This was a physics bonus assignment. I have been trying to solve it for the last 2 hours. Let's say that the density of The Moon and The Earth are the same. Moon radius:Earth radius is $0.273:1$! How long does the pendulum have to be for the period ($T$) to be $2$ seconds. We have the following constants available.
$$g(\text{Earth})=9.81m/s^2$$ $$G=6.67\times10^{-11}$$ I would really appreciate your help! This is not an obligatory assignment.
Lets take the approximation $$ T = 2\pi \sqrt{\frac{L}{g}} $$ for the moon we have to compute $g(\text{moon})$ $$ g(\text{moon}) = \frac{GM_{\text{moon}}}{r^2} $$ this is the magnitude of the acceleration. $$ M_{\text{moon}} = \rho V_{\text{moon}} = \rho \frac{4}{3}\pi r^3 $$ now since we know that density ($\rho$) is consistent between earth and moon we also have $$ g(\text{earth}) = \frac{4}{3}\pi G\rho R_{e} \implies \rho = \frac{g(\text{earth})}{\frac{4}{3}\pi G R_{e}} $$ thus $$ g(\text{moon}) = \frac{4}{3}\pi G r \frac{g(\text{earth})}{\frac{4}{3}\pi G R_{e}} \implies g(\text{moon}) = g(\text{earth})\frac{r}{R_e} $$ thus we get $$ T_{\text{moon}} = 2\pi\sqrt{\frac{L}{g(\text{moon})}} = 2\pi \sqrt{\frac{L}{g(\text{earth})\frac{r}{R_e}}} $$ thus $$
$$ L = g(\text{earth})\frac{r}{R_e}\frac{T_{\text{moon}}^2}{4\pi^2}\approx 0.0277 m $$
