# Pendulum on the moon, (Highschool)

A simple pendulum used as a clock, set with the correct time at earth, was sent to moon, it was noticed that it is late 36mins for each hour on earth. Calculate the ratio between acceleration of gravity on moon, and acceleration of gravity on earth.

Since gravity on moon is less, its period should be bigger than earth's, therefore its period is going to be 96mins to earth 60mins, by applying the law $T=2\pi \sqrt{\frac{l}{g}}$ for both, yet when the teacher came to solve it, he set the ratio where $T_M = 24, T_E = 60$ where the period on the moon is shorter, which is physically against the law... I am a bit confused, if someone would kindly explain it, it'd be most appreciated.

• I'd wager that it was just a mistake by the teacher (the period on the moon should be larger). Without seeing more of the derivation, I'm not sure anyone could say much about what was done. That is to say, you should consult with him about it. – Kyle Kanos May 11 '15 at 21:15
• I did consult it with him, and he said give me time to think about it, the next class he told me that I was wrong, and if I wanted to combine periods, earth's period should take the value 96. To be honest I tried to solve it after making the moon's period larger, however the answer was incorrect... – Black Crescent May 12 '15 at 17:14
• The easiest way to show him the larger period is the one on the moon is to ask him what the period is for a $l=1\,\rm m$ pendulum (doing it right should give $T_m\sim5\,{\rm sec}>T_e\sim2\,{\rm sec}$). – Kyle Kanos May 12 '15 at 17:17
• @KyleKanos he admits that the period on moon should be larger and agrees, however his explaination is that we should be taking the difference for us to realize the true period on moon – Black Crescent May 12 '15 at 18:00

## 3 Answers

You are right that $$\frac{T_M}{T_E} = \sqrt\frac{g_E}{g_M}>1,$$ which means that a pendulum period on the Moon is longer than the period of the same pendulum on the Earth.

The rest, in my opinion, is just not precise enough. I think it is very confusing to use the early/late terminology when discussing two modes of operation of the same clock. I suggest the following (less subtle) formulation of the problem statement:

A pendulum clock designed for the Moon swings once an hour. However, its period is $24$ minutes on Earth. What is the ratio of acceleration of gravity on the Moon and on Earth?

From such a formulation it is clear that $T_M = 60$ and $T_E = 24$ in minutes. It therefore is obvious, that the ratio $g_E/g_M \simeq 6.25$.

• I wrote the question in the exact way it was written..., so I wouldn't know better. – Black Crescent May 12 '15 at 17:19
• Then it obviously is just a question about the words used. What matters here is the physics. Hence, you should feel good about yourself figuring the physical content of the question right away :) Kudos! – jarm May 12 '15 at 19:13
• I appreciate your kind compliment, I will consult with my teacher again and try to better understand the question. – Black Crescent May 12 '15 at 20:49

For a pendulum clock to be 36 minutes late, that would mean that in the same amount of time that the clock on the Earth completes 60 cycles (where 1 cycles demarcates 1 minute), the clock on the Moon would only complete 24 cycles. This leads to the period of the pendulum clock on the moon being 60 minutes/24 cycles. Use this idea to create a ratio, and you should be able to calculate a rough value for the acceleration due to gravity on the moon.

• What is your question specifically. Make sure to state exactly what you would like to know. – Jaywalker Mar 10 '16 at 10:14

Standard formula for a period of a pendulum $T$ in terms of its length $L$ and free fall acceleration $g$ is
$$T=2\pi \sqrt{\frac{L}{g}}$$ If you know that that free fall acceleration $g$ on the Moon is about 6 times less than on the Earth, it gives you the answer: on the Moon the same pendulum will have a period about $\sqrt{6}\approx 2.45$ longer than on the Earth.

If you don't know the ratio of free fall accelerations on the Moon and the Earth, you can resort to expressing free fall acceleration on the Moon in terms of gravitational constant $G$, mass of the Moon $M$ and its radius $R$:$$g = \frac{GM}{R^2}$$ This gives the formula for a pendulum's period:$$T = 2\pi R\sqrt{\frac{L}{GM}}$$

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