Question

If the gravity of the moon is about `1/6`th the gravity on earth, how will a clock's period on the moon compare with that on earth?

Answer 1

Will run slow

Explanation:

For small-ish oscillations, we know that period, T, of an oscillation is `T = 2 pi sqrt (l/g)` where l is the length of the pendulum and g is prevailing gravity

as `g_(mo on) approx (g_(earth))/6`

then `T_(mo on) approx sqrt6 * T_(earth)=2.45*T_(earth)`

So, if you like, a tick and a tock should take 2.45 times as long and the clock should run slow.