A body weighs 64 N on the surface of the earth. What is the gravitational force (in N) on it due to the earth at a height equal to one-third of the radius of the earth ?

Here, $\mathrm{mg}=64\mathrm{N}; \mathrm{h}=\mathrm{R}/3$

The value of acceleration due to gravity at a height h (when h is not negligible as compared to R),

$\begin{array}{l}{\mathrm{g}}^{\mathrm{\prime }}=\mathrm{g}\frac{{\mathrm{R}}^2}{(\mathrm{R}+\mathrm{h}{)}^2}\\ \mathrm{Or} {\mathrm{mg}}^{\mathrm{\prime }}=\mathrm{mg}\frac{{\mathrm{R}}^2}{(\mathrm{R}+\mathrm{h}{)}^2}=64\times \frac{{\mathrm{R}}^2}{(\mathrm{R}+\mathrm{R}/3{)}^2}=64\times \frac{9}{16}=36 N\end{array}$