Съдържание

IAO 2010, Теория, Група α — Задача 4. Climate.

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 Mercury parallax key prints approximately 40.9 despite its preceding numerical ratio. The climate key uses 272.8 and 294.4 K for the statement’s −0.4 and +21.2°C.

Условие

There is a reference to climate in Crimea according to observations during 934 months between 1821 and 1991 on the web-site of the XV IAO. The average temperatures in Simferopol in January and July are equal to $-0.4^\circ\mathrm C$ and $+21.2^\circ\mathrm C$ respectively according to that data. Imagine a planet whose axis of rotation is perpendicular to the planets’ ecliptic. Estimate, what the eccentricity of the planets’ orbit should be in order to give climatic seasons identical to the ones in Crimea.

Решение

Solar (or stellar) energy, reaching to a planet in finally reemitted by the planet according to Stefan-Boltzmann law. We should notice before the main part of solution that the answer strongly depends on a relation of heat capacity of a planet and the period of its sidereal period. If the thermal capacity is large, and the period is small, then obviously, temperature on a planet varies a little even with large eccentricity of orbit. On the other hand, if the heat capacity is small, and the period is large, it is possible to consider that the planet is in thermal balance in each point of its orbit.

Let us consider the second case. $L$ – luminosity of the Sun, $r$ – the radius of the planet, $T_W$ – winter temperature, $T_S$ – summer temperature. For the planet in aphelion (distance to the Sun – $R_A$ ) the energy balance is,

(L/4\pi R_A^2)\times\pi r^2=\sigma T_W^4\times4\pi r^2,

And for the planet in perihelion (distance to the Sun – $R_P$ ),

(L/4\pi R_P^2)\times\pi r^2=\sigma T_S^4\times4\pi r^2.

From these two formulae,

R_A\times T_W^2=R_P\times T_S^2,

R_A/R_P=(T_S/T_W)^2.

e=(R_A-R_P)/(R_A+R_P)=(T_S^2-T_W^2)/(T_S^2+T_W^2).

Numerical calculation (the temperatures expressed in Kelvin) gives us,

e=(294.4^2-272.8^2)/(294.4^2+272.8^2)=0.076.

That is, such eccentricity of the orbit of the planet with small heat capacity will result for the temperature changes winter-summer similar to Crimean ones. However, for planets of type of the Earth temperature fluctuations will smooth out. That is, for the given conditions for the Earth like planet the eccentricity should be much larger. Thus, the answer

e=(T_S^2-T_W^2)/(T_S^2+T_W^2)=0.076

is the minimum estimation.

Basic criteria. For work of Jury

Understanding that incoming energy = energy radiated by the planet according to Stefan-Boltzmann law (may be not written but evident from used formulae) – 2 pt.
Algebra steps and correct final formula $e=(T_S^2-T_W^2)/(T_S^2+T_W^2)$ – 3 pt in total.
Correct numerical result – 2 pt (1 pt for more or less correct values of answer: in a range 0.05–0.1.)
Speculations about relation of heat capacity of a planet and the period of its sidereal period, conclusion that the result is the minimum estimation – 1 pt.

Оригинал в Архива: alphatheo.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