Съдържание

IAO 2010, Теория, Група β — Задача 4. White dwarf.

Оригинал на условието (изображение)Съобщи за грешка

Всички материали от IAO 2010

Theoretical round. Problems to solve · 16 октомври 2010 г. · 8 т.

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): «impossible situation – 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.

The answers «Да-Yes» or «Нет-No» has to be written in English or Russian.

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.

Some constants and formulae

Constant	Value
Speed of light in vacuum, $c$ (m/s)	299792458
Constant of gravitation, $G$ (N·m²/kg²)	$6.674\cdot10^{-11}$
Solar constant, $A$ (W/m²)	1367
Hubble constant, $H_0$ (km/s/Mpc), mean value	70
Hubble constant, diapason of values	50–100
Plank constant, $h$ (J·s)	$6.626\cdot10^{-34}$
Charge of electron, $e$ (C)	$1.602\cdot10^{-19}$
Mass of electron, $m_e$ (kg)	$9.109\cdot10^{-31}$
Proton-to-electron ratio	1836.15
Faraday constant, $F$ (C/mol)	96485
Magnetic constant, $\mu_0$ (H/m)	$1.257\cdot10^{-6}$
Universal gas constant, $R$ (J/mol/K)	8.314
Boltzmann constant, $k$ (J/K)	$1.381\cdot10^{-23}$
Standard atmosphere (Pa)	101325
Stefan-Boltzmann constant, $\sigma$ (W/m²/K⁴)	$5.670\cdot10^{-8}$
Wien’s displacement constant, $b$ (m·K)	0.002897
Laboratory wavelength of H $\alpha$ (Å)	6563
Refractive index of water for 20°C, $n$	1.334
Area of sphere	$S=4\pi R^2$
$\pi$	3.14159265

Data of some stars

Star	Designation	RA	DEC	$p$	m	SC
Arcturus	$\alpha$ Boo	14h 15m 40s	19° 10′ 57″	0″.089	−0.05	K1
Vega	$\alpha$ Lyr	18h 36m 56s	38° 47′ 01″	0″.129	0.03	A0
Deneb	$\alpha$ Cyg	20h 41m 26s	45° 16′ 49″	0″.001	1.25	A2
Polaris	$\alpha$ UMi	02h 31m 51s	89° 15′ 51″	0″.007	2.02	F7
Sirius	$\alpha$ CMa	06h 45m 09s	−16° 42′ 58″	0″.379	−1.46	A1

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

Body, planet	Average distance to central body (astr. units)	Average distance (mln. km)	Sidereal (or analogous) period (troph. years)	Period (days)	Eccentricity $e$	Equat. diameter (km)	Mass ( $10^{24}$ kg)	Av. density (g/cm³)	Grav. acceler. at surf. (m/s²)	Max. magn. from Earth **)	Albedo
Sun	$1.6\cdot10^9$	$2.5\cdot10^{11}$	$2.2\cdot10^8$	$8\cdot10^{10}$		1392000	1989000	1.409		$-26.8^m$
Mercury	0.387	57.9	0.241	87.97	0.206	4879	0.3302	5.43	3.70		0.06
Venus	0.723	108.2	0.615	224.70	0.007	12104	4.8690	5.24	8.87		0.78
Earth	1.000	149.6	1.000	365.26	0.017	12756	5.9742	5.515	9.81		0.36
Moon	0.00257	0.38440	0.0748	27.3217	0.055	3475	0.0735	3.34	1.62	$-12.7^m$	0.07
Mars	1.524	227.9	1.880	686.98	0.093	6794	0.6419	3.94	3.71	$-2.0^m$	0.15
Jupiter	5.204	778.6	11.862	4332.59	0.048	142984	1899.8	1.33	24.86	$-2.7^m$	0.66
Saturn	9.584	1433.7	29.458	10759.20	0.054	120536	568.50	0.70	10.41	$0.7^m$	0.68
Eris	68.01			204862	0.434	2600	0.0167		0.8		0.86

**) For Moon – in mean opposition.

Source discrepancies

Printed data and formulae are retained. The official key uses several ungrouped slash expressions. The Venus key initially prints divisor 1.406 before using 19.6 later. The ISS key prints 6371 + 350 = 6720 km.

