Consider a point mass $m$ at a height $h$ above the surface of the earth as shown in figure. This body placed at point $\mathrm{P}$ from the distance $h\left(r>\mathrm{R}_{\mathrm{E}}\right)$ from the surface of earth.

Radius of earth is $\mathrm{R}_{\mathrm{E}}$.

Its distance from the centre of the earth is $r=\mathrm{R}_{\mathrm{E}}+h$

The magnitude of force on the body,

$\mathrm{F}(h) =\frac{\mathrm{GM}_{\mathrm{E}} m}{\left(\mathrm{R}_{\mathrm{E}}+h\right)^{2}}$

$\therefore \frac{\mathrm{F}(h)}{m} =\frac{\mathrm{GM}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)^{2}}$

$\therefore$ Acceleration due to gravity at height $h$ from the surface of earth,

$g(h)=\frac{\mathrm{GM}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{F}}+h\right)^{2}}$ $……(1)$

Acceleration due to gravity on the surface of earth,

$g\left(\mathrm{R}_{\mathrm{E}}\right)=\frac{\mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}^{2}}$ $……(2)$

Taking ratio of equation $( 1 )$ and $(2)$,

$\frac{g(h)}{g\left(\mathrm{R}_{\mathrm{E}}\right)}=\frac{\mathrm{GM}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)^{2}} \times \frac{\mathrm{R}_{\mathrm{E}}^{2}}{\mathrm{GM}_{\mathrm{E}}}$

$\therefore g(h)=g\left(\frac{\mathrm{R}_{\mathrm{E}}^{2}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)^{2}}\right)\left[\because g\left(\mathrm{R}_{\mathrm{E}}\right)=g\right]$

$\therefore g(h) =g \frac{1}{\left(1+\frac{h}{\mathrm{R}_{\mathrm{E}}}\right)^{2}}$ $……(3)$

$=g\left(1+\frac{h}{\mathrm{R}_{\mathrm{E}}}\right)^{-2}$

$\therefore g(h) =g\left(1-\frac{2 h}{\mathrm{R}_{\mathrm{E}}}\right)$ $……(4)$

( $\because$ Taking two terms in using binomial expression)

Equation $(4)$ tells that for small heights $h$ above the value of $g$ decreased by a factor $\left(1-\frac{2 h}{\mathrm{R}_{\mathrm{E}}}\right)$

Equation $(3)$ can be used for any height, while

equation $(4)$ used only for $h<\mathrm{R}_{\mathrm{E}}$.