Question

A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth ?

Answer 1

Weight of the body, W = 63 N
Acceleration due to gravity at height h from the Earth’s surface is given by the relation:
$g' = \frac{g}{(\frac{1+h}{R_e})^2}$
Where,

g = Acceleration due to gravity on the Earth’s surface
Re = Radius of the Earth

for $h = \frac{R_e}{2}$
$g' =\frac{ g}{ (1+\frac{R_e}{2 x R_e})^2} =\frac{ g}{ (1+\frac{1}{2})^2} =\frac{ 4}{9} g$

Weight of a body of mass m at height h is given as:
W' = mg
= $m\times\frac{4}{9} g = \frac{4}{9}\times$ mg
= $\frac{4}{9}$ W
$= \frac{4}{9} \times 63 = 29N$