IAO 2013, Теория, Група α — Задача 1. Star rise in Moletai
XVIII International Astronomy Olympiad. Theoretical round. Problems to solve. Group α · 6 септември 2013 г. · 8 т.
Бележка
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.
Parameters of orbits and physical characteristics
The bilingual source sheet gives the following values. Blank cells reproduce blank source cells.
| Body / planet | Average distance (AU) | Average distance ( km) | Sidereal period (tropical years) | Sidereal period (days) | Orbital inclination | Eccentricity | Equatorial diameter (km) | Mass ( kg) | Average density (g/cm³) | Gravity at surface (m/s²) | Axial tilt | Maximum magnitude from Earth | Albedo |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Sun | 1392000 | 1989000 | 1.409 | −26.74 | |||||||||
| Mercury | 0.387 | 57.9 | 0.241 | 87.969 | 7.00° | 0.206 | 4879 | 0.3302 | 5.43 | 3.70 | 0.01° | 0.06 | |
| Venus | 0.723 | 108.2 | 0.615 | 224.7007 | 3.40 | 0.007 | 12104 | 4.8690 | 5.24 | 8.87 | 177.36 | 0.78 | |
| Earth | 1.000 | 149.6 | 1.000 | 365.2564 | 0.00 | 0.017 | 12756 | 5.9742 | 5.515 | 9.81 | 23.44 | 0.36 | |
| Moon | 0.00257 | 0.38440 | 0.0748 | 27.3217 | 5.15 | 0.055 | 3475 | 0.0735 | 3.34 | 1.62 | 6.7 | −12.7 | 0.07 |
| Mars | 1.524 | 227.9 | 1.880 | 686.98 | 1.85 | 0.093 | 6794 | 0.6419 | 3.94 | 3.71 | 25.19 | −2.0 | 0.15 |
| Jupiter | 5.204 | 778.6 | 11.862 | 4332.59 | 1.30 | 0.048 | 142984 | 1899.8 | 1.33 | 24.86 | 3.13 | −2.7 | 0.66 |
| Saturn | 9.584 | 1433.7 | 29.458 | 10759.20 | 2.48 | 0.054 | 120536 | 568.50 | 0.70 | 10.41 | 26.73 | 0.7 | 0.68 |
For outer planets and 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) | |
| Solar constant, A (W/m²) | 1367 |
| Hubble parameter, (km/s/Mpc), mean value | 71 |
| Hubble parameter, diapason of values | 50–100 |
| Planck constant, h (J·s) | |
| Charge of electron, e (C) | |
| Mass of electron, (kg) | |
| Proton-to-electron ratio | 1836.15 |
| Faraday constant, F (C/mol) | 96485 |
| Magnetic constant, (H/m) | |
| Universal gas constant, R (J/mol/K) | 8.314 |
| Boltzmann constant, k (J/K) | |
| Stefan–Boltzmann constant, (W/m²/K⁴) | |
| Wien’s displacement constant, b (m·K) | 0.002897 |
| Laboratory wavelength of Hα (Å) | 6562.81 |
| Tropical year length, T (days) | 365.242199 |
| Standard atmosphere (Pa) | 101325 |
| Visible light extinction by the terrestrial atmosphere in zenith (minimum) | 19%, 0.23 mag |
| Refractive index of water for 20°C, n | 1.334 |
| Moment of inertia of a solid ball | |
| Volume of ball | |
| Area of sphere | |
| 3.14159265 | |
| e | 2.71828183 |
| Golden ratio, | 1.61803399 |
Data of some stars
| Star | Designation | RA | DEC | Parallax | Apparent magnitude | Spectral class | Mass () |
|---|---|---|---|---|---|---|---|
| Sun | ⊙ | 0h–24h | −23°26′ to +23°26′ | 8.794″ | vis −26.74; bol −26.82 | G2 | 1 |
| Proxima Centauri | V645 Cen, α Cen C | 14h 29m 43s | −62°40′46″ | 0.769″ | 11.05 | M5.5 | 0.123 |
| Alpha Centauri A | α Cen A | 14h 39m 37s | −60°50′02″ | 0.747″ | −0.01 | G2 | 1.1 |
| Alpha Centauri B | α Cen B | 14h 39m 35s | −60°50′14″ | 1.34 | K1 | 0.9 | |
| Beta Centauri | β Cen | 14h 03m 49s | −60°22′23″ | 0.009″ | 0.61 | B1 | 21 |
| Epsilon Eridani | ε Eri | 03h 32m 56s | −09°27′30″ | 0.311″ | 3.74 | K2 | 0.82 |
| Gliese 581 | HO Lib | 15h 19m 27s | −07°43′20″ | 0.16″ | vis 10.57; bol 8.0 | M3V | 0.31 |
The Alpha Centauri parallax is printed once beside the pair; the second cell is left blank here as in the source.
Historical units
Units of power
1 horse-power (hp) = 735.49875 W.
Local units of length in the end of XVIII century
- 1 arshin (arsh) = 0.711187 m
- 1 span (sp) = 1/4 arshin
- 1 vershok (vrsh) = 1/4 span
- 1 sajene (sj) = 3 arshin
- 1 verst (vrst) = 500 sajene
Условие
An observer in Moletai recorded that a star culminated at 02:54 and set at 05:45 on September 8, 2013. Effects of irregularities of the horizon should not be taken into account.
1.1. At what time will the star rise on September 9, 2013? [5 т.]
1.2. In approximately which direction do you need to wait for the rising of the star? Choose one of the alternatives: N, NE, E, SE, S, SW, W, NW. Draw a picture with an explanation. [3 т.]
Решение
Official English solution sketch. 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.
1.1. Time of rising
The time elapsed from the rising of the star to culmination is equal to the time elapsed from culmination to set. Therefore, on September 8, this star rose at 02:54 − 02:51 = 0:03. The next time the star will rise after a sidereal day, equal to 23:56, that is, at 23:59 but still on September 8. On September 9 the star will rise after one more sidereal day, that is at 23:55.
1.2. Direction
Very roughly one can say that the star is above the horizon ¼ days, and under the horizon — ¾ days. The star culminates at south. Accordingly, it is above the horizon: ⅛ of rotation before the culmination and ⅛ of rotation after it, that is, it rises approximately at south-east and sets approximately at south-west. At the picture that should be drawn by the participants, it can be seen even better. Answer: SE. (ЮВ in Russian).
Official basic criteria
1.1. Totally 5 pt, including:
- Understanding approach that culmination is the middle point between star rise and set — 1 pt.
- Time of the first star rise on September 8 calculation, result 0:03 — 1 pt.
- Using sidereal approach for the future calculations — 1 pt.
- Issue that next sunrise will be still on September 8 — 1 pt.
- On September 9 the star will rise after one more sidereal day, that is at 23:55 — 1 pt.
1.2. 3 pt, including:
- Argumentation about the star is above the horizon ¼ days, ⅛ of rotation before the culmination and ⅛ of rotation after — 1 pt.
- Correct illustration — 1 pt.
- Rises approximately at south-east, answer: SE — 1 pt.
Official solution sketches: 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.
Basic criteria: note for jury
Jury members should evaluate the students’ 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.
Оригинал в Архива: alphatheo.pdf · официални решения: theosol.pdf · допълнителни данни: theodata.pdf · допълнителни данни: theo_cr_1.pdf · допълнителни данни: theo_cr_2.pdf · допълнителни данни: theo_cr_3.pdf · допълнителни данни: constants.pdf · допълнителни данни: hrd.pdf