A body weighs 250,N on the surface of the ear | vedclass

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(A) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula $g' = g(1 - \frac{d}{R})$,where $g$ is the a

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$125$

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(A) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula $g' = g(1 - \frac{d}{R})$,where $g$ is the acceleration due to gravity at the surface,$R$ is the radius of the earth,and $d$ is the depth.Given that the body is halfway down to the center of the earth,the depth $d = \frac{R}{2}$.Substituting this value into the formula,we get $g' = g(1 - \frac{R/2}{R}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.The weight of the body at depth $d$ is $W' = mg' = m(\frac{g}{2}) = \frac{mg}{2}$.Since the weight at the surface is $W = mg = 250\,N$,the weight at half the depth is $W' = \frac{250}{2} = 125\,N$.

$A$ body weighs $250\,N$ on the surface of the earth. How much will it weigh halfway down to the center of the earth (in $,N$)?

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