IAO 2010, Теория, Група α — Задача 2. Observation of a star.
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, (m/s) | 299792458 |
| Constant of gravitation, (N·m²/kg²) | |
| Solar constant, (W/m²) | 1367 |
| Hubble constant, (km/s/Mpc), mean value | 70 |
| Hubble constant, diapason of values | 50–100 |
| Plank constant, (J·s) | |
| Charge of electron, (C) | |
| Mass of electron, (kg) | |
| Proton-to-electron ratio | 1836.15 |
| Faraday constant, (C/mol) | 96485 |
| Magnetic constant, (H/m) | |
| Universal gas constant, (J/mol/K) | 8.314 |
| Boltzmann constant, (J/K) | |
| Standard atmosphere (Pa) | 101325 |
| Stefan-Boltzmann constant, (W/m²/K⁴) | |
| Wien’s displacement constant, (m·K) | 0.002897 |
| Laboratory wavelength of H (Å) | 6563 |
| Refractive index of water for 20°C, | 1.334 |
| Area of sphere | |
| 3.14159265 |
Data of some stars
| Star | Designation | RA | DEC | m | SC | |
|---|---|---|---|---|---|---|
| Arcturus | Boo | 14h 15m 40s | 19° 10′ 57″ | 0″.089 | −0.05 | K1 |
| Vega | Lyr | 18h 36m 56s | 38° 47′ 01″ | 0″.129 | 0.03 | A0 |
| Deneb | Cyg | 20h 41m 26s | 45° 16′ 49″ | 0″.001 | 1.25 | A2 |
| Polaris | UMi | 02h 31m 51s | 89° 15′ 51″ | 0″.007 | 2.02 | F7 |
| Sirius | 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 | 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 | 87.97 | 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 | 0.07 | |
| Mars | 1.524 | 227.9 | 1.880 | 686.98 | 0.093 | 6794 | 0.6419 | 3.94 | 3.71 | 0.15 | |
| Jupiter | 5.204 | 778.6 | 11.862 | 4332.59 | 0.048 | 142984 | 1899.8 | 1.33 | 24.86 | 0.66 | |
| Saturn | 9.584 | 1433.7 | 29.458 | 10759.20 | 0.054 | 120536 | 568.50 | 0.70 | 10.41 | 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
A star is observed from the Earth. It is found that its magnitude is when observed at Zenith and when observed at above the horizon. What would the apparent magnitude of the star be if observed from above the atmosphere (from a satellite, for example)?
Решение
Light from a star at a height of above the horizon passes layers of atmosphere at an angle of , that is, the thickness of each layer will be 1.414 times larger. That is, light will pass effectively 1.414 thickness of atmosphere. Thus, part of all atmosphere increases the stellar magnitude by . It means, the whole of atmosphere increases the stellar magnitude by . The stellar magnitude of the star observed from above the atmosphere will be less by this value from the value measured while observed at zenith from the Earth,
Basic criteria. For work of Jury
- Understanding nature – 2 pt.
- 0.414 part of all atmosphere increases the stellar magnitude on – 2 pt.
- All atmosphere: – 2 pt.
- Correct calculation – 2 pt.
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