l=\int_{r_{2}}^{r_{1}}\left(1-\frac{2 r_{g}}{r}\right)^{-1 / 2} \mathrm{~d} r $$b. Assuming Sgr A* is a non-spinning black hole with mass $4.1 \times 10^{6} \mathrm{M}_{\odot}$ and at a distance of $8.34 \mathrm{kpc}$ : i. Derive the (unlensed) radius of the photon sphere, $r_{\mathrm{ph}}$, in units of $\mathrm{r}_{\mathrm{g}}$, by considering a balance between the centripetal and (Newtonian) gravitational forces, but with the relativistic correction $v^{\prime}=v \sqrt{1-2 r_{g} / r}$ where $\mathrm{v}_{0}$ is the classical velocity and $\mathrm{r}=\mathrm{r}_{\mathrm{ph}}$ when $\mathrm{v}=\mathrm{c}$.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: The Event Horizon Telescope (EHT) is a project to use many widely-spaced radio telescopes as a Very Long Baseline Interferometer (VBLI) to create a virtual telescope as big as the Earth. This extraordinary size allows sufficient angular resolution to be able to image the space close to the event horizon of a super massive black hole (SMBH), and provide an opportunity to test the predictions of Einstein's theory of General Relativity (GR) in a very strong gravitational field. In April 2017 the EHT collaboration managed to co-ordinate time on all of the telescopes in the array so that they could observe the SMBH (called M87*) at the centre of the Virgo galaxy, M87, and they plan to also image the SMBH at the centre of our galaxy (called Sgr A*). [figure1] Figure 3: Left: The locations of all the telescopes used during the April 2017 observing run. The solid lines correspond to baselines used for observing M87, whilst the dashed lines were the baselines used for the calibration source. Credit: EHT Collaboration. Right: A simulated model of what the region near an SMBH could look like, modelled at much higher resolution than the EHT can achieve. The light comes from the accretion disc, but the paths of the photons are bent into a characteristic shape by the extreme gravity, leading to a 'shadow' in middle of the disc - this is what the EHT is trying to image. The left side of the image is brighter than the right side as light emitted from a substance moving towards an observer is brighter than that of one moving away. Credit: Hotaka Shiokawa. Some data about the locations of the eight telescopes in the array are given below in 3-D cartesian geocentric coordinates with $X$ pointing to the Greenwich meridian, $Y$ pointing $90^{\circ}$ away in the equatorial plane (eastern longitudes have positive $Y$ ), and positive $Z$ pointing in the direction of the North Pole. This is a left-handed coordinate system. | Facility | Location | $X(\mathrm{~m})$ | $Y(\mathrm{~m})$ | $Z(\mathrm{~m})$ | | :--- | :--- | :---: | :---: | :---: | | ALMA | Chile | 2225061.3 | -5440061.7 | -2481681.2 | | APEX | Chile | 2225039.5 | -5441197.6 | -2479303.4 | | JCMT | Hawaii, USA | -5464584.7 | -2493001.2 | 2150654.0 | | LMT | Mexico | -768715.6 | -5988507.1 | 2063354.9 | | PV | Spain | 5088967.8 | -301681.2 | 3825012.2 | | SMA | Hawaii, USA | -5464555.5 | -2492928.0 | 2150797.2 | | SMT | Arizona, USA | -1828796.2 | -5054406.8 | 3427865.2 | | SPT | Antarctica | 809.8 | -816.9 | -6359568.7 | The minimum angle, $\theta_{\min }$ (in radians) that can be resolved by a VLBI array is given by the equation $$ \theta_{\min }=\frac{\lambda_{\mathrm{obs}}}{d_{\max }}, $$ where $\lambda_{\text {obs }}$ is the observing wavelength and $d_{\max }$ is the longest straight line distance between two telescopes used (called the baseline), assumed perpendicular to the line of sight during the observation. An important length scale when discussing black holes is the gravitational radius, $r_{g}=\frac{G M}{c^{2}}$, where $G$ is the gravitational constant, $M$ is the mass of the black hole and $c$ is the speed of light. The familiar event horizon of a non-rotating black hole is called the Schwartzschild radius, $r_{S} \equiv 2 r_{g}$, however this is not what the EHT is able to observe - instead the closest it can see to a black hole is called the photon sphere, where photons orbit in the black hole in unstable circular orbits. On top of this the image of the black hole is gravitationally lensed by the black hole itself magnifying the apparent radius of the photon sphere to be between $(2 \sqrt{3+2 \sqrt{2}}) r_{g}$ and $(3 \sqrt{3}) r_{g}$, determined by spin and inclination; the latter corresponds to a perfectly spherical non-spinning black hole. The area within this lensed image will appear almost black and is the 'shadow' the EHT is looking for. [figure2] Figure 4: Four nights of data were taken for M87* during the observing window of the EHT, and whilst the diameter of the disk stayed relatively constant the location of bright spots moved, possibly indicating gas that is orbiting the black hole. Credit: EHT Collaboration. The EHT observed M87* on four separate occasions during the observing window (see Fig 4), and the team saw that some of the bright spots changed in that time, suggesting they may be associated with orbiting gas close to the black hole. The Innermost Stable Circular Orbit (ISCO) is the equivalent of the photon sphere but for particles with mass (and is also stable). The total conserved energy of a circular orbit close to a non-spinning black hole is given by $$ E=m c^{2}\left(\frac{1-\frac{2 r_{g}}{r}}{\sqrt{1-\frac{3 r_{g}}{r}}}\right) $$ and the radius of the ISCO, $r_{\mathrm{ISCO}}$, is the value of $r$ for which $E$ is minimised. We expect that most black holes are in fact spinning (since most stars are spinning) and the spin of a black hole is quantified with the spin parameter $a \equiv J / J_{\max }$ where $J$ is the angular momentum of the black hole and $J_{\max }=G M^{2} / c$ is the maximum possible angular momentum it can have. The value of $a$ varies from $-1 \leq a \leq 1$, where negative spins correspond to the black hole rotating in the opposite direction to its accretion disk, and positive spins in the same direction. If $a=1$ then $r_{\text {ISCO }}=r_{g}$, whilst if $a=-1$ then $r_{\text {ISCO }}=9 r_{g}$. The angular velocity of a particle in the ISCO is given by $$ \omega^{2}=\frac{G M}{\left(r_{\text {ISCO }}^{3 / 2}+a r_{g}^{3 / 2}\right)^{2}} $$ [figure3] Figure 5: Due to the curvature of spacetime, the real distance travelled by a particle moving from the ISCO to the photon sphere (indicated with the solid red arrow) is longer than you would get purely from subtracting the radial co-ordinates of those orbits (indicated with the dashed blue arrow), which would be valid for a flat spacetime. Relations between these distances are not to scale in this diagram. Credit: Modified from Bardeen et al. (1972). The spacetime near a black hole is curved, as described by the equations of GR. This means that the distance between two points can be substantially different to the distance you would expect if spacetime was flat. GR tells us that the proper distance travelled by a particle moving from radius $r_{1}$ to radius $r_{2}$ around a black hole of mass $M$ (with $r_{1}>r_{2}$ ) is given by $$ \Delta l=\int_{r_{2}}^{r_{1}}\left(1-\frac{2 r_{g}}{r}\right)^{-1 / 2} \mathrm{~d} r $$ problem: b. Assuming Sgr A* is a non-spinning black hole with mass $4.1 \times 10^{6} \mathrm{M}_{\odot}$ and at a distance of $8.34 \mathrm{kpc}$ : i. Derive the (unlensed) radius of the photon sphere, $r_{\mathrm{ph}}$, in units of $\mathrm{r}_{\mathrm{g}}$, by considering a balance between the centripetal and (Newtonian) gravitational forces, but with the relativistic correction $v^{\prime}=v \sqrt{1-2 r_{g} / r}$ where $\mathrm{v}_{0}$ is the classical velocity and $\mathrm{r}=\mathrm{r}_{\mathrm{ph}}$ when $\mathrm{v}=\mathrm{c}$. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of r_{g}, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

