Съдържание

IAO 2013, Теория — Задача 2. α-2. Глизе 581 g

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IAO 2013

XVIII Международная астрономическая олимпиада. XVIII International Astronomy Olympiad. Литва, Вильнюс, 6 – 14. IX. 2013, Vilnius, Lithuania. Round Theo · 6 септември 2013 г. · 8 т.

Внимание: Бележка към темата

Ръкописни български условия за групи α, β и общи αβ; решенията и критериите са на английски. В αβ-6 ръкописът дава 3400 km, а английският оригинал и ключът — 3400 pc. Източниковите разминавания са запазени и отбелязани.

Общо указание

Общо указание. Възможно е не във всички задачи въпросите да са поставени коректно. Някои въпроси (възможно е това да е основният въпрос в задачата, а може и да е някой подвъпрос) може да нямат смисъл. В такива случаи в отговора трябва да се напише (на руски или на английски): „ситуация невозможна - impossible situation“. Разбира се, отговорът трябва да е подкрепен с пресмятания или логически разсъждения.

Данните от таблиците (Слънчева система, звезди, константи) могат да се използват във всички задачи.

Отговорите „Да - Yes“ или „Нет - No“ трябва да бъдат написани на руски или на английски.

Общо указание

Данните от таблиците (Слънчева система, звезди, константи) могат да се използват във всички задачи.

Отговорите „Да - Yes“ или „Нет - No“ трябва да бъдат написани на руски или на английски.

Parameters of orbits and physical characteristics

The bilingual source sheet gives the following values. Blank cells reproduce blank source cells.

Body / planet	Average distance (AU)	Average distance ( $10^6$ km)	Sidereal period (tropical years)	Sidereal period (days)	Orbital inclination $i$	Eccentricity $e$	Equatorial diameter (km)	Mass ( $10^{24}$ kg)	Average density (g/cm³)	Gravity at surface (m/s²)	Axial tilt	Maximum magnitude from Earth	Albedo
Sun	$1.6\cdot10^9$	$2.5\cdot10^{11}$	$2.2\cdot10^8$	$8\cdot10^{10}$			1392000	1989000	1.409			−26.74
Mercury	0.387	57.9	0.241	87.969	7.00°	0.206	4879	0.3302	5.43	3.70	0.01°		0.06
Venus	0.723	108.2	0.615	224.7007	3.40	0.007	12104	4.8690	5.24	8.87	177.36		0.78
Earth	1.000	149.6	1.000	365.2564	0.00	0.017	12756	5.9742	5.515	9.81	23.44		0.36
Moon	0.00257	0.38440	0.0748	27.3217	5.15	0.055	3475	0.0735	3.34	1.62	6.7	−12.7	0.07
Mars	1.524	227.9	1.880	686.98	1.85	0.093	6794	0.6419	3.94	3.71	25.19	−2.0	0.15
Jupiter	5.204	778.6	11.862	4332.59	1.30	0.048	142984	1899.8	1.33	24.86	3.13	−2.7	0.66
Saturn	9.584	1433.7	29.458	10759.20	2.48	0.054	120536	568.50	0.70	10.41	26.73	0.7	0.68

For outer planets and Moon — in mean opposition.

Some constants and formulae

Constant or formula	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 parameter, $H_0$ (km/s/Mpc), mean value	71
Hubble parameter, diapason of values	50–100
Planck 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)	$1.381\cdot10^{-23}$
Stefan–Boltzmann constant, $\sigma$ (W/m²/K⁴)	$5.670\cdot10^{-8}$
Wien’s displacement constant, b (m·K)	0.002897
Laboratory wavelength of Hα (Å)	6562.81
Tropical year length, T (days)	365.242199
Standard atmosphere (Pa)	101325
Visible light extinction by the terrestrial atmosphere in zenith (minimum)	19%, 0.23 mag
Refractive index of water for 20°C, n	1.334
Moment of inertia of a solid ball	$I=\frac25MR^2$
Volume of ball	$V=\frac43\pi R^3$
Area of sphere	$S=4\pi R^2$
$\pi$	3.14159265
e	2.71828183
Golden ratio, $\varphi$	1.61803399

Data of some stars

Star	Designation	RA	DEC	Parallax $p$	Apparent magnitude $m$	Spectral class	Mass ( $\mathcal M_{\odot}$ )
Sun	⊙	0h–24h	−23°26′ to +23°26′	8.794″	vis −26.74; bol −26.82	G2	1
Proxima Centauri	V645 Cen, α Cen C	14h 29m 43s	−62°40′46″	0.769″	11.05	M5.5	0.123
Alpha Centauri A	α Cen A	14h 39m 37s	−60°50′02″	0.747″	−0.01	G2	1.1
Alpha Centauri B	α Cen B	14h 39m 35s	−60°50′14″		1.34	K1	0.9
Beta Centauri	β Cen	14h 03m 49s	−60°22′23″	0.009″	0.61	B1	21
Epsilon Eridani	ε Eri	03h 32m 56s	−09°27′30″	0.311″	3.74	K2	0.82
Gliese 581	HO Lib	15h 19m 27s	−07°43′20″	0.16″	vis 10.57; bol 8.0	M3V	0.31

The Alpha Centauri parallax is printed once beside the pair; the second cell is left blank here as in the source.

