A body weighs $72 N$ on the surface of 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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Answer 1

At the surface of Earth, weight $W = 72 N$
$∴ F = m g = 72 N$
$m = \frac{7 2}{1 0} = 7.2 k g$
Now, at some height, h acceleration due to gravity is given as
$g^{'} = \frac{g}{( 1 + \frac{h}{R} ) ^{2}}$
Here, $h = \frac{R}{2}$
$g^{'} = \frac{g}{( 1 + \frac{1}{2} ) ^{2}} = \frac{4}{9} g = 4.44 m / s^{2}$
Now, the force experienced by the body at half of the radius of the earth is given by:
$F^{'} = m g^{'} = 7.2 \times 4.44 = 31.97 N \approx 32 N$

Answer 2

At the surface of Earth, weight

W = 72 N

∴ F = m g = 72 N

m = \frac{7 2}{1 0} = 7.2 k g

Now, at some height, h acceleration due to gravity is given as

g^{'} = \frac{g}{( 1 + \frac{h}{R} ) ^{2}}

Here,

h = \frac{R}{2}

g^{'} = \frac{g}{( 1 + \frac{1}{2} ) ^{2}} = \frac{4}{9} g = 4.44 m / s^{2}

Now, the force experienced by the body at half of the radius of the earth is given by:

F^{'} = m g^{'} = 7.2 \times 4.44 = 31.97 N \approx 32 N