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Astronomy_704

“天问一号” 火星探测器已经被火星捕获。若探测器在距离火星表面高为 $h$ 的圆形轨道上绕火星飞行, 环绕 $n$ 周飞行总时间为 $t$, 已知引力常量为 $G$, 火星半径为 $R$, 则下列给出的火星表面重力加速度 $g$ (忽略自转) 和平均密度 $\rho$ 的表达式正确的是 ( ) A: $g=\frac{4 \pi^{2}(R+h)^{3}}{R^{2} t^{2}}, \rho=\frac{3 \pi(R+h)^{3}}{G t^{2} R^{3}}$ B: $g=\frac{4 \pi^{2} n^{2}(R+h)^{3}}{R^{2} t^{2}}, \rho=\frac{3 \pi n^{2}(R+h)^{3}}{G t^{2} R^{3}}$ C: $g=\frac{4 \pi^{2} t^{2}(R+h)^{3}}{R^{2} n^{2}}, \quad \rho=\frac{3 \pi t^{2}(R+h)^{3}}{G n^{2} R^{3}}$ D: $g=\frac{4 \pi^{2} n^{2}(R+h)^{3}}{R^{2} t^{2}}, \quad \rho=\frac{3 \pi(R+h)^{3}}{G t^{2} R^{3}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: “天问一号” 火星探测器已经被火星捕获。若探测器在距离火星表面高为 $h$ 的圆形轨道上绕火星飞行, 环绕 $n$ 周飞行总时间为 $t$, 已知引力常量为 $G$, 火星半径为 $R$, 则下列给出的火星表面重力加速度 $g$ (忽略自转) 和平均密度 $\rho$ 的表达式正确的是 ( ) A: $g=\frac{4 \pi^{2}(R+h)^{3}}{R^{2} t^{2}}, \rho=\frac{3 \pi(R+h)^{3}}{G t^{2} R^{3}}$ B: $g=\frac{4 \pi^{2} n^{2}(R+h)^{3}}{R^{2} t^{2}}, \rho=\frac{3 \pi n^{2}(R+h)^{3}}{G t^{2} R^{3}}$ C: $g=\frac{4 \pi^{2} t^{2}(R+h)^{3}}{R^{2} n^{2}}, \quad \rho=\frac{3 \pi t^{2}(R+h)^{3}}{G n^{2} R^{3}}$ D: $g=\frac{4 \pi^{2} n^{2}(R+h)^{3}}{R^{2} t^{2}}, \quad \rho=\frac{3 \pi(R+h)^{3}}{G t^{2} R^{3}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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ZH

text-only

Astronomy_109

在天体运动中, 把两颗相距很近的恒星称为双星。已知组成某双星系统的两颗恒星质量分别为 $m_{l}$ 和 $m_{2}$ 相距为 $L$ 。在万有引力作用下各自绕它们连线上的某一点, 在同平面内做匀速同周运动, 运动过程中二者之间的距离始终不变。已知万有引力常量为 $G$ 。 $m_{1}$ 的动能为 $E_{k}$ 则 $m_{2}$ 的动能为 A: $G \frac{m_{1} m_{2}}{l}-E_{k}$ B: $G \frac{m_{1} m_{2}}{2 l}-E_{k}$ C: $\frac{m_{1}}{m_{2}} E_{k}$ D: $\frac{m_{2}}{m_{1}} E_{k}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 在天体运动中, 把两颗相距很近的恒星称为双星。已知组成某双星系统的两颗恒星质量分别为 $m_{l}$ 和 $m_{2}$ 相距为 $L$ 。在万有引力作用下各自绕它们连线上的某一点, 在同平面内做匀速同周运动, 运动过程中二者之间的距离始终不变。已知万有引力常量为 $G$ 。 $m_{1}$ 的动能为 $E_{k}$ 则 $m_{2}$ 的动能为 A: $G \frac{m_{1} m_{2}}{l}-E_{k}$ B: $G \frac{m_{1} m_{2}}{2 l}-E_{k}$ C: $\frac{m_{1}}{m_{2}} E_{k}$ D: $\frac{m_{2}}{m_{1}} E_{k}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_364

如图所示, 在发射地球同步卫星的过程中, 卫星首先进入粗圆轨道 $\mathrm{I}$, 然后在 $\mathrm{Q}$ 点通过改变卫星速度, 让卫星进入地球同步轨道II, 则 ( ) [图1] A: 该卫星的发射速度必定大于 $11.2 \mathrm{~km} / \mathrm{s}$ B: 卫星在轨道上运行不受重力 C: 在轨道 $\mathrm{I}$ 上, 卫星在 $\mathrm{P}$ 点的速度大于在 $\mathrm{Q}$ 点的速度 D: 卫星在 $\mathrm{Q}$ 点通过加速实现由轨道 I 进入轨道II

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 如图所示, 在发射地球同步卫星的过程中, 卫星首先进入粗圆轨道 $\mathrm{I}$, 然后在 $\mathrm{Q}$ 点通过改变卫星速度, 让卫星进入地球同步轨道II, 则 ( ) [图1] A: 该卫星的发射速度必定大于 $11.2 \mathrm{~km} / \mathrm{s}$ B: 卫星在轨道上运行不受重力 C: 在轨道 $\mathrm{I}$ 上, 卫星在 $\mathrm{P}$ 点的速度大于在 $\mathrm{Q}$ 点的速度 D: 卫星在 $\mathrm{Q}$ 点通过加速实现由轨道 I 进入轨道II 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_940

Forces of Nature In the BBC programme Forces of Nature, Brian Cox uses a Eurofighter Typhoon to try and overtake the spin of the Earth such that the setting Sun appears to rise instead. [figure1] A Eurofighter sets off from the equator just as the top edge of the Sun has gone below the horizon, and rapidly accelerates due west up to a speed of $500 \mathrm{~m} \mathrm{~s}^{-1}$. Given that the Sun has an angular diameter of $0.5^{\circ}$ as viewed from Earth, what is the minimum amount of time the fighter jet needs to fly for in order to see the whole of the Sun above the horizon?

