Physics - Gravitation Question with Solution | TestHub

Acceleration due to gravity at surface, $g=\left(\frac{GM}{R^2}\right)$ where, $\mathrm{G}$ is the universal gravitational constant, $\mathrm{R}$ is the radius of earth. At a depth $\mathrm{h}$ below the surface, considering force balance: $\begin{array}{rl}& \frac{G\left\{M{\left(\frac{R-h}{R}\right)}^3\right\}m}{(R-h{)}^2}=m{g}^{\prime }\\ & \Rightarrow g^{\prime }=\left(\frac{GM}{R^2}\right)\frac{(R-h)}{R}=g\left(\frac{R-h}{R}\right)=\frac{g}{n}\\ & \Rightarrow \frac{R-h}{R}=\frac{1}{n}\\ & \Rightarrow \frac{-h}{R}=\frac{1-n}{n}\end{array}$ $\Rightarrow \mathrm{h}=\frac{(\mathrm{n}-1)\mathrm{R}}{\mathrm{n}}$