A body weighs 72 N on the surface of the Earth. What will be the force of gravit

Question

A body weighs 72 N 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?

Options:

A . 32 N

B . 20 N

C . 10 N

D . 0 N

Answer: Option A
:
A
Weight of the body at earth's surface is:

W_{e} = \frac{G M m}{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}^{'} = \frac{G M m}{{(R + h)}^{2}}

According to the question,

h = \frac{R}{2}

W_{e}^{'} = \frac{G M m}{{(R + \frac{R}{2})}^{2}}

W_{e}^{'} = \frac{4 G M m}{9 R^{2}}

....(2)
Dividing equation (2) by equation(1), we get

\frac{W_{e}^{'}}{W_{e}} = \frac{4 G M m / 9 R^{2}}{G M m / R^{2}}

\frac{W_{e}^{'}}{W_{e}} = \frac{4}{9}

Given weight on earth's surface is,

W_{e} = 72 N

\frac{W_{e}^{'}}{72} = \frac{4}{9}

W_{e} = 32 N

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