Съдържание

IAO 2010, Теория, Група α — Задача 5. Cosmonaut.

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

Imagine a small space station with a total mass of $M=50$ ton placed far away from any celestial body. A cosmonaut with a total mass (with ammunitions) of $m=100$ kg appears $L=80$ m away from the station without any fuel. Calculate (estimate) the time $\tau$ after which the cosmonaut reaches the station due to gravitational attraction. Consider the initial velocity of the cosmonaut to be zero.

Решение

For estimation we may consider the space station and cosmonaut as pointlike particles. The time $\tau_{1/2}$ after which these particles collide due to gravitational attraction is approximately equal to half of period of revolution of mass $m$ around mass $M$ on the orbit with the major semiaxis $L/2$ . And the time $\tau$ that we should find approximately equals $\tau_{1/2}$ (neglecting the last part of orbit comparable with the sizes of station and cosmonaut). According to general III Kepler law,

4\pi^2/GM=T^2/(L/2)^3,\qquad\text{so}\qquad T=2\pi(L/2)^{3/2}(GM)^{-1/2},\qquad\text{and}\qquad\tau_{1/2}=T/2=\pi(L^3/8GM)^{1/2}

\tau\approx\tau_{1/2}=4.35\cdot10^5\;\mathrm{s}\approx7250\;\mathrm{min}\approx121\;\mathrm{h}\approx5\;\text{days}.

Basic criteria. For work of Jury

General note:

Only solutions using approximation of the limit for half of period of rotation on the elliptic orbits or using correct integral calculations $\tau$ may be evaluated in frames of “max 8 pt” criteria.

Solutions using formulae like $gt^2/2$ with using some average $g$ – not more than 2 pt in total.

Solutions using formulae like $gt^2/2$ with taking $g$ as at the initial or final point – not more than 1 pt in total.

Max 8 pt criteria for solutions using approach with period of rotation and III Kepler law:

General understanding of the physical process (using approach with period of rotation) – 1 pt.
$\tau\approx\tau_{1/2}$ – 1 pt.
Algebra steps and correct finding $\tau_{1/2}=T/2=\pi(L^3/8GM)^{1/2}$ – 4 pt in total.
Correct calculation of numerical answer – 2 pt (1 pt for more or less correct values of answer: in a range of “a few days”).

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