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: Forces of Nature In the BBC programme Forces of Nature, Brian Cox uses a Eurofighter Typhoon to try and overtake the spin of the Earth such that the setting Sun appears to rise instead. [figure1] A Eurofighter sets off from the equator just as the top edge of the Sun has gone below the horizon, and rapidly accelerates due west up to a speed of $500 \mathrm{~m} \mathrm{~s}^{-1}$. Given that the Sun has an angular diameter of $0.5^{\circ}$ as viewed from Earth, what is the minimum amount of time the fighter jet needs to fly for in order to see the whole of the Sun above the horizon? All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of minutes, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

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Astronomy

EN

multi-modal

Astronomy_879

Abhay looks at the light curves for two main sequence stars A and B, which you can assume are blackbodies. A has its peak at a frequency two times as high as that of B. By looking at the depth of spectral lines, Abhay can also determine that A has higher metallicity than B. Abhay makes the following statements: $\mathrm{P}$ : A has higher absolute magnitude than $\mathrm{B}$. Q: A is older than B. Which of the following is true? A: P and Q are true. B: $\mathrm{P}$ and $\mathrm{Q}$ are false. C: $\mathrm{P}$ is true and $\mathrm{Q}$ is false. D: $\mathrm{P}$ is false and $\mathrm{Q}$ is true. E: We don't have sufficient information for one or more of these statements.

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Abhay looks at the light curves for two main sequence stars A and B, which you can assume are blackbodies. A has its peak at a frequency two times as high as that of B. By looking at the depth of spectral lines, Abhay can also determine that A has higher metallicity than B. Abhay makes the following statements: $\mathrm{P}$ : A has higher absolute magnitude than $\mathrm{B}$. Q: A is older than B. Which of the following is true? A: P and Q are true. B: $\mathrm{P}$ and $\mathrm{Q}$ are false. C: $\mathrm{P}$ is true and $\mathrm{Q}$ is false. D: $\mathrm{P}$ is false and $\mathrm{Q}$ is true. E: We don't have sufficient information for one or more of these statements. You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

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Astronomy

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text-only

Astronomy_424

2009 年 10 月 7 日电，美国宇航局（NASA）的斯皮策（Spitzer）太空望远镜近期发现土星外环绕着一个巨大的漫射环. 该环比已知的由太空尘埃和冰块组成的土星环要大得多. 据悉, 这个由细小冰粒及尘埃组成的土星环温度接近 $-157^{\circ} \mathrm{C}$, 结构非常松散,难以反射光线, 所以此前一直未被发现, 而仅能被红外探测仪检测到. 这一暗淡的土星环由微小粒子构成, 环内侧距土星中心约 600 万公里, 外侧距土星中心约 1800 万公里. 若忽略微粒间的作用力，假设土环上的微粒均绕土星做圆周运动，则土环内侧、外侧微粒的 A: 线速度之比为 $\sqrt{3}: 1$ B: 角速度之比为 $1: 1$ C: 周期之比为 $1: 1$ D: 向心加速度之比为 9:1

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 2009 年 10 月 7 日电，美国宇航局（NASA）的斯皮策（Spitzer）太空望远镜近期发现土星外环绕着一个巨大的漫射环. 该环比已知的由太空尘埃和冰块组成的土星环要大得多. 据悉, 这个由细小冰粒及尘埃组成的土星环温度接近 $-157^{\circ} \mathrm{C}$, 结构非常松散,难以反射光线, 所以此前一直未被发现, 而仅能被红外探测仪检测到. 这一暗淡的土星环由微小粒子构成, 环内侧距土星中心约 600 万公里, 外侧距土星中心约 1800 万公里. 若忽略微粒间的作用力，假设土环上的微粒均绕土星做圆周运动，则土环内侧、外侧微粒的 A: 线速度之比为 $\sqrt{3}: 1$ B: 角速度之比为 $1: 1$ C: 周期之比为 $1: 1$ D: 向心加速度之比为 9:1 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

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Astronomy

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text-only

Astronomy_1064

In science fiction films the asteroid belt is typically portrayed as a region of the Solar System where the spacecraft needs to dodge and weave its way through many large asteroids that are rather close together. However, if this image were true then very few probes would be able to pass through the belt into the outer Solar System. [figure1] Figure 1 Artist conceptual illustration of the asteroid belt (left). Schematic of the Solar System with the asteroid belt between Mars and Jupiter (right). This question will look at the real distances between asteroids.b. The main part of the asteroid belt extends from $2.1 \mathrm{AU}$ to $3.3 \mathrm{AU}$, and has an average angular width of $16.0^{\circ}$, as viewed from the Sun. Calculate the average thickness of the belt, and hence its total volume, $V_{\text {belt }}$.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: In science fiction films the asteroid belt is typically portrayed as a region of the Solar System where the spacecraft needs to dodge and weave its way through many large asteroids that are rather close together. However, if this image were true then very few probes would be able to pass through the belt into the outer Solar System. [figure1] Figure 1 Artist conceptual illustration of the asteroid belt (left). Schematic of the Solar System with the asteroid belt between Mars and Jupiter (right). This question will look at the real distances between asteroids. problem: b. The main part of the asteroid belt extends from $2.1 \mathrm{AU}$ to $3.3 \mathrm{AU}$, and has an average angular width of $16.0^{\circ}$, as viewed from the Sun. Calculate the average thickness of the belt, and hence its total volume, $V_{\text {belt }}$. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \mathrm{~km}^{3}, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_295

如图所示, $\mathrm{A}$ 为地球赤道表面上的物体, $\mathrm{B}$ 为地球赤道平面内的圆轨道卫星, $\mathrm{C}$ 为地球同步卫星。已知卫星 $\mathrm{B} 、 \mathrm{C}$ 的轨道半径之比为 $1: 3$, 下列说法正确的是 ( ) [图1] A: 卫星 B 的向心加速度大于物体 A 的向心加速度 B: 卫星 B 的向心力大于卫星 C 的向心力 C: 卫星 $\mathrm{B} 、 \mathrm{C}$ 的线速度大小之比为 3: 1 D: 卫星 $\mathrm{B} 、 \mathrm{C}$ 的运行周期之比为 1: 9

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 如图所示, $\mathrm{A}$ 为地球赤道表面上的物体, $\mathrm{B}$ 为地球赤道平面内的圆轨道卫星, $\mathrm{C}$ 为地球同步卫星。已知卫星 $\mathrm{B} 、 \mathrm{C}$ 的轨道半径之比为 $1: 3$, 下列说法正确的是 ( ) [图1] A: 卫星 B 的向心加速度大于物体 A 的向心加速度 B: 卫星 B 的向心力大于卫星 C 的向心力 C: 卫星 $\mathrm{B} 、 \mathrm{C}$ 的线速度大小之比为 3: 1 D: 卫星 $\mathrm{B} 、 \mathrm{C}$ 的运行周期之比为 1: 9 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_206

神舟六号载人航天飞船经过 115 小时 32 分钟的太空飞行, 绕地球飞行 77 圈, 飞船返回舱终于在 2005 年 10 月 17 日凌晨 4 时 33 分成功着陆, 航天员费俊龙、聂海胜安全返回, 已知万有引力常量 $G$, 地球表面的重力加速度 $g$, 地球的半径 $R$, 神舟六号飞船太空飞行近似为圆周运动, 则下列论述正确的是( ) A: 可以计算神舟六号飞船绕地球的太空飞行离地球表面的高度 $h$ B: 可以计算神舟六号飞船在绕地球的太空飞行的加速度 C: 可以计算神舟六号飞船在绕地球的太空飞行的线速度 D: 飞船返回舱打开减速伞下降的过程中，飞船中的宇航员处于失重状态

