Mercury rotates once on its axis with respect to the distant stars in 58 2/3 days and has a sidereal period of 88 days. A star is overhead to an observer on Mercury at midnight. How many times will Mercury orbit the Sun before this star is again directly over the observer's head at midnight? (Draw a diagram to help with this exercise.)
A)It will never again reach this specific position.
B)one
C)two
D)three

Question

Quizplus · Accepted Answer

Let's first calculate how many times Mercury rotates on its axis in 88 days (its sidereal period).

Rotation per day = (1 rotation / 58 2/3 days) = (3 rotations / 176 days)
Rotations in 88 days = 88 * (3/176) = 1.5 rotations

This means that after one full orbit around the Sun (which takes 88 days), Mercury will have rotated 1.5 times on its axis with respect to the distant stars.

Now, let's consider the star that is overhead at midnight. After one orbit around the Sun (88 days), Mercury will have completed 1.5 rotations on its axis, but the star will have completed 2 rotations on its own axis (since it is much further away and has a much longer period).

So, in order for the star to be overhead at midnight again, Mercury must complete another 1/2 rotation on its own axis (since 1.5 rotations + 1/2 rotation = 2 rotations).

Therefore, the star will be overhead again after Mercury completes two full orbits around the Sun (since each orbit takes 88 days, and we need to wait for the star to complete two rotations on its own axis).

So the answer is choice C, two orbits.

Mercury Rotates Once on Its Axis with Respect to the Distant