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

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

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Answer 1 · 2016-08-13T08:43:14

Will run slow

Explanation:

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

as $g_{m o o n} \approx \frac{g_{e a r t h}}{6}$ $g_(mo on) approx (g_(earth))/6$

then $T_{m o o n} \approx \sqrt{6} \cdot T_{e a r t h} = 2.45 \cdot T_{e a r t h}$ $T_(mo on) approx sqrt6 * T_(earth)=2.45*T_(earth)$

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

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

1 Answer

Explanation:

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