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

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Question

$W_{s}=m g_{s}=72 N$

$W_{h}=m g_{h}=\frac{m g_{s}}{\left(1+\frac{h}{R}\right)^{2}}=\frac{72 N}{\left(1+\frac{R / 2}{R}\right)^{2}}=\frac{72}{9 / 4}

Vedclass · Accepted Answer

$32$

Vedclass · Answer

$W_{s}=m g_{s}=72 N$

$W_{h}=m g_{h}=\frac{m g_{s}}{\left(1+\frac{h}{R}\right)^{2}}=\frac{72 N}{\left(1+\frac{R / 2}{R}\right)^{2}}=\frac{72}{9 / 4}$

$W_{h}=32 N$

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

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