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 神舟六号载人航天飞船经过 115 小时 32 分钟的太空飞行, 绕地球飞行 77 圈, 飞船返回舱终于在 2005 年 10 月 17 日凌晨 4 时 33 分成功着陆, 航天员费俊龙、聂海胜安全返回, 已知万有引力常量 $G$, 地球表面的重力加速度 $g$, 地球的半径 $R$, 神舟六号飞船太空飞行近似为圆周运动, 则下列论述正确的是( ) A: 可以计算神舟六号飞船绕地球的太空飞行离地球表面的高度 $h$ B: 可以计算神舟六号飞船在绕地球的太空飞行的加速度 C: 可以计算神舟六号飞船在绕地球的太空飞行的线速度 D: 飞船返回舱打开减速伞下降的过程中，飞船中的宇航员处于失重状态 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_1040

In order to move a spacecraft between orbits we must apply a thrust using rockets, which changes the velocity of the spacecraft by $\Delta v$. In this question we will ignore changes in the mass of the spacecraft due to the burning of fuel. For an object of mass $m$ in a circular orbit of radius $r$ around an object with mass $M$ (where $m \ll M$ ) the orbital velocity, $v_{\text {orb }}$, is given by the formula $v_{\text {orb }}=\sqrt{\frac{G M}{r}}$.d. Derive an expression for the total time spent on the transfer orbit, $t_{H}$, and calculate it for an Earth to Mars transfer. Give your answer in months. (Use 1 month = 30 days).

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is an expression. Here is some context information for this question, which might assist you in solving it: In order to move a spacecraft between orbits we must apply a thrust using rockets, which changes the velocity of the spacecraft by $\Delta v$. In this question we will ignore changes in the mass of the spacecraft due to the burning of fuel. For an object of mass $m$ in a circular orbit of radius $r$ around an object with mass $M$ (where $m \ll M$ ) the orbital velocity, $v_{\text {orb }}$, is given by the formula $v_{\text {orb }}=\sqrt{\frac{G M}{r}}$. problem: d. Derive an expression for the total time spent on the transfer orbit, $t_{H}$, and calculate it for an Earth to Mars transfer. Give your answer in months. (Use 1 month = 30 days). All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is an expression without equals signs, e.g. ANSWER=\frac{1}{2} g t^2

EX

Astronomy

EN

text-only

Astronomy_1053

It is often said that the Sun rises in the East and sets in the West, however this is only true twice a year at the equinoxes. In the Northern hemisphere, the Sun will rise northwards of East on the June solstice, and southwards of East on the December solstice; this is directly tied in with the varying length of day too, since the Sun either has a greater or shorter distance to travel across the sky (see Figure 1). [figure1] Figure 1: Left: The path of the Sun across the sky during the equinoxes and solstices, as viewed by an observer in the Northern hemisphere at a latitude of $\sim 40^{\circ}$. Credit: Daniel V. Schroeder / Weber State University. Right: The same idea but viewed from Iceland at a latitude of $65^{\circ}$, where by being so close to the Artic circle the day length can get close to 24 hours in June and almost no daylight in December. Credit: Kristn Bjarnadttir / University of Iceland. During the equinox, the Sun travels along the projection of the Earth's equator. In this question, we will assume a circular orbit for the Earth, and all angles will be calculated in degrees. A simple model for the vertical angle between the Sun and the horizon (known at the altitude), $h$, as a function of the bearing on the horizon, $A$ (measured clockwise from North, also called the azimuth), the latitude of the observer, $\phi$ (positive in Northern hemisphere, negative in Southern hemisphere), and the vertical angle of the Sun relative to the celestial equator (known as the solar declination), $\delta$, is given as: $$ h=-\left(90^{\circ}-\phi\right) \cos (A)+\delta $$ The solar declination can be considered to vary sinusoidally over the year, going from a maximum of $\delta=+23.44^{\circ}$ at the June solstice (roughly $21^{\text {st }}$ June) to a minimum of $\delta=-23.44^{\circ}$ on the December solstice (roughly $21^{\text {st }}$ December). It can be shown using spherical trigonometry that the precise model connecting $\delta, h, \phi$ and $A$ is: $$ \sin (\delta)=\sin (h) \sin (\phi)+\cos (h) \cos (\phi) \cos (A) . $$ Using the precise model, the path of the Sun across the sky forms a shape that is not quite the cosine shape of the simple model, and is shown in Figure 2. [figure2] Figure 2: The altitude of the Sun as a function of bearing during the equinoxes and solstices, as viewed by an observer at a latitude of $+56^{\circ}$. Whilst it resembles the cosine shape of the simple model well at this latitude, there are small deviations. Credit: Wikipedia. By using further spherical trigonometry, we can derive a second helpful equation in the precise model: $$ \sin (h)=\sin (\phi) \sin (\delta)+\cos (\phi) \cos (\delta) \cos (H) $$ Here, $H$ is the solar hour angle, which measures the angle between the Sun and solar noon as measured along the projection of the Earth's equator on the sky. Conventionally, $H=0^{\circ}$ at solar noon, is negative before solar noon, and is positive afterwards. Since the sun's hour angle increases at an approximately constant rate due to the rotation of the Earth, we can convert this angle into a time using the conversion $360^{\circ}=24^{\mathrm{h}}$.c. Reconsider the Oxford observer at the June solstice, but this time use the two equations of the precise model. Ignore any atmospheric effects. i. Calculate the bearing of sunrise and the duration of the day (in hours and minutes), taking the Sun to be a point source.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is an expression. Here is some context information for this question, which might assist you in solving it: It is often said that the Sun rises in the East and sets in the West, however this is only true twice a year at the equinoxes. In the Northern hemisphere, the Sun will rise northwards of East on the June solstice, and southwards of East on the December solstice; this is directly tied in with the varying length of day too, since the Sun either has a greater or shorter distance to travel across the sky (see Figure 1). [figure1] Figure 1: Left: The path of the Sun across the sky during the equinoxes and solstices, as viewed by an observer in the Northern hemisphere at a latitude of $\sim 40^{\circ}$. Credit: Daniel V. Schroeder / Weber State University. Right: The same idea but viewed from Iceland at a latitude of $65^{\circ}$, where by being so close to the Artic circle the day length can get close to 24 hours in June and almost no daylight in December. Credit: Kristn Bjarnadttir / University of Iceland. During the equinox, the Sun travels along the projection of the Earth's equator. In this question, we will assume a circular orbit for the Earth, and all angles will be calculated in degrees. A simple model for the vertical angle between the Sun and the horizon (known at the altitude), $h$, as a function of the bearing on the horizon, $A$ (measured clockwise from North, also called the azimuth), the latitude of the observer, $\phi$ (positive in Northern hemisphere, negative in Southern hemisphere), and the vertical angle of the Sun relative to the celestial equator (known as the solar declination), $\delta$, is given as: $$ h=-\left(90^{\circ}-\phi\right) \cos (A)+\delta $$ The solar declination can be considered to vary sinusoidally over the year, going from a maximum of $\delta=+23.44^{\circ}$ at the June solstice (roughly $21^{\text {st }}$ June) to a minimum of $\delta=-23.44^{\circ}$ on the December solstice (roughly $21^{\text {st }}$ December). It can be shown using spherical trigonometry that the precise model connecting $\delta, h, \phi$ and $A$ is: $$ \sin (\delta)=\sin (h) \sin (\phi)+\cos (h) \cos (\phi) \cos (A) . $$ Using the precise model, the path of the Sun across the sky forms a shape that is not quite the cosine shape of the simple model, and is shown in Figure 2. [figure2] Figure 2: The altitude of the Sun as a function of bearing during the equinoxes and solstices, as viewed by an observer at a latitude of $+56^{\circ}$. Whilst it resembles the cosine shape of the simple model well at this latitude, there are small deviations. Credit: Wikipedia. By using further spherical trigonometry, we can derive a second helpful equation in the precise model: $$ \sin (h)=\sin (\phi) \sin (\delta)+\cos (\phi) \cos (\delta) \cos (H) $$ Here, $H$ is the solar hour angle, which measures the angle between the Sun and solar noon as measured along the projection of the Earth's equator on the sky. Conventionally, $H=0^{\circ}$ at solar noon, is negative before solar noon, and is positive afterwards. Since the sun's hour angle increases at an approximately constant rate due to the rotation of the Earth, we can convert this angle into a time using the conversion $360^{\circ}=24^{\mathrm{h}}$. problem: c. Reconsider the Oxford observer at the June solstice, but this time use the two equations of the precise model. Ignore any atmospheric effects. i. Calculate the bearing of sunrise and the duration of the day (in hours and minutes), taking the Sun to be a point source. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is an expression without equals signs, e.g. ANSWER=\frac{1}{2} g t^2

