Съдържание

IAO 2010, Теория, Група β — Задача 5. International Space Station.

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 graph presents the change in height of the ISS orbit depending on time. Estimate the average density of the atmosphere at heights around 340–360 km. Consider the orbit to be circular. The mass of ISS is 362 ton. Consider the cross-section of the ISS-complex to be $S=500\;\mathrm{m^2}$ (including the sections of the solar cell array).

ISS height in km, October 2009 to October 2010. — Graph for problem 5

Решение

Let us consider the orbit of station as circular, and its average radius was

R_0+h=6371\;\mathrm{km}+350\;\mathrm{km}=6720\;\mathrm{km}\approx6.7\cdot10^6\;\mathrm m,

where $R_0$ – radius of the Earth, and $h$ – the height of the orbit). It is obvious, that the upward jumps are corrections of the orbit by engines, and its slow descent during the lowering is due to loss of energy by the station due to friction in the upper atmosphere. We see that all these parts have an average inclination of approximate 2-2.5 kilometers per month or about 75 meters per day, that is, the change in the altitude of the orbit per day is

\Delta h=-75\;\mathrm m.

To consider correctly all the events, it is better to use energetic approach, calculating the total energy of the station

E=\Pi+K=-GMm/(R_0+h)+mV^2/2,

and not only kinetic $K$ or potential $\Pi$ . (Here $G$ is gravitational constant, $M$ is mass of the Earth, $V$ is orbital speed of station.) By the way, note that speed of station increases with lowering of the height (the station is braked, and speed increases!). From conditions of moving on circular orbit

GMm/(R_0+h)^2=mV^2/(R_0+h),

Whence

GMm/(R_0+h)=mV^2,\quad\Pi=-2K,\quad E=-K=\Pi/2.

Note. The formulae written above are fair for the values of potential energy of gravitational interaction counted from infinity (bodies are removed from each other on infinite distance). Generally speaking, the absolute value of potential energy is not meaningful in physics, only the change of potential energy is important.

Thus,

E=-GMm/2(R_0+h).

The process of loss of energy by station and simultaneous increase in its speed with lowering of radius of the orbit can be imagined as follows. Let us imagine a quasistationary process with circular orbit all the time, and the work done by forces of resistance of friction

A=F\cdot L

changes the parameters of this circular orbit. Here $F$ – force of resistance, $L$ – the distance moved. The force $F=\Delta P/\Delta t$ may be found from the following arguments: during each period of time $\Delta t$ a mass $\mu=\rho\cdot S\cdot V\cdot\Delta t$ of averagely motionless molecules collides with the station ( $\rho$ – density of atmosphere at height of orbit of the station). As result of elastic collisions their speed relative to the station varies from $-V$ to $+V$ , and relative to the Earth – from 0 to $2V$ . That is, during the period of time $\Delta t$ the station transmits to the molecules the momentum

\Delta P=\mu\cdot2V=2V\cdot\rho\cdot S\cdot V\cdot\Delta t=2\cdot\rho\cdot S\cdot V^2\cdot\Delta t,

Whence

F=\Delta P/\Delta t=2\cdot\rho\cdot S\cdot V^2,

Thus, during time $\tau=24$ hours the station moves a distance $L=V\cdot\tau$ , and the work of forces of friction (and, accordingly, loss of energy by station) is

A=F\cdot L=2\cdot\rho\cdot S\cdot V^3\cdot\tau,

Change in energy

\Delta E=-A=-2\cdot\rho\cdot S\cdot V^3\cdot\tau.

On the other hand, change in energy of the station in this time is

\Delta E=-GMm/2(R_0+h+\Delta h)-\{-GMm/2(R_0+h)\}\approx\Delta h\cdot GMm/2(R_0+h)^2,

where $\Delta h$ – change in height of the orbit (value of $\Delta h$ is negative!).

-2\cdot\rho\cdot S\cdot V^3\cdot\tau=\Delta h\cdot GMm/2(R_0+h)^2,

\rho=-\Delta h\cdot GMm/\{4(R_0+h)^2\cdot S\cdot V^3\cdot\tau\}

and, taking into account that $V^2=GM/(R_0+h)$ ,

\rho=-m\Delta h/\{4\tau S\cdot(GM)^{1/2}(R_0+h)^{1/2}\}.

Numerical calculations:

\rho=3.12\cdot10^{-12}\;\mathrm{kg/m^3}\approx3\cdot10^{-12}\;\mathrm{kg/m^3}.

So the numerical answer is:

\rho\approx3\cdot10^{-12}\;\mathrm{kg/m^3}

(more accuracy is not reasonable).

Basic criteria. For work of Jury

General note: the frame work of school physics using only the energetic approach to obtain the solution, and usage the total energy $E=\Pi+K$ , not only $\Pi$ or $K$ is correct.

Max 8 pt criteria for solutions using total energy approach:

Understanding that using $E=\Pi+K$ is necessary – 2 pt.
Correct understanding that the slow descent during the lowering is due to loss of energy, correct finding that the rate $\Delta h/\Delta\tau$ is about 75 m/day – 1 pt.
Algebra steps and correct finding: total energy $E=-GMm/2(R_0+h)$ – 1 pt.
Correct finding work of forces – 1 pt.
Algebra steps and final correct formula – 2 pt in total.
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