The percentage decrease in the weight of a bo | vedclass

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(A) The acceleration due to gravity at a height $h$ above the surface of the earth is given by $g_h = g(1 - 2h/R)$, where $R$ is the radius of the ear

Vedclass · Accepted Answer

$1.5$

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(A) The acceleration due to gravity at a height $h$ above the surface of the earth is given by $g_h = g(1 - 2h/R)$, where $R$ is the radius of the earth.The weight of a body is $W = mg$, so the weight at height $h$ is $W_h = mg_h = mg(1 - 2h/R)$.The decrease in weight is $\Delta W = W - W_h = mg - mg(1 - 2h/R) = mg(2h/R)$.The percentage decrease in weight is $(\Delta W / W) 	imes 100 = (2h/R) 	imes 100$.Given $h = 48 	ext{ km}$ and $R = 6400 	ext{ km}$.Percentage decrease $= (2 	imes 48 / 6400) 	imes 100 = (96 / 6400) 	imes 100 = 96 / 64 = 1.5\%$.

The percentage decrease in the weight of a body when taken to a height of $48 \text{ km}$ above the surface of the earth is (Radius of the earth is $6400 \text{ km}$) (in $\%$)

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