Mercury rotates once with respect to the distant stars in 59 days and has a sidereal period of 88 days.If a star is seen to be overhead to an observer on Mercury at midnight on a particular day,how many times will Mercury orbit the Sun before this star is again above the observer's head in a nighttime sky? (See Figure 6-44,Comins,Discovering the Essential Universe,6th ed.)

Question

Quizplus · Accepted Answer

Since Mercury has a sidereal period of 88 days, it takes 88 days for Mercury to complete one orbit around the Sun with respect to the distant stars. If a star is overhead to an observer on Mercury at midnight on a particular day, then it must be at the same position with respect to the distant stars after 88 days have passed.

However, during these 88 days, Mercury also rotates on its own axis, completing 1.5 rotations (i.e., 59 days/rotation) with respect to the distant stars. So after 88 days have passed, the star will not be overhead at midnight, but rather after an additional 29.5 days (i.e., halfway into the next rotation).

Therefore, it will take another 88 days for Mercury to complete another orbit around the Sun with respect to the distant stars and for the star to be overhead at midnight again. This means that the star will be overhead twice in the nighttime sky before Mercury completes two orbits around the Sun, making the answer B) twice.

Mercury Rotates Once with Respect to the Distant Stars in 59