Derive an expression for the variation of acceleration due to gravity
a) above and
b) below the surface of the earth

a) Variation of “g” with above the surface of earth : Consider a body of mass “m” at a height ‘h’ from the surface of the earth of mass
“M” and radius “R”
The acceleration due to gravity on the surface of earth $\mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{R}}^2}$.... (1)

The acceleration due to gravity at height from the earth ${\mathrm{g}}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h}{)}^2}$.... (2)

$\begin{array}{l}\frac{\left(2\right)}{\left(1\right)}\Rightarrow \frac{{\mathrm{g}}_{\mathrm{h}}}{\mathrm{g}}=\frac{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h}{)}^2}}{\frac{\mathrm{GM}}{{\mathrm{R}}^2}}=\frac{{\mathrm{R}}^2}{(\mathrm{R}+\mathrm{h}{)}^2}\\ \frac{\mathrm{g}}{\mathrm{g}}=\frac{1}{{\left(\frac{\mathrm{R}+\mathrm{h}}{\mathrm{R}}\right)}^2}\Rightarrow \mathrm{g}=\mathrm{g}{\left[\frac{\mathrm{R}+\mathrm{h}}{\mathrm{R}}\right]}^{-2}\\ =\mathrm{g}{\left[1+\frac{\mathrm{h}}{\mathrm{R}}\right]}^{-2}=\mathrm{g}\left(1-\frac{2\mathrm{h}}{\mathrm{R}}\right)\end{array}$

The “g” decreases with height.

b) Variation of “g” with Depth: Consider a body of mass “m” is at a depth ‘d’ from the surface of the earth as shown in fig
The acceleration due to gravity $\left(\mathrm{g}\right)=\frac{\mathrm{GM}}{{\mathrm{R}}^2}$

 $\begin{array}{l}\text{ Mass }=\mathrm{V}\times \mathrm{d}=\frac{4}{3}{\mathrm{\pi R}}^3\times \mathrm{\rho }\\ \mathrm{g}=\frac{\mathrm{G}\frac{4}{3}{\mathrm{\pi R}}^3\mathrm{\rho }}{{\mathrm{R}}^2}=\mathrm{G}\frac{4}{3}\mathrm{\pi R\rho }...........\left(2\right)\end{array}$

The acceleration due to gravity in distance R–d from center of earth is

$\begin{array}{l}{\mathrm{g}}_{\mathrm{d}}=\mathrm{G}\frac{4}{3}\mathrm{\pi }(\mathrm{R}-\mathrm{d}).........\left(2\right)\\ \frac{{\mathrm{g}}_{\mathrm{d}}}{\mathrm{g}}=\frac{\mathrm{G}\frac{\mathrm{A}}{3}\mathrm{\pi }(\mathrm{R}-\mathrm{d})\mathrm{\varnothing }}{\mathrm{G}\frac{4}{3}\mathrm{\pi R\varphi }}\\ \frac{{\mathrm{g}}_{\mathrm{d}}}{\mathrm{g}}=\frac{\mathrm{R}-\mathrm{d}}{\mathrm{R}}\Rightarrow {\mathrm{g}}_{\mathrm{d}}=\mathrm{g}\left[\frac{\mathrm{R}-\mathrm{d}}{\mathrm{R}}\right]\\ {\mathrm{g}}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)\text{ "g" decreases with depth }\end{array}$