Question

A body weighs $72N$ on the surface of the Earth. What will be the force of gravity acting on the body at a height equal to half the radius of the Earth above the Earth's surface?

• A. $32N$
• B. $20N$
• C. $10N$
• D. $0N$

Answer 1

The correct option is A $32N$

Weight of the body at earth's surface is:
$W_e=\frac{GMm}{R^2}$ .......(1)
where, G-universal gravitational constant, M-mass of earth and R-radius of earth, m- Mass of the body

Weight of the body at a height $h$ above the earth's surface is given by
$W_e^{\prime }=\frac{GMm}{{(R+h)}^2}$

According to the question, $h=\frac{R}{2}$
$W_e^{\prime }=\frac{GMm}{{(R+\frac{R}{2})}^2}$
$W_e^{\prime }=\frac{4GMm}{9R^2}$ ....(2)

Dividing equation (2) by equation(1), we get
$\frac{W_e^{\prime }}{W_e}=\frac{4GMm/9R^2}{GMm/R^2}$
$\frac{W_e^{\prime }}{W_e}=\frac{4}{9}$
Given weight on earth's surface is, $W_e=72N$
$\frac{W_e^{\prime }}{72}=\frac{4}{9}$
$W_e^{\prime }=32N$