A body weighs 72N on the surface of the earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth from the surface ?

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32 N

16 N

28 N

72 N

answer is A.

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Detailed Solution

$mg = 72 N$ (body weight on the surface)
$g = \frac{GM}{R^{2}}$
At a height $H = \frac{R}{2}, g^{'} = \frac{GM}{{(R + \frac{R}{2})}^{2}} = \frac{4 GM}{9 R^{2}}$

Body weight at height $H = \frac{R}{2}$
$mg = m \times \frac{4}{9} \frac{GM}{R^{2}} = m \times \frac{4}{9} \times g = \frac{4}{9} mg = \frac{4}{9} \times 72 = 32 N$

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