Historical units

Units of power

1 horse-power (hp) = 735.49875 W.

Local units of length in the end of XVIII century

1 arshin (arsh) = 0.711187 m
1 span (sp) = 1/4 arshin
1 vershok (vrsh) = 1/4 span
1 sajene (sj) = 3 arshin
1 verst (vrst) = 500 sajene

Условие

Това небесно тяло в системата на звездата Глизе 581 е най-подобната на Земята планета, открита извън Слънчевата система, и екзопланетата, призната за най-пригодна за развитие на белтъчен живот.

Оценете орбиталния период на планетата Глизе 581g. Орбитата ѝ считайте за кръгова.

Отговори

Решение

Официалното решение и критериите по-долу са на английски, както в източника.

Official English solution sketch. 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.

Orbital period

It is obvious that there should be about the same conditions as on the Earth for the existence of albuminous based life on the planet.

(Following in the solution indexes G refer to the system Gliese 581, and indexes $\odot$ and E — to the Sun and Earth).

In the “data of some stars” table one can find the mass of the star Gliese 581, which is equal to $\mathcal M_G=0.31\mathcal M_{\odot}$ , its parallax $p=0.16^{\prime\prime}$ , and the apparent bolometric magnitude $m=8^{\mathrm m}.0$ .

Knowing the parallax and the apparent bolometric magnitude of the star, one can find its absolute bolometric magnitude:

M=m+5^{\mathrm m}+5^{\mathrm m}\lg p=8^{\mathrm m}.0+5^{\mathrm m}+(-3^{\mathrm m}.98)\approx9^{\mathrm m}.0.

It is $9^{\mathrm m}-4^{\mathrm m}.8=4^{\mathrm m}.2$ greater than the absolute magnitude of the Sun. Thus, the radiation from Gliese 581 is $I_{\odot}/I_G=10^{4.2/2.5}\approx48$ times less than that of the Sun.

Taking into account the Stefan–Boltzmann law, let us compare the temperature regimes on Earth ( $T_E$ ) and on the planet Gliese 581 g ( $T_G$ ).

It is known that the balance temperature T on the planet is defined by the balance of incoming and radiated energy. The radiated energy depends only on the surface area and the temperature (the dependence on the temperature is described by the Stefan–Boltzmann $E\sim T^4$ , but is not necessary to know the exact relationship for the task). In absence of internal sources of energy (we make this assumption) the radiated energy is equal to the energy coming from the central luminary. Both the radiated and incoming energies are proportional to the surface area of the planet, hence the balance does not depend on the size of the planet. The incoming energy is proportional to the radiation of the central star and inversely proportional to the distance to the star:

E_E\sim I_{\odot}/R_E^2,\qquad E_G\sim I_G/R_G^2.

Thus, for the temperatures on Earth and Gliese 581 g to be equal, it is necessary that

E_E=E_G\quad\text{or}\quad R_G/R_E=(I_G/I_{\odot})^{1/2}.

According to the generalized Kepler’s third law

\tau^2\mathcal M/a^3=\mathrm{const},\quad\text{that is}\quad\tau\sim a^{3/2}/\mathcal M^{1/2},

or as the orbits are circular,

\tau\sim R^{3/2}/\mathcal M^{1/2},\qquad\tau_G/\tau_E=(R_G/R_E)^{3/2}/(\mathcal M_G/\mathcal M_{\odot})^{1/2},

\tau_G=\tau_E(R_G/R_E)^{3/2}/(\mathcal M_G/\mathcal M_{\odot})^{1/2}=\tau_E(I_G/I_{\odot})^{3/4}/(\mathcal M_G/\mathcal M_{\odot})^{1/2},

\tau_G=365\ \mathrm{days}\times48^{-3/4}/0.31^{1/2},\qquad\tau_G\approx36\ \mathrm{days}.

Official basic criteria

Conditions as on the Earth, the same temperature — 1 pt.
Correct taking the necessary data from the tables — 0.5 pt.
Finding (any way) that the radiation from Gliese 581 is about 48 times less than that of the Sun — 2.5 pt.
Dependence on R — 1 pt.
Correct using the generalized Kepler’s third law, calculations — 2 pt.
Final calculations and correct result — 1 pt.

Official solution sketches: 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.

Basic criteria: note for jury

Jury members should evaluate the students’ solutions in essence, and not by looking on formal existence the mentioned sentences or formulae. The formal presence of the mentioned positions in the text is not necessary to give the respective points. Points should be done if the following steps de facto using these positions.

Оригинал в Архива: bgtheo.pdf · официални решения: theosol.pdf · допълнителни данни: theodata.pdf · допълнителни данни: theo_cr_1.pdf · допълнителни данни: theo_cr_2.pdf · допълнителни данни: theo_cr_3.pdf · допълнителни данни: constants.pdf · допълнителни данни: hrd.pdf · допълнителни данни: alphatheo.pdf · допълнителни данни: betatheo.pdf · допълнителни данни: theo_cr_1.png · допълнителни данни: theo_cr_2.png · допълнителни данни: theo_cr_3.png