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

Question

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Answer 1

Weight of body on the surface of the earth $= m g = 72 N$
Acceleration due to gravity at height h is
$g_{h} = \frac{g R _{E}^{2}}{( R _{E} + h ) ^{2}}$

Substitute $h = \frac{R _{E}}{2}$ in above expression:
$g_{h} = \frac{g R _{E}^{2}}{( R _{E} + \frac{R _{E}}{2} ) ^{2}}$
$= \frac{4}{9} g$

Gravitational force on body at height h is
$F = m g_{h}$

$= m \times \frac{4}{9} g = \frac{4}{9} \times m g$

$= \frac{4}{9} \times 72 N = 32 N$

Answer 2

Weight of body on the surface of the earth

= m g = 72 N

Acceleration due to gravity at height h is

g_{h} = \frac{g R _{E}^{2}}{( R _{E} + h ) ^{2}}

Substitute

h = \frac{R _{E}}{2}

in above expression:

g_{h} = \frac{g R _{E}^{2}}{( R _{E} + \frac{R _{E}}{2} ) ^{2}}

= \frac{4}{9} g

Gravitational force on body at height h is

F = m g_{h}

= m \times \frac{4}{9} g = \frac{4}{9} \times m g

= \frac{4}{9} \times 72 N = 32 N