Assuming the Earth to be a sphere of uniform | vedclass

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(D) The weight of a body of mass $m$ at the Earth's surface is given by $W = mg = 250 \; N$.The acceleration due to gravity at a depth $d$ below the E

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$125$

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(D) The weight of a body of mass $m$ at the Earth's surface is given by $W = mg = 250 \; N$.The acceleration due to gravity at a depth $d$ below the Earth's surface is given by the formula $g' = g(1 - \frac{d}{R_e})$,where $R_e$ is the radius of the Earth.Given that the body is halfway down to the center,the depth is $d = \frac{R_e}{2}$.Substituting this value into the formula,we get:$g' = g(1 - \frac{R_e/2}{R_e}) = g(1 - \frac{1}{2}) = \frac{1}{2}g$.The weight of the body at depth $d$ is $W' = mg'$.Substituting $g' = \frac{1}{2}g$,we get $W' = m(\frac{1}{2}g) = \frac{1}{2}mg = \frac{1}{2}W$.Since $W = 250 \; N$,the new weight is $W' = \frac{1}{2} 	imes 250 = 125 \; N$.

Assuming the Earth to be a sphere of uniform mass density,how much would a body weigh (in $N$) halfway down to the centre of the Earth if it weighed $250 \; N$ on the surface?

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