Question

if R is the radius of earth and g is the acceleration due to gravity on earths's surface then mean density of earth is

Answer 1

Acceleration due to gravity, radius and density are related as,
$$\begin{gathered} g=\frac{4}{3}\mathrm{\pi G\rho R} \\ \mathrm{Density}\mathrm{of}\mathrm{earth}, \\ \mathrm{\rho }=\frac{3}{4}\frac{\mathrm{g}}{\mathrm{\pi GR}} \\ \mathrm{We}\mathrm{can}\mathrm{also}\mathrm{derive}\mathrm{as}\mathrm{below}, \\ \mathrm{We}\mathrm{know}\mathrm{that}, \\ \mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{R}}^2}----\left(\mathrm{i}\right) \\ \mathrm{And}\mathrm{density},\mathrm{\rho }=\frac{\mathrm{Mass}}{\mathrm{Volume}}=\frac{\mathrm{M}}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3} \\ \mathrm{Or}\mathrm{M}=\frac{4}{3}{\mathrm{\pi R}}^3\mathrm{\rho } \\ \mathrm{Putting}\mathrm{in}\left(\mathrm{i}\right), \\ \mathrm{g}=\frac{\mathrm{G}\left(\frac{4}{3}{\mathrm{\pi R}}^3\mathrm{\rho }\right)}{{\mathrm{R}}^2} \\ \mathrm{or}\mathrm{g}=\frac{4}{3}\mathrm{\pi G\rho R} \\ \mathrm{or}\mathrm{\rho }=\frac{3}{4}\frac{\mathrm{g}}{\mathrm{\pi GR}} \end{gathered}$$