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Условие Отговори Решение

IOAA 2014, Теория, окончателен вариант — Задача 11. Apparent magnitude of the Moon

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Всички материали от IOAA 2014

8th International Olympiad on Astronomy and Astrophysics — Theoretical Test, 12 short problems (final edition)

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Marking scheme

ALL THE CORRECT SOLUTIONS WHICH ARE DIFFERENT FROM THE AUTHOR’S SOLUTION WILL BE EVALUATED AND MARKED ACORDINGLY TO THE MARKING SCHEME.

Условие

The apparent magnitude of the Moon as seen from the Sun is $M_M=0.25^{\mathrm m}$ .

Calculate the values of the apparent magnitudes of the Moon (as seen from the Earth) corresponding to the following Moon – phases: full-moon and the first quarter. Assume: the Moon – Earth distance - $d_{ME}=385000\ \mathrm{km}$ , the Earth – Sun distance - $d_{ES}=1\ \mathrm{AU}$ , the Moon –Sun distance, $d_{MS}=1\ \mathrm{AU}$ . For terrestrial observers, following phase factor must be used to correct the lunar brightness for curvature of lunar surface and phase of the moon

p(\Psi)=\frac23\left[\left(1-\frac{\Psi}{\pi}\right)\cos\Psi+\frac1\pi\sin\Psi\right],

where $\Psi$ is the phase angle.

Отговори

Решение

Problem 11. Marking scheme Apparent magnitude of the Moon

General analysis of the problem — 6 points
The analysis of the 2 particular situations — 4 points

The apparent magnitude of a planet from the Solar System depends on the phase angle $M = M\left(\Psi\right)$ .

The apparent magnitude of the body is given by the relation:

m = M + 2{,}5\cdot\log\frac{d_{\mathrm{C,S}}^2\cdot d_{\mathrm{C,O}}^2}{d_0^4\cdot p\left(\Psi\right)},

unde: $d_{B,\mathrm{S}} -$ the distance between the body and the Sun; $d_{B,\mathrm{O}} -$ distance between the body and observer; $d_0 = 1\ \mathrm{AU};\ \Psi -$ the phase angle ; $p\left(\Psi\right) -$ the phase function :

p\left(\Psi\right) = \frac{2}{3}\cdot\left[\left(1 - \frac{\Psi}{\pi}\right)\cos\Psi + \frac{1}{\pi}\sin\Psi\right],

$\Psi$ as seen in the figure below is given by the cosine law.

\cos\Psi = \frac{d_{\mathbf{B}\mathrm{O}}^2 + d_{B\mathrm{S}}^2 - d_{\mathrm{OS}}^2}{2d_{B\mathrm{O}}\cdot d_{B\mathrm{S}}}.

In particularly for the Moon

\cos\Psi = \frac{d_{ME}^2 + d_{M\mathrm{S}}^2 - d_{E\mathrm{S}}^2}{2d_{ME}\cdot d_{M\mathrm{S}}};

p\left(\Psi\right) = \frac{2}{3}\cdot\left[\left(1 - \frac{\Psi}{\pi}\right)\cos\Psi + \frac{1}{\pi}\sin\Psi\right];

m_M = M_M + 2{,}5\cdot\log\frac{d_{M,\mathrm{S}}^2\cdot d_{ME}^2}{d_0^4\cdot p\left(\Psi\right)}.

Sun–Moon–Earth geometry and lunar phase angle.

Particular cases:

Full moon

\Psi = 0;

\cos\Psi = 1;\ \sin\Psi = 0;

p\left(\Psi\right) = \frac{2}{3};

d_{M\mathrm{S}} = 1\ \mathrm{AU};\ d_{ME} = 385000\ \mathrm{km} \approx 0{,}00256\ \mathrm{AU} = 256\cdot 10^{-5}\ \mathrm{SU};\ d_0 = 1\ \mathrm{S}U;

m_M = M_M - 12{,}5^{\mathrm{m}} = 0{,}25^{\mathrm{m}} - 12{,}5^{\mathrm{m}} = -12{,}25^{\mathrm{m}}.

First Quarter

\Psi = 90^{\circ};

\cos\Psi = 0;\ \sin\Psi = 1;

p\left(\Psi\right) = \frac{2}{3\pi} \approx 0{,}2;

\frac{d_{M,\mathrm{S}}^2\cdot d_{ME}^2}{d_0^4\cdot p\left(\Psi\right)} = \frac{65536\cdot 10^{-10}}{0{,}2} = 491520\cdot 10^{-10};

m_M = M_M - 10{,}75^{\mathrm{m}} = 0{,}25^{\mathrm{m}} - 10{,}75^{\mathrm{m}} = -10{,}5^{\mathrm{m}}.

Fig. General Sun–body–observer geometry defining the phase angle Psi.

Оригинал в Архива: Short Theoretical.pdf · официални решения: short_final_sol.pdf · допълнителни данни: ioaa1.jpg · допълнителни данни: ioaa2.jpg · допълнителни данни: ioaa3.jpg · допълнителни данни: ioaa4.jpg · допълнителни данни: ioaa5.jpg · допълнителни данни: ioaa6.jpg · допълнителни данни: ioaa7.jpg