Question

The weight of a body on the surface of the earth is $ \, 63 \, N$ . What is the gravitational force on it due to the earth at a height equal to half the radius of earth?

• A. 28 N
• B. 23 N
• C. 12 N
• D. 67 N

Answer 1

Answer: (A)

Given h $=\frac{R_{e}}{2}$
Acceleration due to gravity at altitude $h$ is given by
$( g )^{\prime}=\frac{ g }{\left(1+\frac{h}{(R)_{e}}\right)^{2}}$
$=\frac{ g }{\left(1+\frac{(R)_{e} / 2}{(R)_{e}}\right)^{2}} $
$=\frac{ g }{\left(1+\frac{1}{2}\right)^{2}} $
$=\frac{ g }{(3 / 2)^{2}}= g ^{\prime}=\frac{4}{9} g \ldots$ (i)
Weight of the body at earth's surface
$w=m g=63 \,N \ldots $..(ii)
Weight of the body at altitude $h=R_{e} / 2$
$w^{\prime}= m g^{\prime}=\frac{4}{9} mg \ldots $ (iii)
Using Eq. (ii), we get
$w^{\prime}=\frac{4}{9} \times 63$
$w^{\prime}=28 \,N$