A body weighs 72 N on the surface of the ear | vedclass

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(C) The weight of a body on the surface of the earth is given by $W_{s} = m g_{s} = 72 \ N$.The acceleration due to gravity at a height $h$ above the

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(C) The weight of a body on the surface of the earth is given by $W_{s} = m g_{s} = 72 \ N$.The acceleration due to gravity at a height $h$ above the surface of the earth is given by $g_{h} = \frac{g_{s}}{(1 + \frac{h}{R})^{2}}$,where $R$ is the radius of the earth.Given $h = \frac{R}{2}$,we substitute this into the formula:$g_{h} = \frac{g_{s}}{(1 + \frac{R/2}{R})^{2}} = \frac{g_{s}}{(1 + 0.5)^{2}} = \frac{g_{s}}{(1.5)^{2}} = \frac{g_{s}}{2.25} = \frac{g_{s}}{9/4} = \frac{4}{9} g_{s}$.The gravitational force (weight) at height $h$ is $W_{h} = m g_{h} = m 	imes (\frac{4}{9} g_{s}) = \frac{4}{9} W_{s}$.Substituting $W_{s} = 72 \ N$:$W_{h} = \frac{4}{9} 	imes 72 = 4 	imes 8 = 32 \ N$.

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

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