Suppose Mercury had 5-to-3 spin-orbit coupling and that its sidereal period remained 88 days, as now. What would be the time from noon to noon on Mercury, in days?

Question

Quizplus · Accepted Answer

In a 5-to-3 spin-orbit coupling, Mercury would rotate 5 times for every 3 orbits around the Sun. Given its sidereal period of 88 days, the synodic period (noon to noon) can be calculated using the formula: synodic period = (sidereal period * orbital period) / (orbital period - sidereal period). With a 5-to-3 coupling, the orbital period is 88 days * (3/5) = 52.8 days. Thus, the synodic period is (88 * 52.8) / (52.8 - 88) = 264 days.

Suppose Mercury Had 5-To-3 Spin-Orbit Coupling and That Its Sidereal