# Question #4b882

Oct 7, 2016

I found:
${l}_{\text{Earth}} = 1 m$
${l}_{\text{Moon}} = 20 c m$

#### Explanation:

Here you could use the fact that the period $T$ of a pendulum depends upon its length $l$ and the acceleration of gravity $g$ as:
$T = 2 \pi \sqrt{\frac{l}{g}}$

and rearranging:

$l = {\left(\frac{T}{2 \pi}\right)}^{2} \cdot g$

1) on Earth $g = 9.8 \frac{m}{s} ^ 2$
so you get:
${l}_{\text{Earth}} = {\left(\frac{2}{2 \pi}\right)}^{2} \cdot 9.8 = 0.99 \approx 1 m$

2) On the Moon we have ${g}_{\text{Moon}} = 1.6 \frac{m}{s} ^ 2$ (from literature) and so:
${l}_{\text{Moon}} = {\left(\frac{2}{2 \pi}\right)}^{2} \cdot 1.6 = 0.16 \approx 0.2 m = 20 c m$