Съдържание

IAO 2008, Теория, Група β — Задача 1. Temperature of star’s core.

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Всички материали от IAO 2008

XIII International Astronomy Olympiad · Theoretical round. Problems to solve · 13 октомври 2008 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.
Data from the table of planetary data may be used for solving every problem.

Note for jury and team leaders.

The proposed sketches are not full; the team leaders have to give more detailed explanations for students. But the correct solutions in the students’ papers (enough for 8 pts) may be shorter.

Элементы орбит. Физические характеристики некоторых планет, Луны и Солнца

Parameters of orbits.
Physical characteristics of some planets, Moon and Sun

Небесное тело, планета / Body, planet	Среднее расстояние от центрального тела / Average distance to central body, астр. ед. / astr. units	Среднее расстояние от центрального тела / Average distance to central body, млн. км / mln. km	Сидерический (или аналогичный) период обращения / Sidereal (or analogous) period, тропич. годах / troph. years	Сидерический (или аналогичный) период обращения / Sidereal (or analogous) period, средних сутках / days	Эксцентриситет / Eccentricity, e	Экваториальн. диаметр / Equat. diameter, km	Масса / Mass, $10^{24}$ kg	Средняя плотность / Av. density, g/cm³	Ускор. своб. пад. у пов. / Grav. acceler. at surf., m/s²	Макс. блеск, вид. с Земли / Max. magn. from Earth **)	Альбедо / Albedo
Солнце / Sun	1,6·10⁹	2,5·10¹¹	2,2·10⁸	8·10¹⁰		1392000	1989000	1,409		−26,8ᵐ
Меркурий / Mercury	0,387	57,9	0,241	87,97	0,206	4 879	0,3302	5,43	3,70		0,06
Венера / Venus	0,723	108,2	0,615	224,70	0,007	12 104	4,8690	5,24	8,87		0,78
Земля / Earth	1,000	149,6	1,000	365,26	0,017	12 756	5,9742	5,515	9,81		0,36
Луна / Moon	0,00257	0,38440	0,0748	27,3217	0,055	3 475	0,0735	3,34	1,62	−12,7ᵐ	0,07
Марс / Mars	1,524	227,9	1,880	686,98	0,093	6 794	0,6419	3,94	3,71		0,15
Юпитер / Jupiter	5,204	778,6	11,862	4 332,59	0,048	142 984	1899,8	1,33	24,86		0,66
Сатурн / Saturn	9,584	1433,7	29,458	10 759,20	0,054	120 536	568,50	0,70	10,41		0,68

**) Для Луны – в среднем противостоянии.
**) For Moon – in mean opposition.

Координаты на эпоху 2000
Epoch 2000 coordinates

	α	δ
Полярная звезда / Polar star	2ʰ 31ᵐ 48ˢ.7	89° 15′ 51″

Условие

We may consider, that the solar core consists of a mix of completely ionized hydrogen and helium, the part of atoms of helium is equal to $\alpha=0.08$ (that is, the numbers of atoms of hydrogen and helium is 92% : 8%). The temperature in the centre of solar core is equal to 15 million degrees and the density is $150\ \mathrm{g/cm^3}$ . Let us assume that there is a carbon star (100% of completely ionized carbon) with the same parameters (the same density and pressure in the center and the same mass). Find the temperature in the center of the core of this carbon star. Atomic numbers and atomic masses of hydrogen, helium and carbon are equal accordingly to: $Z_H=1$ , $A_H=1$ , $Z_{He}=2$ , $A_{He}=4$ , $Z_C=6$ , $A_C=12$ . One may assume the gas in cores as ideal.

Отговори

Решение

1β. Temperature of star’s core. Gas in the cores of stars consists of ions (completely ionized atoms) and electrons. The number of protons of each atom number is equal to number of electrons. As all atoms in the cores are ionized, the number of free electrons corresponds to the number of protons in its nucleus, that is Z. If $n_H$ , $n_{He}$ and $n_C$ are concentrations of ions of hydrogen, helium and carbon, the concentrations of corresponding electrons are equal to $Z_H\cdot n_H$ , $Z_{He}\cdot n_{He}$ and $Z_C\cdot n_C$ .

The total pressure of the gas in stars is equal to the sum of pressure of the gas of ions $p_i$ and the pressure of gas of electrons $p_e$ ,

p=p_i+p_e.

According to formulae for ideal gas

p_i=n_i kT\qquad\text{and}\qquad p_e=n_e kT,

where $n_i$ and $n_e$ are concentrations of ions and electrons respectively. The total pressure is equal to

p=(n_i+n_e)kT.

Then we write equations for the total pressures in the stars.

For the Sun:

p_S=(n_H+Z_Hn_H+n_{He}+Z_{He}n_{He})kT_S=(2n_H+3n_{He})kT_S=(2(1-\alpha)n_S+3\alpha n_S)kT_S=(2+\alpha)n_SkT_S.

For $\alpha=0.08$ :

p_S=2.08n_SkT_S.

For the carbon star:

p_C=(n_C+Z_Cn_C)kT_C=7n_CkT_C.

Since these pressures are equal to each other (condition of the problem)

p_S=p_C,

we may write

7n_CkT_C=2.08n_SkT_S,

T_C=0.297(n_S/n_C)T_S.

Further we shall find a relation between the concentrations of ions in the solar core and the core of the carbon star, $n_S$ and $n_C$ . Practically all mass of matter is in the nuclei of atoms, therefore the densities in the cores of each star are equal to the concentration of ions multiplied by the masses of ions which are proportional to the nuclear masses of the elements. As the density of substance in the solar core and the core of the carbon star are identical,

A_Hn_H+A_{He}n_{He}=A_Cn_C,

A_H(1-\alpha)n_S+A_{He}\alpha n_S=A_Cn_C,

(1\cdot0.9+4\cdot0.1)n_S=12n_C,\qquad1.3n_S=12n_C,\qquad n_S=9.23n_C.

Thus,

T_C=0.297\cdot9.23T_S=2.74T_S.

T_C=2.74\cdot1.5\cdot10^7\,\mathrm K\approx4.1\cdot10^7\,\mathrm K.

(Since we talk about millions of degrees, especially, about an estimations, it is without importance whether the temperature given is in degrees Celsius or Kelvin.)

Answer: the temperature in the core of the carbon star is about 40 million degrees.

Теми и трудност

Оценена трудност: 4/5.

Звезди · Топлинни явления · Строеж на звездите · Идеални газове

Предпоставки: Разширена олимпиадна подготовка

Оригинал в Архива: betatheo.pdf · официални решения: theosol.pdf · допълнителни данни: IAO2008-data.pdf