The Sun moving backwards in Mercury's sky
- Round
- Theory
- Points
- 20
- Level
- Silver
- Taught in
- Section 6.2
Adapted from IOAA 2022 (Kutaisi), theory problem 9. We have restated it in our own words. The original gives no orbital elements, so we add a = 0.387 au and e = 0.206, the values the official solution uses. The official solution sets v/r equal to the spin rate, which counts the small outward part of Mercury's velocity, and gets 8.28 days. The angular speed of the Sun round Mercury is the sideways speed divided by r, which gives 8.2 days. For the real Mercury the answer is 8.1 days. The solution was last checked on 23 October 2026.
Mercury's orbit is unusually eccentric, with a semi-major axis of 0.387 au and an eccentricity of 0.206, and Mercury spins once on its axis, measured against the stars, in two thirds of its orbital period. As a result, the Sun seen from Mercury moves backwards across the sky for a short time near perihelion. Find the total time in each orbit for which the Sun moves backwards, in Earth days.
- By Kepler's third law the orbital period is
- Mercury's angular speed round the Sun is
- By (3.10),
- The mean anomaly grows steadily with time, and the backward motion runs from
AnswerThe Sun moves backwards in Mercury's sky for about 8.2 Earth days in each orbit, centred on perihelion.
This problem also appears among the olympiad problems of Section 6.2, Elongation, retrograde motion and visibility, which teaches the ideas it uses. Equation and figure numbers refer to the course.