EX

Astronomy

EN

multi-modal

Astronomy_1030

The Universe consists of three main components: radiation (including neutrinos), matter (both atoms and dark matter), and dark energy. The overall density of the universe has been dominated by the density of each of those in turn at different times in its history, leading to three different epochs (shown in Fig 3). [figure1] Figure 3: A outline of the three main epochs in the history of the Universe. You do not need to read any data off this graph to answer this question. Credit: Pearson Education, Inc. The scale factor, $a$, describes how the Universe has expanded (i.e. a measure of the relative radius of the Universe), and the current value is defined as $a_{0} \equiv 1$ where the subscript ' 0 ' indicates it is as measured today. At earlier times $a<1$ and at the Big Bang $a=0$. The redshift of an object, $z$, is related to the scale factor as $a=(1+z)^{-1}$ and so the redshift corresponding to now is $z=0$, and far away objects have higher redshift $(z>0)$ since we observe them as they were long ago when the scale factor was smaller. For a Universe to be flat (i.e. zero curvature), its average density must be equal to the critical density, $$ \rho_{\text {crit }, 0}=\frac{3 H_{0}^{2}}{8 \pi G}, $$ where $H_{0}$ is the Hubble constant, measured in 2018 from the cosmic microwave background by the Planck spacecraft to be $67.36 \mathrm{~km} \mathrm{~s}^{-1} \mathrm{Mpc}^{-1}$. The density of the $i^{\text {th }}$ component of the Universe can be expressed relative to the critical density as the density parameter, $$ \Omega_{i}=\frac{\rho_{i}}{\rho_{\text {crit }}} . $$ Planck measured the current density parameters of dark energy and matter as $\Omega_{\Lambda, 0}=0.6847$ and $\Omega_{m, 0}=0.3153$ respectively. In each epoch, the scale factor increases at a different rate with time, $t$, as the density also varies differently with scale factor. - Radiation-dominated epoch: The Universe's early history, where $\rho \propto a^{-4}$ and so $a \propto t^{1 / 2}$ - Matter-dominated epoch: This represents much of the history of the Universe, where $\rho \propto$ $a^{-3}$ and so $a \propto t^{2 / 3}$ - Dark-energy-dominated epoch: This is an era we have recently entered and will remain in for the rest of time, where $\rho$ doesn't vary with scale factor (i.e. is a constant) and so $a \propto e^{H_{0} t}$ Assuming the Universe has always been flat, find the time $t_{e q}$, corresponding to when the densities of matter and radiation were equal, given that data from Planck has allowed us to calculate the redshift of this to be $z_{e q}=3402$, and find the average density of the Universe at $t_{e q}$. Again, do not try and read any data off the graph.

You are participating in an international Astronomy competition and need to solve the following question. This question involves multiple quantities to be determined. problem: The Universe consists of three main components: radiation (including neutrinos), matter (both atoms and dark matter), and dark energy. The overall density of the universe has been dominated by the density of each of those in turn at different times in its history, leading to three different epochs (shown in Fig 3). [figure1] Figure 3: A outline of the three main epochs in the history of the Universe. You do not need to read any data off this graph to answer this question. Credit: Pearson Education, Inc. The scale factor, $a$, describes how the Universe has expanded (i.e. a measure of the relative radius of the Universe), and the current value is defined as $a_{0} \equiv 1$ where the subscript ' 0 ' indicates it is as measured today. At earlier times $a<1$ and at the Big Bang $a=0$. The redshift of an object, $z$, is related to the scale factor as $a=(1+z)^{-1}$ and so the redshift corresponding to now is $z=0$, and far away objects have higher redshift $(z>0)$ since we observe them as they were long ago when the scale factor was smaller. For a Universe to be flat (i.e. zero curvature), its average density must be equal to the critical density, $$ \rho_{\text {crit }, 0}=\frac{3 H_{0}^{2}}{8 \pi G}, $$ where $H_{0}$ is the Hubble constant, measured in 2018 from the cosmic microwave background by the Planck spacecraft to be $67.36 \mathrm{~km} \mathrm{~s}^{-1} \mathrm{Mpc}^{-1}$. The density of the $i^{\text {th }}$ component of the Universe can be expressed relative to the critical density as the density parameter, $$ \Omega_{i}=\frac{\rho_{i}}{\rho_{\text {crit }}} . $$ Planck measured the current density parameters of dark energy and matter as $\Omega_{\Lambda, 0}=0.6847$ and $\Omega_{m, 0}=0.3153$ respectively. In each epoch, the scale factor increases at a different rate with time, $t$, as the density also varies differently with scale factor. - Radiation-dominated epoch: The Universe's early history, where $\rho \propto a^{-4}$ and so $a \propto t^{1 / 2}$ - Matter-dominated epoch: This represents much of the history of the Universe, where $\rho \propto$ $a^{-3}$ and so $a \propto t^{2 / 3}$ - Dark-energy-dominated epoch: This is an era we have recently entered and will remain in for the rest of time, where $\rho$ doesn't vary with scale factor (i.e. is a constant) and so $a \propto e^{H_{0} t}$ Assuming the Universe has always been flat, find the time $t_{e q}$, corresponding to when the densities of matter and radiation were equal, given that data from Planck has allowed us to calculate the redshift of this to be $z_{e q}=3402$, and find the average density of the Universe at $t_{e q}$. Again, do not try and read any data off the graph. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Your final quantities should be output in the following order: [when the densities of matter and radiation were equal, the average density of the Universe]. Their units are, in order, [years, $\text{kg m}^{-3}$], but units shouldn't be included in your concluded answer. Their answer types are, in order, [expression, expression]. Please end your response with: "The final answers are \boxed{ANSWER}", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5

