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(D) Let $R$ be the radius of the Earth. The diameter of the Earth is $2R$.Given that the height $h$ of point $P$ above the surface is equal to the dia

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$g / 9$

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(D) Let $R$ be the radius of the Earth. The diameter of the Earth is $2R$.Given that the height $h$ of point $P$ above the surface is equal to the diameter of the Earth,so $h = 2R$.The distance of point $P$ from the center of the Earth is $r = R + h = R + 2R = 3R$.The acceleration due to gravity at the surface of the Earth is $g = \frac{GM}{R^2}$.The acceleration due to gravity at point $P$ at a distance $r$ from the center is $g' = \frac{GM}{r^2}$.Substituting $r = 3R$ into the equation,we get:$g' = \frac{GM}{(3R)^2} = \frac{GM}{9R^2}$.Since $g = \frac{GM}{R^2}$,we can write:$g' = \frac{1}{9} \left( \frac{GM}{R^2} \right) = \frac{g}{9}$.

The height of any point $P$ above the surface of the Earth is equal to the diameter of the Earth. The value of acceleration due to gravity at point $P$ will be: (Given $g = $ acceleration due to gravity at the surface of the Earth)

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