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(C) The weight of a body at the surface of the earth is $W = mg$,where $g$ is the acceleration due to gravity at the surface.At a height $h$ above the

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$4W/9$

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

$A$ body has weight $W$ newton at the surface of the earth. Its weight at a height equal to half the radius of the earth will be:

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