If the value of g at the surface of the earth is $9.8{\text{ms}}^{-2}$ then the value of g at a place 480 km above them surface of the earth will be (radius of earth = 6400 km)

Given:
${\text{g}}_{\text{e}}=9.8{\text{m/s}}^{\text{2}}$ 

$\text{h}=48\text{0km,}\text{R}=6400\text{km,}{\text{g}}_{\text{n}}=\text{?}$

We know, 

$g=\frac{GM}{R^2}$

Since G and M are constant,

$g \propto \frac{1}{R^2}$

$\therefore \frac{g_h}{g_e}=\frac{{R_e}^2}{{R_h}^2}$

$\because {\mathrm{R}}_{\mathrm{h}}= \mathrm{R} + \mathrm{h}$

${\text{g}}_h={\text{g}}_{\text{e}}{\left(\frac{\text{R}}{\text{R}+\text{h}}\right)}^{\text{2}}$

$=9.8{\left(\frac{6400}{6400+480}\right)}^2$

$=9.8{\left(\frac{6400}{6880}\right)}^2$

$=9.8(0.86)\Rightarrow 8.4{\text{m/s}}^{\text{2}}$