Question

A particle weighs 120 N on the surface of the earth. At what height above the earth's surface will be its weights be 30N ? Radius of the earth is 6400 km.

Answer 1

The acceleration due to gravity ($g$) at a height ($h$) above the surface of the earth is given by,

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

where $g$ = Acceleration due to gravity at the surface of the earth

$R$ = Radius of the earth

As weight gets one fourth the value of $g$ should also change accordingly since the mass of the particle remains constant i.e. $g^{\prime }=\frac{g}{4}$.

$⟹\frac{g}{4}=g\left(1-\frac{2h}{R}\right)$

$⟹\frac{2h}{R}=\frac{3}{4}$

$⟹h=\frac{3R}{8}$

$=\frac{3\times 6400}{8}$

$=2400km$.