Съдържание

IOAA 2014, Теория, предварителен вариант — Задача 14. Apparent magnitude of the Moon

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IOAA 2014

8th International Olympiad on Astronomy and Astrophysics, Suceava - Gura Humorului - August 2014 — THEORETICAL TEST, 15 short problems

Indications

The problems were elaborated concerning two aspects: a) To cover merely all the subjects from the syllabus;
b) The average time for solving the items is about 15 minutes per a short problem;
In your folder you will find out the following: a) Answer sheets
b) Draft sheets
c) The envelope with the subjects in English and the translated version of them in your mother tongue;
The solutions of the problems will be written down only on the answer sheets you receive on your desk. PLEASE WRITE ONLY ON THE PRINTED SIDE OF THE PAPER SHEET. DON’T USE THE REVERSE SIDE. The evaluator will not take into account what is written on the reverse of the answer sheet.
The draft sheets is for your own use to try calculation, write some numbers etc. BEWARE: These papers are not taken into account in evaluation, at the end of the test they will be collected separately . Everything you consider as part of the solutions have to be written on the answer sheets.
Each problem have to be started on a new distinct answer sheet.
On each answer sheet please fill in the designated boxes as follows: a) In PROBLEM NO. box write down only the number of the problem: i.e. 1 – 15 for short problems, 16 – 19 for long problems. Each sheet containing the solutions of a certain problem, should have in the box the number of the problem;
b) In Student ID – fill in your ID you will find on your envelope, consisted of 3 letters and 2 digits.
c) In page no. box you will fill in the number of page, starting from 1. We advise you to fill this boxes after you finish the test
We don’t understand your language, but the mathematic language is universal, so use as more relationships as you think that your solution will be better understand by the evaluator. If you want to explain in words we kindly ask you to use short English propositions.
Use the pen you find out on the desk. It is advisable to use a pencil for the sketches.
At the end of the test: a) Don’t forget to put in order your papers;
b) Put the answer sheets in the folder 1. Please verify that all the pages contain your ID, correct numbering of the problems and all pages are in the right order and numbered. This is an advantage of ease of understanding your solutions.
c) Verify with the assistant the correct number of answer sheets used fill in this number on the cover of the folder and sign.
d) Put the draft papers in the designated folder, Put the test papers back in the envelope.
e) Go to swim

GOOD LUCK !

Условие

You know that the absolute magnitude of the Moon is $M_L = 0{,}25^{\mathrm{m}}$ . Calculate the values of the apparent magnitudes of the Moon corresponding to the following Moon –phases : full-moon and the first quarter. You know: the Moon – Earth distance - $d_{LP} = 385000\ \mathrm{km}$ , the Earth – Sun distance - $d_{PS} = 1\ \mathrm{AU}$ , the Moon –Sun distance, $d_{LS} = 1\ \mathrm{AU}$

Отговори

Решение

Problem 14. 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 bellow is given by the cosine law.

Triangle Sun – body – observer: the Sun at upper left, “The body fro Solar System” as a black dot at right, and an observer standing on a grey planet below; dashed sides d_B,S (Sun–body), d_B,O (body–observer) and d_O,S (observer–Sun); the phase angle Ψ at the body. — Fig.

\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

Sun, Moon and Earth: the Sun at left, the Moon at right as a circle with centre L, points A (top), B (bottom), C, D and B′, radius R, its right half shaded grey; the angle ψ at L between the directions to the Sun and to the Earth, and again between LB and LD; dashed lines d_MS (Sun–Moon), d_ES (Sun–Earth) and d_ME (Moon–Earth); an observer standing on the Earth below.

\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)}.

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}}.

Оригинал в Архива: th_short_ver1.pdf · официални решения: th_short_ver1_sol.pdf