A body weighs 48 N on the surface of the ear | vedclass

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Question

(B) The weight of a body on the surface of the earth is $W = mg = 48 \ N$.The acceleration due to gravity at a height $h$ above the surface of the ear

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(B) The weight of a body on the surface of the earth is $W = mg = 48 \ N$.The acceleration due to gravity at a height $h$ above the surface of the earth is given by $g' = g \left( \frac{R}{R+h} ight)^2$.Given $h = \frac{R}{3}$,we substitute this into the formula:$g' = g \left( \frac{R}{R + \frac{R}{3}} ight)^2 = g \left( \frac{R}{\frac{4R}{3}} ight)^2 = g \left( \frac{3}{4} ight)^2 = g \left( \frac{9}{16} ight)$.The weight at height $h$ is $W' = mg' = mg \left( \frac{9}{16} ight)$.Substituting $mg = 48 \ N$:$W' = 48 	imes \frac{9}{16} = 3 	imes 9 = 27 \ N$.

$A$ body weighs $48 \ N$ on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is (in $N$)

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