IAO 2015, Теория, Група α — Задача 5. Motion of a satellite
XX International Astronomy Olympiad. Theoretical round. Group α · 15 октомври 2015 г.
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General note. Maybe not all problems have correct questions. Some questions (maybe the main question of the problem, maybe one of the subquestions) may make 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 tables (Planetary data, stars, constants, etc.) may be used for solving every problem.
The answers «Да-Yes» or «Нет-No» have to be written in English or Russian.
Ещё раз прочитайте перед началом работы.
Общее замечание. Не исключено, что не во всех задачах вопросы поставлены корректно. Некоторые вопросы (возможно, главный вопрос задачи, возможно – подвопрос) могут не иметь смысла. В этом случае следует написать в ответе (по-русски или по-английски): «ситуация невозможна – impossible situation». Естественно, ответ должен быть подкреплён вычислениями или логическими рассуждениями.
Данные из таблиц (Солнечная система, звёзды, константы) могут быть использованы в любой задаче.
Ответы «Да-Yes» или «Нет-No» должны быть написаны по-русски или по-английски.
Условие
An artificial satellite, moving in equatorial, slightly elliptical orbit passed the perigee point at the height of km from the sea level; and its speed at the point was 0.6% higher than the circumferential velocity of the given point. Find the time after which the satellite will reach the height km?
Отговори
- Approximately 1160 s (19.4 min), expressly an estimate valid only for a small orbital segment.
Решение
α-5. Motion of a satellite.
To solve the problem let's work in the frame of reference rotating around the Earth along with the satellite. In this frame the satellite is affected by two forces: the centrifugal force and the gravitational force , where and are the circular and increased by 0.6% velocities of the satellite at the height H = 428 km, and – the radius of this orbit. Just we see that in this frame of reference the first minutes after the perigee passage the satellite simply moved from the Earth with a constant acceleration.
, where is the acceleration of gravity at the height H.
Then from the formula the time to reach the height may be found, , which is about 1160 seconds or 19.4 minutes. It is much smaller than the period (about 95 minutes), but already comparable with the quarter of the period. So this result is estimation.
Note: As an exact solution this method is applicable only to a small part of the orbit of the satellite after the perigee passage, in the case of (the period of the satellite).
Parameters of orbits and physical characteristics of planets, Sun and Moon
| Body, planet | Average distance to central body, in astr. units | Average distance to central body, in mln. km | Sidereal (or analogous) period, in tropical years | Sidereal (or analogous) period, in days | Eccentricity, e | Equat. diameter, km | Mass, kg | Av. density, g/cm³ | Grav. acceler. at surf., m/s² | Max. magn. from Earth **) | Albedo |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Sun | 1392000 | 1989000 | 1,409 | ||||||||
| Mercury | 0,387 | 57,9 | 0,241 | 87,969 | 0 ,206 | 4 879 | 0,3302 | 5,43 | 3,70 | 0,06 | |
| Venus | 0,723 | 108,2 | 0,615 | 224,701 | 0,007 | 12 104 | 4,8690 | 5,24 | 8,87 | 0,78 | |
| Earth | 1,000 | 149,6 | 1,000 | 365,256 | 0,017 | 12 756 | 5,9742 | 5,515 | 9,81 | 0,36 | |
| Moon | 0,00257 | 0,38440 | 0,0748 | 27,3217 | 0,055 | 3 475 | 0,0735 | 3,34 | 1,62 | 0,07 | |
| Mars | 1,524 | 227,9 | 1,880 | 686,980 | 0,093 | 6 794 | 0,6419 | 3,94 | 3,71 | 0,15 | |
| Jupiter | 5,204 | 778,6 | 11,862 | 4 332,59 | 0,048 | 142 984 | 1899,8 | 1,33 | 24,86 | 0,66 | |
| Saturn | 9,584 | 1433,7 | 29,458 | 10 759,20 | 0,054 | 120 536 | 568,50 | 0,70 | 10,41 | 0,68 | |
| Uranus | 19,191 | 2871,0 | 84,015 | 30 685,93 | 0,046 | 51 118 | 86,625 | 1,30 | 8,44 | 0,74 | |
| Neptune | 30,071 | 4498,6 | 164,778 | 60 187,64 | 0,008 | 49 532 | 102,78 | 1,76 | 11,20 | 0,58 |
