Съдържание

IOAA 2014, Теория, предварителен вариант — Задача 8. Gradient temperatures

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.
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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.
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d) Put the draft papers in the designated folder, Put the test papers back in the envelope.
e) Go to swim

GOOD LUCK !

Условие

The spectra of two stars with different temperatures $T_1$ and respectively $T_2$ were compared. In the spectrum of each star, two very close spectral lines corresponding to the wavelength with values $\lambda_1$ and respectively $\lambda_2$ were found. For each line of this spectral lines, the difference between the corresponding visual apparent magnitudes of the stars are $\Delta m_{\lambda_1} = m_{1,\lambda_1} - m_{2,\lambda_1}$ and $\Delta m_{\lambda_2} = m_{1,\lambda_2} - m_{2,\lambda_2}$ . $m_{1,\lambda_1}$ is the apparent magnitude of the star 1 for the wavelength $\lambda_1$ , $m_{1,\lambda_2}$ is the apparent magnitude of the star 1 for the wavelength $\lambda_2$ , $m_{2,\lambda_1}$ is the apparent magnitude of the star 2 for the wavelength $\lambda_1$ , $m_{2,\lambda_2}$ is the apparent magnitude of the star 2 for the wavelength $\lambda_2$ .

Determine the temperature $T_1$ of one of the two stars, if the temperature $T_2$ of the other star is already known, by using the Plank expression of black body radiation:

r\left(\lambda\right) = \frac{2\pi hc^2}{\lambda^5}\left(e^{hc/k\lambda c} - 1\right)^{-1},

where: $h -$ Planck’s constant; $k -$ Boltzmann’s constant; $c -$ speed of light in vacuum. You will consider that $hc \gg k\lambda T$ .

Отговори

Решение

Problem 8. Marking scheme . Gradient temperatures

Finding out the relations (1) and (2) by the correct using of the approximation $hc \gg k\lambda T$ . — 4 points
Correct using of the Pogson’s formula — 3 points
For correct conduct of calculations and obtaining the final formula (5) — 3points

Detailed solution

By using the Plank expression the spectral intensities for the two wavelengths in the doublet emitted by the star number 1 are:

r_1\left(\lambda_1\right) = \frac{2\pi hc^2}{\lambda_1^5\left(e^{hc/\lambda_1kT_1} - 1\right)};\ r_1\left(\lambda_2\right) = \frac{2\pi hc^2}{\lambda_2^5\left(e^{hc/\lambda_2kT_1} - 1\right)};

And by considering $hc \gg k\lambda T;$

r_1\left(\lambda_1\right) = \frac{2\pi hc^2}{\lambda_1^5}e^{-hc/\lambda_1kT_1};\ r_1\left(\lambda_2\right) = \frac{2\pi hc^2}{\lambda_2^5}e^{-hc/\lambda_2kT_1}. \qquad (1)

Respectively, the spectral intensities for the two wavelengths in the doublet emitted by the star number 2 are:

r_2\left(\lambda_1\right) = \frac{2\pi hc^2}{\lambda_1^5\left(e^{hc/\lambda_1kT_2} - 1\right)};\ r_2\left(\lambda_2\right) = \frac{2\pi hc^2}{\lambda_2^5\left(e^{hc/\lambda_2kT_2} - 1\right)};

And by considering $hc \gg k\lambda T;$

r_2\left(\lambda_1\right) = \frac{2\pi hc^2}{\lambda_1^5}e^{-hc/\lambda_1kT_2};\ r_2\left(\lambda_2\right) = \frac{2\pi hc^2}{\lambda_2^5}e^{-hc/\lambda_2kT2}. \qquad (2)

Using the Pogson formula for star 1 and 2 and for $\lambda_1$ , result:

\log\frac{r_1\left(\lambda_1\right)}{r_2\left(\lambda_1\right)} = -0{,}4\left(m_{1,\lambda_1} - m_{2,\lambda_1}\right) = -0{,}4\cdot\Delta m_{\lambda_1};

\log e^{hc/\lambda_1k\left(1/T_2 - 1/T_1\right)} = -0{,}4\cdot\Delta m_{\lambda_1} \qquad (3)

Similar the Pogson formula for $\lambda_2$ ,:

\log\frac{r_1\left(\lambda_2\right)}{r_2\left(\lambda_2\right)} = -0{,}4\left(m_{1,\lambda_2} - m_{2,\lambda_2}\right) = -0{,}4\cdot\Delta m_{\lambda_2}

\log e^{hc/\lambda_2k\left(1/T_2 - 1/T_1\right)} = -0{,}4\cdot\Delta m_{\lambda_2} \qquad (4)

Using the relations (3) and (4)

\log e^{hc/\lambda_2k\left(1/T_2 - 1/T_1\right)} - \log e^{hc/\lambda_1k\left(1/T_2 - 1/T_1\right)} = -0{,}4\cdot\Delta m_{\lambda_2} + 0{,}4\cdot\Delta m_{\lambda_1};

\log e^{hc/\lambda_2k\left(1/T_2 - 1/T_1\right)} + \log e^{-hc/\lambda_1k\left(1/T_2 - 1/T_1\right)} = -0{,}4\cdot\Delta m_{\lambda_2} + 0{,}4\cdot\Delta m_{\lambda_1};

\left[\frac{hc}{k}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\cdot\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right)\right]\cdot\log e = 0{,}4\left(\Delta m_{\lambda_1} - \Delta m_{\lambda_2}\right);

Because $\log e \approx 0{,}43;\ \dfrac{0{,}4}{\log e} \approx 0{,}93;$

\frac{1}{T_2} - \frac{1}{T_1} = -0{,}93\cdot\frac{\left(\Delta m_{\lambda_1} - \Delta m_{\lambda_2}\right)}{\frac{hc}{k}\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right)} = -0{,}93\cdot\frac{k\lambda_1\lambda_2\left(\Delta m_{\lambda_1} - \Delta m_{\lambda_2}\right)}{hc\left(\lambda_1 - \lambda_2\right)};

T_1 = T_2\cdot\frac{hc\left(\lambda_1 - \lambda_2\right)}{hc\left(\lambda_1 - \lambda_2\right) + 0{,}93\cdot k\lambda_1\lambda_2T_2\left(\Delta m_{\lambda_1} - \Delta m_{\lambda_2}\right)}.\ \textbf{(5)}

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