Question

At which height from the earth's surface, acceleration due to gravity is decreased by $75\%$ of its value at earth's surface.

Answer 1

Acceleration due to gravity at height $h$ from earth surface is:

$g^{\prime }=g\left(1-\frac{2h}{R}\right)$

$\frac{g^{\prime }}{g}=\left(1-\frac{2h}{R}\right)$

$\frac{75}{100}=\left(1-\frac{2h}{R}\right)$

$\frac{25}{100}=\frac{2h}{R}$

$h=\frac{R}{8}=\frac{6.4\times {10}^6}{8}=8\times {10}^{5}m=800km$

Thus, at $800km$ above earth surface $g$ becomes $75\%$ of its value on earth surface.