MPV

Astronomy

EN

multi-modal

Astronomy_81

2020 年 7 月 23 日, 我国“天宫一号”探测器在中国文昌航天发射场发射升空。设未来的某天, 该探测器在火星表面完成探测任务返回地球, 探测器在控制系统的指令下,离开火星表面坚直向上做加速直线运动; 探测器的内部有一固定的压力传感器, 质量为 $m$ 的物体水平放置在压力传感器上, 当探测器上升到距火星表面高度为火星半径的 $\frac{1}{4}$ 时,探测器的加速度为 $a$, 压力传感器的示数为 $F$, 引力常量为 $G$ 。忽略火星的自转, 则火星表面的重力加速度为 ( ) A: $\frac{5}{4}\left(\frac{F}{m}-a\right)$ B: $\frac{25}{16}\left(\frac{F}{m}-a\right)$ C: $\frac{4}{5}\left(\frac{F}{m}-a\right)$ D: $\frac{16}{25}\left(\frac{F}{m}-a\right)$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2020 年 7 月 23 日, 我国“天宫一号”探测器在中国文昌航天发射场发射升空。设未来的某天, 该探测器在火星表面完成探测任务返回地球, 探测器在控制系统的指令下,离开火星表面坚直向上做加速直线运动; 探测器的内部有一固定的压力传感器, 质量为 $m$ 的物体水平放置在压力传感器上, 当探测器上升到距火星表面高度为火星半径的 $\frac{1}{4}$ 时,探测器的加速度为 $a$, 压力传感器的示数为 $F$, 引力常量为 $G$ 。忽略火星的自转, 则火星表面的重力加速度为 ( ) A: $\frac{5}{4}\left(\frac{F}{m}-a\right)$ B: $\frac{25}{16}\left(\frac{F}{m}-a\right)$ C: $\frac{4}{5}\left(\frac{F}{m}-a\right)$ D: $\frac{16}{25}\left(\frac{F}{m}-a\right)$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_1152

All stars lose mass during their lifetimes due to two main routes: particles escaping their surface (referred to as the stellar wind), and the mass defect of the nuclear reactions occurring in their cores. In practice, the mass loss rate can vary quite considerably during a star's lifetime, particularly once it has left the main sequence when the stellar wind can become much more substantial. Wolf-Rayet stars are massive stars near the end of their lives, presumed to be the in the stage just before a supernova, and are losing substantial amounts of mass due to very fast stellar winds. This deposits considerable energy into the surrounding interstellar medium (ISM) and can sweep up material into a thin bubble around the star, visible as a type of planetary nebula. [figure1] Figure 1: The nebula NGC 2359 around the Wolf-Rayet star WR7. The nebula is known as Thor's Helmet due to its resemblance to the helmet worn by the character from the Marvel Comics series. Credit: Star Shadows Remote Observatory and PROMPT/UNC.c. The Wolf-Rayet star WR7 is in the constellation of Canis Major and its strong winds are responsible for the nebula known as Thor's Helmet. The star has a mass of $16 M_{\odot}$, a radius of $1.41 R_{\odot}$ and a surface temperature of $112000 \mathrm{~K}$, with a measured $v_{\infty}$ of $1545 \mathrm{~km} \mathrm{~s}^{-1}$. Using your new value for $\dot{M}$, calculate the total mass expelled from the star and hence the total kinetic energy the stellar wind has so far deposited into the ISM during this stage of the star's life.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: All stars lose mass during their lifetimes due to two main routes: particles escaping their surface (referred to as the stellar wind), and the mass defect of the nuclear reactions occurring in their cores. In practice, the mass loss rate can vary quite considerably during a star's lifetime, particularly once it has left the main sequence when the stellar wind can become much more substantial. Wolf-Rayet stars are massive stars near the end of their lives, presumed to be the in the stage just before a supernova, and are losing substantial amounts of mass due to very fast stellar winds. This deposits considerable energy into the surrounding interstellar medium (ISM) and can sweep up material into a thin bubble around the star, visible as a type of planetary nebula. [figure1] Figure 1: The nebula NGC 2359 around the Wolf-Rayet star WR7. The nebula is known as Thor's Helmet due to its resemblance to the helmet worn by the character from the Marvel Comics series. Credit: Star Shadows Remote Observatory and PROMPT/UNC. problem: c. The Wolf-Rayet star WR7 is in the constellation of Canis Major and its strong winds are responsible for the nebula known as Thor's Helmet. The star has a mass of $16 M_{\odot}$, a radius of $1.41 R_{\odot}$ and a surface temperature of $112000 \mathrm{~K}$, with a measured $v_{\infty}$ of $1545 \mathrm{~km} \mathrm{~s}^{-1}$. Using your new value for $\dot{M}$, calculate the total mass expelled from the star and hence the total kinetic energy the stellar wind has so far deposited into the ISM during this stage of the star's life. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \mathrm{~km}^{3}, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_1151

