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
\(= \frac{4}{9} \times 63 = 29N\)