**) For outer planets and Moon – in mean opposition.
| Coordinates | Observatory | Kazan kremlin | Petrovskoye |
|---|---|---|---|
| λ ( E / в.д. ) | 48° 49′ | 49° 06′ | 49° 06′ |
| φ ( N / с.ш. ) | 55° 50′ | 55° 48′ | 55° 41′ |
| Timezone | UT+3 | UT+3 | UT+3 |
Some constants and formulae
| Constant or formula | Value |
|---|---|
| Speed of light in vacuum c (m/s) | 299792458 |
| Constant of gravitation G (N·m²/kg²) | |
| Solar constant A (W/m²) | 1367 |
| Hubble parameter H₀ (km/s/Mpc), mean | 68 |
| Hubble parameter, range | 50–100 |
| Planck constant h (J·s) | |
| Charge of electron e (C) | |
| Mass of electron mₑ (kg) | |
| Proton-to-electron ratio | 1836.15 |
| Faraday constant F (C/mol) | 96485 |
| Magnetic constant μ₀ (H/m) | |
| Universal gas constant R (J/mol/K) | 8.314 |
| Boltzmann constant k (J/K) | |
| Stefan–Boltzmann constant σ (W/m²/K⁴) | |
| Wien displacement constant b (m·K) | 0.002897 |
| Laboratory wavelength Hα (Å) | 6562.81 |
| Tropical year length T (days) | 365.242199 |
| Nodal period of lunar orbit (years) | −18.6 |
| Standard atmosphere (Pa) | 101325 |
| Visible light extinction by terrestrial atmosphere in zenith (minimum) | 19%, 0.23 magnitudes |
| Height of homogeneous atmosphere (m) | 7991 |
| Refractive index of water for 20°C n | 1.334 |
| Moment of inertia of a solid ball | |
| Volume of a ball | |
| Area of sphere | |
| π | 3.14159265 |
| e | 2.71828183 |
| Golden ratio φ | 1.61803399 |
20 brightest stars, other stars and Greek alphabet
| Name | Designation | RA | Dec | Parallax (arcsec) | Magnitude | Spectral class |
|---|---|---|---|---|---|---|
| Altair | α Aql | 19h50m47s | 08°52′06″ | 0.195 | 0.77 | A7 |
| Capella | α Aur | 05h16m41s | 45°59′53″ | 0.073 | 0.08 | G5+G0 |
| Arcturus | α Boo | 14h15m38s | 19°10′57″ | 0.089 | −0.04ᵛ | K1 |
| Canopus | α Car | 06h23m57s | −52°41′45″ | 0.028 | −0.72 | F0 |
| Toliman (Rigel Kent) | α Cen A; B | 14h39m36s | −60°50′07″ | 0.747 | −0.01; 1.33 | G2; K1 |
| Hadar | β Cen | 14h03m49s | −60°22′23″ | 0.009 | 0.61 | B1 |
| Sirius | α CMa | 06h45m09s | −16°42′58″ | 0.375 | −1.46 | A1 |
| Procyon | α CMi | 07h39m18s | 05°13′30″ | 0.288 | 0.38 | F5 |
| Acrux | α Cru | 12h26m36s | −63°05′57″ | 0.010 | 0.77 | B0 |
| Becrux | β Cru | 12h47m43s | −59°41′20″ | 0.009 | 1.30 | B0 |
| Deneb | α Cyg | 20h41m26s | 45°16′49″ | 0.002 | 1.25 | A2 |
| Achernar | α Eri | 01h37m43s | −57°14′12″ | 0.026 | 0.46 | B3 |
| Pollux | β Gem | 07h45m19s | 28°01′35″ | 0.097 | 1.14 | K0 |
| Vega | α Lyr | 18h36m56s | 38°47′01″ | 0.123 | 0.03 | A0 |
| Betelgeuse | α Ori | 05h55m10s | 07°24′25″ | 0.005 | 0.5ᵛ | M2 |
| Rigel | β Ori | 05h14m32s | −08°12′06″ | 0.013 | 0.12 | B8 |
| Fomalhaut | α PsA | 22h57m39s | −29°37′20″ | 0.130 | 1.16 | A3 |
| Antares | α Sco | 16h29m24s | −26°25′55″ | 0.024 | 0.96 | M1+B4 |
| Aldebaran | α Tau | 04h35m55s | 16°30′33″ | 0.048 | 0.85ᵛ | K5 |
| Spica | α Vir | 13h25m12s | −11°09′41″ | 0.023 | 0.98 | B1 |
Some other stars
| Name | Designation | RA | Dec | Parallax (arcsec) | Magnitude | Spectral class |
|---|---|---|---|---|---|---|
| Hamal | α Ari | 02h07m10s | 23°27′45″ | 0.050 | 2.01 | K2 |
| Polaris | α UMi | 02h31m49s | 89°15′51″ | 0.007 | 1.97ᵛ | F7 |
| Kochab | β UMi | 14h50m42s | 74°09′20″ | 0.025 | 2.07 | K4 |
| Proxima Centauri | V645 Cen, α Cen C | 14h29m43s | −62°40′46″ | 0.769 | 11.05 | M5.5 |
Greek alphabet
| Uppercase | Lowercase | Name |
|---|---|---|
| Α | α | alpha |
| Β | β | beta |
| Γ | γ | gamma |
| Δ | δ | delta |
| Ε | ε | epsilon |
| Ζ | ζ | zeta |
| Η | η | eta |
| Θ | θ | theta |
| Ι | ί | iota |
| Κ | κ | kappa |
| Λ | λ | lambda |
| Μ | μ | mu |
| Ν | ν | nu |
| Ξ | ξ | xi |
| Ο | ο | omicron |
| Π | π | pi |
| Ρ | ρ | rho |
| Σ | σ | sigma |
| Τ | τ | tau |
| Υ | υ | upsilon |
| Φ | φ | phi |
| Χ | χ | chi |
| Ψ | ψ | psi |
| Ω | ω | omega |
Theoretical round. Sketches for solutions. For jury job ONLY
Theoretical round. Sketches for solutions. For jury job ONLY
Note. The given sketches are not full; the team leaders have to give more detailed explanations to students. But the correct solutions in the students' papers (enough for 8 pts) may be shorter.
Note. Jury members should evaluate the student's 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.
Theoretical round. Basic criteria. For work of Jury
Note. The given sketches are not full; the team leaders have to give more detailed explanations to students. But the correct solutions in the students' papers (enough for 8 pts) may be shorter.
Note. Jury members should evaluate the student's 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.
Теми и трудност
Трудност: неоценена.
Небесна механика · Орбитално движение · Орбитална енергия и напускане
Оригинал в Архива: theo_alpha.pdf · официални решения: theosol1.jpg · допълнителни данни: theosol2.jpg · допълнителни данни: theosol3.jpg · допълнителни данни: theosol4.jpg · допълнителни данни: theosol5.jpg · допълнителни данни: theo_instr.pdf · допълнителни данни: data_1.pdf · допълнителни данни: data_2.pdf · допълнителни данни: data_3.pdf · допълнителни данни: hrd.pdf · допълнителни данни: sky_map.pdf