GW170817 was the first gravitational wave event arising from a binary neutron star merger to have been detected by the LIGO \& Virgo experiments, and careful localization of the source meant that the electromagnetic counterpart was quickly found in galaxy NGC 4993 (see Figure 2). Such a combination of two completely separate branches of astronomical observation begins a new era of 'multi-messenger astronomy'. Since the gravitational waves allow an independent measurement of the distance to the host galaxy and the light allows an independent measurement of the recessional speed, this observation allows us to determine a new, independent value of the Hubble constant $H_{0}$. Another way of measuring distances to galaxies is to use the Fundamental Plane (FP) relation, which relies on the assumption there is a fairly tight relation between radius, surface brightness, and velocity dispersion for bulge-dominated galaxies, and is widely used for galaxies like NGC 4993. It can be described by the relation $$ \log \left(\frac{D}{(1+z)^{2}}\right)=-\log R_{e}+\alpha \log \sigma-\beta \log \left\langle I_{r}\right\rangle_{e}+\gamma $$ where $D$ is the distance in Mpc, $R_{e}$ is the effective radius measured in arcseconds, $\sigma$ is the velocity dispersion in $\mathrm{km} \mathrm{s}^{-1},\left\langle I_{r}\right\rangle_{e}$ is the mean intensity inside the effective radius measured in $L_{\odot} \mathrm{pc}^{-2}$, and $\gamma$ is the distance-dependent zero point of the relation. Calibrating the zero point to the Leo I galaxy group, the constants in the FP relation become $\alpha=1.24, \beta=0.82$, and $\gamma=2.194$. Figure 2: Left: The GW170817 signal as measured by the LIGO and Virgo gravitational wave detectors, taken from Abbott $e t$ al. (2017). The normalized amplitude (or strain) is in units of $10^{-21}$. The signal is not visible in the Virgo data due to the direction of the source with respect to the detector's antenna pattern. Right: The optical counterpart of GW170817 in host galaxy NGC 4993, taken from Hjorth et al. (2017). By measuring the amplitude (called strain, $h$ ) and the frequency of the gravitational waves, $f_{\mathrm{GW}}$, one can determine the distance to the source without having to rely on 'standard candles' like Cepheid variables or Type Ia supernovae. For two masses, $m_{1}$ and $m_{2}$, orbiting the centre of mass with separation $a$ with orbital angular velocity $\omega$, then the dimensionless strain parameter $h$ is $$ h \simeq \frac{G}{c^{4}} \frac{1}{r} \mu a^{2} \omega^{2} $$ where $r$ is the luminosity distance, $c$ is the speed of light, $\mu=m_{1} m_{2} / M_{\text {tot }}$ is the reduced mass and $M_{\text {tot }}=m_{1}+m_{2}$ is the total mass. The rate of change of frequency of the gravitational waves (called the 'chirp') from a merging binary can be written as $$ \dot{f}_{\mathrm{GW}}=\frac{96}{5} \pi^{8 / 3}\left(\frac{G \mathcal{M}}{c^{3}}\right)^{5 / 3} f_{\mathrm{GW}}^{11 / 3} $$c. Given that the gravitational frequency, $f_{G W}$, is twice the orbital frequency (i.e. $f_{G W}=\omega / \pi$ ) and the 'chirp mass', $\left.\mathcal{M}=\left(\mu^{3} M_{\text {tot }}\right)^{2}\right)^{1 / 5}$, express $h$ in terms of only $\mathcal{M}, r, f_{G W}$, and various fundamental constants.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is an expression. Here is some context information for this question, which might assist you in solving it: GW170817 was the first gravitational wave event arising from a binary neutron star merger to have been detected by the LIGO \& Virgo experiments, and careful localization of the source meant that the electromagnetic counterpart was quickly found in galaxy NGC 4993 (see Figure 2). Such a combination of two completely separate branches of astronomical observation begins a new era of 'multi-messenger astronomy'. Since the gravitational waves allow an independent measurement of the distance to the host galaxy and the light allows an independent measurement of the recessional speed, this observation allows us to determine a new, independent value of the Hubble constant $H_{0}$. Another way of measuring distances to galaxies is to use the Fundamental Plane (FP) relation, which relies on the assumption there is a fairly tight relation between radius, surface brightness, and velocity dispersion for bulge-dominated galaxies, and is widely used for galaxies like NGC 4993. It can be described by the relation $$ \log \left(\frac{D}{(1+z)^{2}}\right)=-\log R_{e}+\alpha \log \sigma-\beta \log \left\langle I_{r}\right\rangle_{e}+\gamma $$ where $D$ is the distance in Mpc, $R_{e}$ is the effective radius measured in arcseconds, $\sigma$ is the velocity dispersion in $\mathrm{km} \mathrm{s}^{-1},\left\langle I_{r}\right\rangle_{e}$ is the mean intensity inside the effective radius measured in $L_{\odot} \mathrm{pc}^{-2}$, and $\gamma$ is the distance-dependent zero point of the relation. Calibrating the zero point to the Leo I galaxy group, the constants in the FP relation become $\alpha=1.24, \beta=0.82$, and $\gamma=2.194$. Figure 2: Left: The GW170817 signal as measured by the LIGO and Virgo gravitational wave detectors, taken from Abbott $e t$ al. (2017). The normalized amplitude (or strain) is in units of $10^{-21}$. The signal is not visible in the Virgo data due to the direction of the source with respect to the detector's antenna pattern. Right: The optical counterpart of GW170817 in host galaxy NGC 4993, taken from Hjorth et al. (2017). By measuring the amplitude (called strain, $h$ ) and the frequency of the gravitational waves, $f_{\mathrm{GW}}$, one can determine the distance to the source without having to rely on 'standard candles' like Cepheid variables or Type Ia supernovae. For two masses, $m_{1}$ and $m_{2}$, orbiting the centre of mass with separation $a$ with orbital angular velocity $\omega$, then the dimensionless strain parameter $h$ is $$ h \simeq \frac{G}{c^{4}} \frac{1}{r} \mu a^{2} \omega^{2} $$ where $r$ is the luminosity distance, $c$ is the speed of light, $\mu=m_{1} m_{2} / M_{\text {tot }}$ is the reduced mass and $M_{\text {tot }}=m_{1}+m_{2}$ is the total mass. The rate of change of frequency of the gravitational waves (called the 'chirp') from a merging binary can be written as $$ \dot{f}_{\mathrm{GW}}=\frac{96}{5} \pi^{8 / 3}\left(\frac{G \mathcal{M}}{c^{3}}\right)^{5 / 3} f_{\mathrm{GW}}^{11 / 3} $$ problem: c. Given that the gravitational frequency, $f_{G W}$, is twice the orbital frequency (i.e. $f_{G W}=\omega / \pi$ ) and the 'chirp mass', $\left.\mathcal{M}=\left(\mu^{3} M_{\text {tot }}\right)^{2}\right)^{1 / 5}$, express $h$ in terms of only $\mathcal{M}, r, f_{G W}$, and various fundamental constants. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is an expression without equals signs, e.g. ANSWER=\frac{1}{2} g t^2

EX

Astronomy

EN

text-only

Astronomy_969

The website what 3 words splits up the Earth's surface into $3 \mathrm{~m} \times 3 \mathrm{~m}$ squares and gives a coordinate of three randomly chosen words (for example the entrance to the Oxford University Physics Department is engage.proud.police). If each of the words is taken from the same list of $n$ words, what value of $n$ is needed? A: $\sim 10000$ B: $\sim 20000$ C: $\sim 30000$ D: $\sim 40000$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: The website what 3 words splits up the Earth's surface into $3 \mathrm{~m} \times 3 \mathrm{~m}$ squares and gives a coordinate of three randomly chosen words (for example the entrance to the Oxford University Physics Department is engage.proud.police). If each of the words is taken from the same list of $n$ words, what value of $n$ is needed? A: $\sim 10000$ B: $\sim 20000$ C: $\sim 30000$ D: $\sim 40000$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

text-only

Astronomy_995

If a planet orbits the Sun with a semi-major axis of $4 \mathrm{AU}$, what is the period of its orbit? A: 4 years B: 8 years C: 12 years D: 64 years

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: If a planet orbits the Sun with a semi-major axis of $4 \mathrm{AU}$, what is the period of its orbit? A: 4 years B: 8 years C: 12 years D: 64 years You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

text-only

Astronomy_160

地球和月球在长期相互作用过程中，形成了“潮汐锁定”月球总是一面正对地球，另一面背离地球, 月球绕地球的运动可看成匀速圆周运动。以下说法正确的是（） [图1] A: 月球的公转周期与自转周期相同 B: 地球对月球的引力大于月球对地球的引力 C: 月球上远地端的向心加速度大于近地端的向心加速度 D: 若测得月球公转的周期和半径可估测月球质量