Условие

The absolute stellar magnitude of a white dwarf is $14^m$ . Humanoids create a planet similar to the Earth by all parameters (including atmosphere and climate) rotating around this white dwarf. Calculate (estimate) the minimal possible sidereal period of the planet.

Решение

The problem is devoted the 100th anniversary of Subrahmanyan Chandrasekhar (Субраманьян Чандрасекар), which will be tomorrow, on October 19, 2010.

For the planet (P) similar to the Earth on all parameters, it is necessary for it to receive from the white dwarf (WD) as much energy as the Earth (E) receives from the Sun (S):

E_P=E_E.

The energy received by a planet from a star is proportional to luminosity (power of radiation) of the star and inversely proportional to square of average distance to a star:

E\sim J/R^2.

For the Earth:

E_E\sim J_S/R_E^2.

For a planet rotating around the white dwarf:

E_P\sim J_{WD}/R_P^2.

Thus,

J_S/R_E^2=J_{WD}/R_P^2,

whence

R_P=R_E\times(J_{WD}/J_S)^{1/2}.

The ratio of luminosities can be found from the difference of absolute stellar magnitudes,

J_{WD}/J_S=10^{-0.4(M_{wd}-M_s)}.

Thus,

R_P=R_E\times(10^{-0.4(M_{wd}-M_s)})^{1/2}=R_E\times10^{-0.2(M_{wd}-M_s)}.

The absolute stellar magnitude of the Sun $M_S$ should be known by a participant of the Olympiad, or it can be calculated that the value of $m_S=-26.8^m$ of the visible stellar magnitude, from the table and formula for distance of 10 pc written in a.u. (10 pc = 2 062 648 a.u.):

M_S=m_S+5^m\cdot\lg2062648=-26.8^m+31.57^m=4.77^m.

According to general III Kepler law,

T^2M/R^3=\mathrm{const}\qquad\text{therefore}\qquad T_P=T_E(R_P/R_E)^{3/2}(M_{WD}/M_S)^{-1/2}.

We know the values $T_E$ and $R_P/R_E$ , then what may be the value of $M_S/M_{WD}$ ? Let us remember the Chandrasekhar (*) limit, that is equal to $1.4M_S$ – the upper limit for mass of a star that may exist as a white dwarf.

(*) Subrahmanyan Chandrasekhar (Субраманьян Чандрасекар) – American scientist of the Indian origin whose 100 anniversary will be tomorrow, on October 19, 2010.

Thus, the minimal sidereal period is equal

T_P=T_E(10^{-0.2(M_{wd}-M_s)})^{3/2}(1.4)^{-1/2}=T_E\times10^{-0.3(M_{wd}-M_s)}\times(1.4)^{-1/2}\approx0.53\;\text{days}.

Basic criteria. For work of Jury

The problem is devoted to the 100th anniversary of Subrahmanyan Chandrasekhar, which will be tomorrow, on October 19, 2010.

Energy received by the planet equals to the one received by the Earth – 1 pt.
Algebra steps and correct finding of coefficient $*10^{-0.3(M_{wd}-M_s)}$ – 3 pt in total.
Since the problem is devoted to Chandrasekhar, using Chandrasekhar limit – 3 pt.
Correct calculation of numerical answer – 1 pt.

Оригинал в Архива: betatheo.jpg · допълнителни данни: theosol1.jpg · допълнителни данни: theosol2.jpg · допълнителни данни: theosol3.jpg · допълнителни данни: theosol4.jpg · допълнителни данни: theosol5.jpg · допълнителни данни: theocriteria1.jpg · допълнителни данни: theocriteria2.jpg · допълнителни данни: constants.jpg · допълнителни данни: orbital-elements.jpg · допълнителни данни: theodata.jpg