IAO 2013, Теория, Група β — Задача 3. Observations from Gliese 581 g
XVIII International Astronomy Olympiad. Vilnius, Lithuania, 6 – 14. IX. 2013. Theoretical round. Problems to solve. Group β · 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
Reference diagram supplied for group β:

3.1. What is the apparent magnitude of our Sun and [3,5 т.]
3.2. what is the approximate constellation in which our Sun will be seen when observed from the planet Gliese 581 g? [2 т.]
3.3. Estimate the angular diameter of the star Gliese 581 when observed from the planet Gliese 581 g. [2,5 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.
3.1. Apparent magnitude
The magnitude of the Sun, visible from any remote location is
where is the apparent magnitude of the Sun visible from Earth, and p is the annual parallax of the location.
The magnitude of the Sun as seen from the system of Gliese 581 is
(And it is not the only way of solution. The value can be found in many other ways, e.g., by formula ).
Source note: the double minus in the intermediate expression is written here as with parentheses for readability; the alternative formula above is retained as printed, including its plus sign before .
3.2. Approximate constellation
While observing from the planet Gliese 581 g, our Sun will be at a point opposite to where the system of Gliese 581 is seen from Earth. Because Gliese 581 is located in the zodiacal constellation of Libra, the sun will be approximately in the opposite zodiacal constellation, that is, in the constellation of Aries. (That solution is correct enough.)
Note: Using a sky-chart one can solve the problem more accurately. The point opposite to the location of the system Gliese 581 at the Earth’s sky, has the coordinates and . This point is located in the constellation of Taurus (Tau), very close to the borders of the constellations of Cetus (Cet) and Aries (Ari). As the students were not provided by a sky-chart, and question was “what is the approximate constellation”, the answer “approximately in the constellation, opposite to the constellation of Libra, that is in the constellation of Aries” is correct.
3.3. Angular diameter
To keep the heat balance a Gliese 581 g should receive the same amount of energy per unit area as is received at Earth.
(As in the previous solution G indices refer to the system of Gliese 581, and indices and E — to the Sun and the Earth).
From the given tables one may find that the spectral class of the star Gliese 581 is M3V. According to the Hertzsprung–Russell diagram the temperature of the star, corresponding to this spectral class, is about . It is times less than the temperature of the Sun.
According to the Stefan–Boltzmann law the power emitted from the star is , and in just the same way we may discourse about the apparent angular area from which the radiation is coming.
Thus, since the apparent angular area of a star is proportional to the square of its apparent angular diameter, the following equations should be correct:
More than three times larger than the Sun!
Official basic criteria
3.1. Totally 3.5 pt, including:
- Formulae for magnitude of the Sun (any way solution) — 2 pt.
- Correct calculations — 1.5 pt.
3.2. Totally 2 pt, including:
- Understanding that Gliese 581 is located in the zodiacal constellation of Libra — 0.5 pt.
- Sun will be at opposite point — 1 pt.
- Answer: constellation of Aries — 0.5 pt.
3.3. Totally 2.5 pt, including:
- Heat balance, the same amount of energy per unit area — 0.5 pt.
- Finding spectral class and temperature (using HR diagram) — 0.5 pt.
- Using Stefan–Boltzmann law — 0.5 pt.
- Final formula — 0.5 pt.
- Final calculations and result — 0.5 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: betatheo.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