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 地球和月球在长期相互作用过程中，形成了“潮汐锁定”月球总是一面正对地球，另一面背离地球, 月球绕地球的运动可看成匀速圆周运动。以下说法正确的是（） [图1] A: 月球的公转周期与自转周期相同 B: 地球对月球的引力大于月球对地球的引力 C: 月球上远地端的向心加速度大于近地端的向心加速度 D: 若测得月球公转的周期和半径可估测月球质量 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_656

79. 关于人造卫星和宇宙飞船，下列说法正确的是（） A: 一艘绕地球运转的宇宙飞船, 宇航员从舱内慢慢走出, 并离开飞船, 飞船因质量减小，所受到的万有引力减小，故飞行速度减小 B: 两颗人造卫星, 只要它们在圆形轨道的运行速度相等, 不管它们的质量、形状差别有多大，它们的运行速度相等，周期也相等 C: 原来在同一轨道上沿同一方向运转的人造卫星一前一后, 若要后一个卫星追上前一个卫星并发生碰撞, 只要将后面一个卫星速率增大一些即可 D: 关于航天飞机与空间站对接问题, 先让航天飞机进入较低的轨道, 然后再对其进行加速，即可实现对接

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 79. 关于人造卫星和宇宙飞船，下列说法正确的是（） A: 一艘绕地球运转的宇宙飞船, 宇航员从舱内慢慢走出, 并离开飞船, 飞船因质量减小，所受到的万有引力减小，故飞行速度减小 B: 两颗人造卫星, 只要它们在圆形轨道的运行速度相等, 不管它们的质量、形状差别有多大，它们的运行速度相等，周期也相等 C: 原来在同一轨道上沿同一方向运转的人造卫星一前一后, 若要后一个卫星追上前一个卫星并发生碰撞, 只要将后面一个卫星速率增大一些即可 D: 关于航天飞机与空间站对接问题, 先让航天飞机进入较低的轨道, 然后再对其进行加速，即可实现对接 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_790

Two asteroids with masses $m_{1}, m_{2}$ and velocities $v_{1}, v_{2}$ collide horizontally and merge into a single object. What is the velocity of the new asteroid? [figure1] A: $\frac{m_{1} v_{1}+m_{2} v_{2}}{2}$ B: $\frac{m_{1} v_{1}+m_{2} v_{2}}{m_{1}+m_{2}}$ C: $\frac{m_{1} v_{1}-m_{2} v_{2}}{2}$ D: $\frac{m_{1} v_{1}-m_{2} v_{2}}{m_{1}+m_{2}}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Two asteroids with masses $m_{1}, m_{2}$ and velocities $v_{1}, v_{2}$ collide horizontally and merge into a single object. What is the velocity of the new asteroid? [figure1] A: $\frac{m_{1} v_{1}+m_{2} v_{2}}{2}$ B: $\frac{m_{1} v_{1}+m_{2} v_{2}}{m_{1}+m_{2}}$ C: $\frac{m_{1} v_{1}-m_{2} v_{2}}{2}$ D: $\frac{m_{1} v_{1}-m_{2} v_{2}}{m_{1}+m_{2}}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

multi-modal

Astronomy_580

“天问一号”的发射为人类探索火星的奥秘打下了坚实的基础, 已知地球与火星的半 径之比为 $a$, 地球与火星的质量之比为 $b$, 地球表面的重力加速度为 $g$ 。现将一汽缸开口向上放在火星表面的水平面上, 质量均为 $m$ 的活塞 $P 、 Q$ 将一定质量的气体密封且分成甲、乙两部分, 活塞 $P 、 Q$ 导热性能良好。已知火星表面的大气压为 $p_{0}$, 温度为 $T_{0}$,活塞的截面积为 $S$, 当系统平衡时, 甲、乙两部分气体的长度均为 $l_{0}, \frac{m a^{2} g}{b}=p_{0} S$, 外界环境温度保持不变。现在活塞 $P$ 上缓慢地添加质量为 $2 \mathrm{~m}$ 的砝码, 经过一段时间系统再次达到平衡。求: 系统平衡时, 将活塞 $P$ 锁定, 同时将活塞 $Q$ 换成不导热的活塞,缓慢地升高气体乙的温度, 当气柱乙的长度再次为 $l_{0}$ 时, 气体乙的温度应为多高? [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: “天问一号”的发射为人类探索火星的奥秘打下了坚实的基础, 已知地球与火星的半 径之比为 $a$, 地球与火星的质量之比为 $b$, 地球表面的重力加速度为 $g$ 。现将一汽缸开口向上放在火星表面的水平面上, 质量均为 $m$ 的活塞 $P 、 Q$ 将一定质量的气体密封且分成甲、乙两部分, 活塞 $P 、 Q$ 导热性能良好。已知火星表面的大气压为 $p_{0}$, 温度为 $T_{0}$,活塞的截面积为 $S$, 当系统平衡时, 甲、乙两部分气体的长度均为 $l_{0}, \frac{m a^{2} g}{b}=p_{0} S$, 外界环境温度保持不变。现在活塞 $P$ 上缓慢地添加质量为 $2 \mathrm{~m}$ 的砝码, 经过一段时间系统再次达到平衡。求: 系统平衡时, 将活塞 $P$ 锁定, 同时将活塞 $Q$ 换成不导热的活塞,缓慢地升高气体乙的温度, 当气柱乙的长度再次为 $l_{0}$ 时, 气体乙的温度应为多高? [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_888

The MIT students have found out about the plans of the Caltech students of sending a rocket to fly a parachute above the "Great Dome" of MIT. They know the rocket will follow an elliptical orbit of semiaxis a with the center of the Earth at one of its foci. To counter this prank, they will aim another rocket at the original rocket, which upon collision will transfer enough energy to the Caltech rocket to make it reach the escape velocity of the Earth at an early point in its trajectory and make it unable to reach back to the Earth's surface again. What is the energy that needs to be transferred to the Caltech rocket in order for the MIT students to reach their goal? Assume that the Earth has mass M, and the Caltech rocket has mass $\mathrm{m}$. A: $G M m /(3 a)$ B: $G M m /\left(3 a^{\wedge} 2\right)$ C: $G M m /(2 a)$ D: $G M m / a$ E: $G M m / a^{\wedge} 2$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: The MIT students have found out about the plans of the Caltech students of sending a rocket to fly a parachute above the "Great Dome" of MIT. They know the rocket will follow an elliptical orbit of semiaxis a with the center of the Earth at one of its foci. To counter this prank, they will aim another rocket at the original rocket, which upon collision will transfer enough energy to the Caltech rocket to make it reach the escape velocity of the Earth at an early point in its trajectory and make it unable to reach back to the Earth's surface again. What is the energy that needs to be transferred to the Caltech rocket in order for the MIT students to reach their goal? Assume that the Earth has mass M, and the Caltech rocket has mass $\mathrm{m}$. A: $G M m /(3 a)$ B: $G M m /\left(3 a^{\wedge} 2\right)$ C: $G M m /(2 a)$ D: $G M m / a$ E: $G M m / a^{\wedge} 2$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

text-only