Съдържание

IAO 2009, Теория, Група α — Задача 4. Catastrophe.

Оригинал на условието (изображение)Съобщи за грешка

Всички материали от IAO 2009

Theoretical round. Problems to solve · 8 ноември 2009 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): “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.

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

Body, planet	Average distance (astr. units)	Average distance (mln. km)	Sidereal period (troph. years)	Period (days)	Eccentricity	Equat. diameter (km)	Mass ( $10^{24}$ kg)	Av. density (g/cm³)	Grav. acceler. (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		$-26.8^{\mathrm m}$
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	$-12.7^{\mathrm m}$	0.07
Mars	1.524	227.9	1.880	686.98	0.093	6794	0.6419	3.94	3.71	$-2.0^{\mathrm m}$	0.15
Jupiter	5.204	778.6	11.862	4332.59	0.048	142984	1899.8	1.33	24.86	$-2.7^{\mathrm m}$	0.66
Saturn	9.584	1433.7	29.458	10759.20	0.054	120536	568.50	0.70	10.41	$0.7^{\mathrm m}$	0.68
Eris	68.01			204862	0.434	2600	0.0167		0.8		0.86

**) For 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 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 for20°C, $n$	1.334
Area of sphere	$S=4\pi R^2$
$\pi$	3.14159265

Data of some stars

Star	Designation	RA	DEC	m	SC
Arcturus	$\alpha$ Boo	$14^{\mathrm h}15^{\mathrm m}40^{\mathrm s}$	$19^\circ10'57''$	-0.05	K1
Vega	$\alpha$ Lyr	$18^{\mathrm h}36^{\mathrm m}56^{\mathrm s}$	$38^\circ47'01''$	0.03	A0
Deneb	$\alpha$ Cyg	$20^{\mathrm h}41^{\mathrm m}26^{\mathrm s}$	$45^\circ16'49''$	1.25	A2
Polaris	$\alpha$ UMi	$02^{\mathrm h}31^{\mathrm m}51^{\mathrm s}$	$89^\circ15'51''$	2.02	F7
Sirius	$\alpha$ CMa	$06^{\mathrm h}45^{\mathrm m}09^{\mathrm s}$	$-16^\circ42'58''$	-1.46	A1

Бележки от източника

The original two-up official key photographs clip some right-hand line endings. Ellipses mark those unreadable endings; visible formulae and printed numerical values are retained. The reference table gives the Eris period as 204862 days, whereas the key uses 204852 days. Original inline slash notation is retained in the key. The photographed alpha masthead is cropped, but the original filename and distinct alpha tasks in the official key establish its group.

Условие

Imagine, that on July 5, 2084 the mass of the Sun suddenly has decreased to half its original value. Calculate a new period of revolution of the Earth around of the new Sun.

Решение

At first we should realise that “suddenly decreasing of solar mass to half its original value” is a hypothetic process and many physical laws of conservation cannot be used to compare parameters before and after since the system is not closed. Since no any other changing done, we should assume that at the moment of the mass decreasing other parameters of the Sun and the Earth have not changed: Positions of the Sun and the Earth, velocity of the Earth. Taking into account this postulate, a first approximation, if mass of a central body is reduced to half its original value, the circular speed becomes parabolic one. That is the Earth will move on a parabolic orbit and will never return. The answer to the problem is meaningless (the period is equal to infinity).

Nevertheless, it is known, that on July 5 the Earth is near the aphelion of its orbit. Its speed in this case is less than circular. Thus, in this case if mass of a central body to reduce twice, the speed of the Earth already is less than parabolic, the Earth will be rotate around the Sun on a prolate elliptic orbit.

For the beginning let us find a relation between the speed of the Earth in aphelions $V_{\mathrm{aph}}$ and circular speed $V_0$ for a motion on a circular orbit with the same semi-axis.

Using the II Kepler law one may write

V_{\mathrm{per}}\cdot R_{\mathrm{per}}=V_{\mathrm{aph}}\cdot R_{\mathrm{aph}},

using the law of conservation of energy

V_{\mathrm{per}}^2/2-GM/R_{\mathrm{per}}=V_{\mathrm{aph}}^2/2-GM/R_{\mathrm{aph}},

and also taking into account, that

R_{\mathrm{per}}=a_0(1-e),\quad R_{\mathrm{aph}}=a_0(1+e),\quad GM=V_0^2\cdot R_0,

it is possible to find

V_{\mathrm{per}}=V_0\cdot\{(1+e)/(1-e)\}^{1/2},

V_{\mathrm{aph}}=V_0\cdot\{(1-e)/(1+e)\}^{1/2}.

The orbit parameters of the Earth after that the mass of Sun has decreased to half its present value are designate as:

$a_N$ – semi-axis of an orbit,

$T_N$ – period of revolution,

$e_N$ – eccentricity,

$V_N$ – speed for a motion on a circular orbit with a semi-axis $a_N$ .

Let us write three additional equations:

distance $R_{\mathrm{aph}}$ , former distance in aphilion, became now the distance in perihelion

a_0(1+e)=R_{\mathrm{aph}}=R_{\mathrm{per}-N}=a_N(1-e_N),

speed $V_{\mathrm{aph}}$ , former speed in aphilion, will now become the speed in perihelion

V_{\mathrm{aph}}=V_{\mathrm{per}-N}=V_N\cdot\{(1+e_N)/(1-e_N)\}^{1/2},

and the relation

V_N=2\pi a_N/T_N.

By solving all equations together it is possible to receive a beautiful expression

a_N=a_0(1+e)/2e.

According to the general III Kepler law

T_N=T_0\cdot2^{1/2}\cdot(a_N/a_0)^{3/2},

one can find

T_N=T_0\cdot2^{1/2}\cdot\{(1+e)/2e\}^{3/2}.

T_N=2^{1/2}\cdot(1.017/0.034)^{3/2}\approx230\ \mathrm{years}.

Оригинал в Архива: alphatheo.png · допълнителни данни: theosol1.jpg · допълнителни данни: theosol2.jpg · допълнителни данни: constants.jpg