# Why is the period of a pendulum on the Moon $\sqrt{6}$ times its period on the Earth? [closed]

I came across this equation: $$T_m= \sqrt{6}T_e$$.

Can anyone tell me how this equation is derived? This is how I tried to, but I got stuck after some time.

So the time period of a simple pendulum on the earth= $$2\pi\sqrt{l/g}$$

The time period of a pendulum on the moon is $$2\pi\sqrt{l/(g/6)}$$

Now how do I create an equation which shows the time period of a pendulum on the moon with respect to the time period of a pendulum on the earth.

And please be as detailed as possible!

• The first time period is $T_e$ and the second is $T_m$, so what problem are you having? – G. Smith Jul 10 at 20:20
• Just divide one expression by the other, keeping in mind that $\sqrt {ab} =\sqrt{a}\sqrt{b}$ – Agnius Vasiliauskas Jul 10 at 20:24
• The 6 is approximate, not exact. Writing $\sqrt 6$ wrongly suggests that it is exact. – G. Smith Jul 10 at 20:33

$$T_m = 2 \pi \sqrt{\frac{l}{\frac{g}{6}}} = 2 \pi \sqrt{\frac{6l}{g}} = 2 \pi \sqrt{6\frac{l}{g}} = \sqrt{6} \left(2 \pi \sqrt{\frac{l}{g}}\right) = \sqrt{6} T_e$$
\begin{align*} T_e&=2\pi\sqrt{\frac lg}\\ T_m&=2\pi\sqrt{\frac l{\frac g6}}=\sqrt6\cdot2\pi\sqrt{\frac lg}=\sqrt6\cdot T_e\\ \end{align*}