IAO 2011, Теория, Група β — Задача 3. Sunrise on Mars

Sunrise duration

For an observer on the North Pole of Mars the sunrise happens once a year, in time close to the day of the martian vernal equinox. The solution of the problem is analogous to the solution about the sunrise on the Earth's poles, but we have to remember that some parameters are different on Mars and on Earth:

• $\alpha$ — the visible angular size of Sun.
• $v$ — the speed of the Sun along the planet's ecliptic.
• $\varepsilon$ — the inclination of the planet's equator plane to the plane of the planet's ecliptic.
• $R$ — radius of the planet.

For the observer on the North Pole of Mars the angle between the ecliptic and the horizon is equal to $\varepsilon=25.2^\circ$. During the sunrise the Sun goes up to the angle equal to its apparent angular diameter, that is

where $D$ is the diameter of the Sun, $L$ is the distance from Mars to the Sun. During this time the Sun moves an angular distance $\varphi=\alpha/\sin\varepsilon$ on the ecliptic. Thus, the duration of the sunset equals to $\tau=\varphi/v=\alpha/(v\sin\varepsilon)$, where $v$ is the speed of the Sun along the martial ecliptic.

Due to the large eccentricity of orbit of Mars both the distance $L$ and speed $v$ may vary significantly ($\pm e=9.3\%$ of mean value). Nevertheless we do not know a priory, and cannot understand from the tables, in which point of orbit Mars is situated at the moments of its vernal equinox. So we shall use mean values, equal to $L=228$ million km and $v=360^\circ/687^{\mathrm d}\approx0.524^\circ$ per day. Thus,

Source note: “sunset” and the discussion of eccentricity are printed in this sunrise sketch; the question explicitly permits a circular orbit. The requested drawing of the Bear on Mars is not supplied in the available sketch.