The value of g on the earth’s surface is $980{\text{cm/s}}^{\text{2}}$ . Its value at a height of 64 km from the earth’s surface is

We know,

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

$M=\rho X V$, where $\rho$ is the density and $V$ is the volume.

$g=\frac{G{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3\mathrm{\rho }}{R^2}$

 $g=\frac{4}{3}\pi G\rho R$ 

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

Let $g_h$ be the acceleration due to gravity at height $h$.

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

$\frac{g_h}{g}={\left(\frac{6400}{6400+64}\right)}^2$

${\text{g}}_h=960.40{\text{cm/s}}^{\text{2}}$