Съдържание

IAO 2006, Теория, Група α — Задача 5. Lunar satellite.

Оригинал на условието (изображение)Съобщи за грешка

Всички материали от IAO 2006

Theoretical round. Problems to solve · 10 ноември 2006 г.

Бележка

Maybe not all problems have correct questions. Some questions (maybe the main question of the problem, maybe one of the subquestions) may have no real sense. In this case you have to write in your answer (in English or Russian): «situation is impossible – ситуация невозможна». Of course, this answer has to be explained numerically or logically.

Source annotations

Printed source inconsistencies are retained: the English telescope key repeats Δx_m on the right of its displacement identity; the galaxy key uses both1.89×10⁻³ and1.86×10⁻³rad; the alpha eccentricity ratio has a sign inconsistency before the later positive0.07 result. The English polar key prints71°06′N/72°48′S; the Russian copy has handwritten70°06′N/71°48′S. No scientific correction has been silently substituted.

Условие

These are two photos of the Moon taken by the same camera mounted on a lunar satellite. The first photo has been made while the satellite was near its periselenion and the second one near the aposelenion. Find from these data the value of the satellite’s orbit eccentricity. Estimate the minimum period between the moments at which these two photos could be taken.

Table of planetary data

Data from the “Table of planetary data” may be used for the solving of every problem.

Решение

The hemisphere of the Moon at photos is about the same as we see it from the Earth.

Taking into account that the Moon is faced to us by one side all the times, it is possible to draw a conclusion that photographing was made from the points of space located approximately between the Earth and the Moon.

The Moon’s image diameter ( $d$ ) at every photo is proportional to its angular size visible from the satellite: $d\sim\alpha=D/L$ , where $D$ – real diameter of Moon and $L$ – the distance to it. $D$ is constant and $L$ is depend on position of the satellite:

$L=L_0(1+e)$ for aposelenion;

$L=L_0(1-e)$ for periselenion.

So if we have diameters of Moon’s image at periselenion ( $d_P$ ) and aposelenion ( $d_A$ ) of the satellite’s orbit we may write:

d_P/d_A=\alpha_P/\alpha_A=[D/(1-e)]/[D/(1+e)]=(1+e)/(1-e).

And: $e=(1-d_P/d_A)/(1+d_P/d_A)$ .

The presented two pictures have been taken for moments near periselenion and aposelenion (not just at periselenion and aposelenion), so $d_1\approx d_P$ and $d_2\approx d_A$ ; instead of “=” in the previous formula we should write “ $\approx$ ”:

e\approx(d_1-d_2)/(d_1+d_2).

By measuring the images one may found $d_1\approx46.5$ mm, $d_2\approx40.5$ mm.

e\approx(d_1-d_2)/(d_1+d_2)\approx0.07.

Taking into account the satellite’s orbit is more or less stable in the space, it will take about half a year to change the situation “full Moon at periselenion” to “full Moon at aposelenion” (the Sun to be at the opposite direction from the Moon).

Note for Jury members

Note for Jury members, problems4 and5. Due to difference in printers the scales of pictures were different in different copies of pictures. So the values of the lengths may be different from those written here. But the ratios, which are necessary for solutions, of the lengths should be the same.

Оригинал в Архива: theo1.jpg · официални решения: theosol1.jpg · допълнителни данни: theo2.jpg · допълнителни данни: theodata1.pdf · допълнителни данни: theodata2.jpg · допълнителни данни: constants.jpg · допълнителни данни: theosol2.jpg · допълнителни данни: theosol3.jpg · допълнителни данни: theosol4.jpg