Question

Calculate the distance below and above the surface of the earth, at which the value of acceleration due to gravity becomes $1/4$th that at earth's surface?

Answer 1

The acceleration due to gravity at depth $x$ below the surface of earth is
$g_x=g(1-\frac{x}{R_e})$ or $\frac{g_x}{g}=(1-\frac{x}{R_e})$
Given $\frac{g_x}{g}=\frac{1}{4}$ $\therefore \frac{1}{4}=1-\frac{x}{R_e}$
or $\frac{x}{R_e}=1-\frac{1}{4}=\frac{3}{4}$ or $x=\frac{3}{4}{R}_e$
The acceleration due to gravity at height $h$ is
$g_h=\frac{g}{{(1+\frac{h}{R_e})}^2}$
or $\frac{g_h}{g}=\frac{1}{{(1+\frac{h}{R_e})}^2}$
Given, $\frac{g_h}{g}=\frac{1}{4}$
$\therefore \frac{1}{4}=\frac{1}{{(1+\frac{h}{R_e})}^2}$
$\Rightarrow 1+\frac{h}{R_e}=2$
or $\frac{h}{R_e}=1$
or $h=R_e$