A body weighs 144 , N at the surface of the e | vedclass

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(D) Given that,the weight of the body at the surface of the earth is $W = mg = 144 \, N$.The acceleration due to gravity at a height $h$ above the sur

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$9$

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(D) Given that,the weight of the body at the surface of the earth is $W = mg = 144 \, N$.The acceleration due to gravity at a height $h$ above the surface of the earth is given by the formula:$g' = g \left( \frac{R}{R + h} \right)^2$Given $h = 3R$,we substitute this into the equation:$g' = g \left( \frac{R}{R + 3R} \right)^2 = g \left( \frac{R}{4R} \right)^2 = g \left( \frac{1}{4} \right)^2 = \frac{g}{16}$The weight of the body at height $h$ is $W' = mg'$.$W' = m \left( \frac{g}{16} \right) = \frac{W}{16}$Substituting the value of $W = 144 \, N$:$W' = \frac{144}{16} = 9 \, N$.Therefore,the weight of the body at a height of $3R$ is $9 \, N$.

$A$ body weighs $144 \, N$ at the surface of the earth. When it is taken to a height of $h = 3R$,where $R$ is the radius of the earth,it would weigh .......... $N$.

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