Съдържание

Условие Отговори Решение

IOAA 2014, Теория, окончателен вариант — Задача 12. Absolute magnitude of a cepheide

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IOAA 2014

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

Указания

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a. Answer sheets

b. Rough work sheets

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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 cepheids are variable stars, whose luminosities vary due to stellar pulsations. The period of the oscillations of a cepheid star is:

P=2\pi R\sqrt{\frac{R}{GM}},

where: R – the mean radius of the cepheid; M – the mass of the cepheid (remains constant during oscillation), you can assume that the temperature is constant during the pulsation;

Express the mean absolute magnitude of the cepheid $M_{\mathrm{cep}}$ , in the following form:

M_{\mathrm{cep}}=-2.5^{\mathrm m}\cdot\log k-\left(\frac{10}{3}\right)^{\mathrm m}\cdot\log P,

where P is the period of cepheid’s pulsation.

Отговори

Решение

Problem 12. Marking scheme Apparent magnitude of the Moon

P = 2\pi R\sqrt{\frac{R}{GM}},

rezults

P^2 = \frac{4\pi^2R^3}{KM};\ R = \sqrt[3]{\frac{KMP^2}{4\pi^2}} = \left(\frac{KM}{4\pi^2}\right)^{1/3}\cdot P^{2/3};

R^2 = \left(\frac{KM}{4\pi^2}\right)^{2/3}\cdot P^{4/3}.

The absolute brightness is:

L_{\mathrm{cef}} = \sigma T_{\mathrm{cef}}^4\cdot 4\pi R^2,

1.5p

And the apparent brightness :

E_{\mathrm{cef}} = \frac{L_{\mathrm{cef}}}{4\pi d_{\mathrm{P,cef}}^2} = \frac{\sigma T_{\mathrm{cef}}^4\cdot 4\pi R^2}{4\pi d_{\mathrm{P,cef}}^2},

1.5p

$d_{\mathrm{P,cef}}$ is the distance between the observer on Erath and the cepheide

E_{\mathrm{cef}} = \frac{\sigma T_{\mathrm{cef}}^4\cdot 4\pi\cdot\left(\frac{KM}{4\pi^2}\right)^{2/3}\cdot P^{4/3}}{4\pi d_{\mathrm{P,cef}}^2}.

Similarly for Sun

E_{\mathrm{S}} = \frac{L_{\mathrm{S}}}{4\pi d_{\mathrm{PS}}^2} = \frac{\sigma T_{\mathrm{S}}^4\cdot 4\pi R_{\mathrm{S}}^2}{4\pi d_{\mathrm{PS}}^2}.

By using the Pogson formula:

\log\frac{E_{\mathrm{cef}}}{E_{\mathrm{S}}} = -0{,}4\left(m_{\mathrm{cef}} - m_{\mathrm{S}}\right);

M_{\mathrm{cef}} = M_{\mathrm{S}} - 5^{\mathrm{m}}\log\frac{\left|d_{\mathrm{P,cef}}\right|}{\left|d_{\mathrm{PS}}\right|} - 2{,}5\cdot\log\frac{E_{\mathrm{cef}}}{E_{\mathrm{S}}};

\frac{T_{\mathrm{cef}}^4\cdot\left(\frac{KM}{4\pi^2}\right)^{3/2}\cdot d_{\mathrm{PS}}^2}{T_{\mathrm{S}}^4\cdot R_{\mathrm{S}}^2\cdot d_{\mathrm{P,cef}}^2} = k_1 = \mathrm{constant};

M_{\mathrm{cef}} = M_{\mathrm{S}} - 5^{\mathrm{m}}\log\frac{\left|d_{\mathrm{P,cef}}\right|}{\left|d_{\mathrm{PS}}\right|} - 2{,}5\cdot\log k_1 - \frac{10}{3}\cdot\log P;

M_{\mathrm{S}} - 5^{\mathrm{m}}\log\frac{\left|d_{\mathrm{P,cef}}\right|}{\left|d_{\mathrm{PS}}\right|} - 2{,}5\cdot\log k_1 = -2{,}5\cdot\log k;

k = \mathrm{constant};

M_{\mathrm{cef}} = -2{,}5\cdot\log k - \frac{10}{3}\cdot\log P.

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