A body weighs 63 ; N on the surface of the ea | vedclass

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(C) Weight of the body on the surface of the earth,$W = mg = 63 \; N$.The acceleration due to gravity at a height $h$ from the Earth's surface is give

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(C) Weight of the body on the surface of the earth,$W = mg = 63 \; N$.The acceleration due to gravity at a height $h$ from the Earth's surface is given by the formula:$g' = g \left( 1 + \frac{h}{R_e} ight)^{-2}$Given that the height $h = \frac{R_e}{2}$,we substitute this into the formula:$g' = g \left( 1 + \frac{R_e/2}{R_e} ight)^{-2} = g \left( 1 + \frac{1}{2} ight)^{-2} = g \left( \frac{3}{2} ight)^{-2} = g \left( \frac{2}{3} ight)^2 = \frac{4}{9} g$The gravitational force (weight) at height $h$ is $W' = mg'$.$W' = m \left( \frac{4}{9} g ight) = \frac{4}{9} (mg)$Substituting the value of $W = 63 \; N$:$W' = \frac{4}{9} 	imes 63 = 4 	imes 7 = 28 \; N$.

$A$ body weighs $63 \; N$ on the surface of the earth. What is the gravitational force (in $N$) on it due to the earth at a height equal to half the radius of the earth?

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