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(D) The weight of a body at height $h$ is given by $W_h = m g_h$ and on the surface is $W = m g$.Given that $W_h = \frac{W}{9}$,we have $m g_h = \frac

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$2$

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(D) The weight of a body at height $h$ is given by $W_h = m g_h$ and on the surface is $W = m g$.Given that $W_h = \frac{W}{9}$,we have $m g_h = \frac{m g}{9}$,which implies $g_h = \frac{g}{9}$.The acceleration due to gravity at height $h$ is $g_h = \frac{GM}{(R+h)^2}$ and on the surface is $g = \frac{GM}{R^2}$.Substituting these into the equation $g_h = \frac{g}{9}$:$\frac{GM}{(R+h)^2} = \frac{1}{9} \cdot \frac{GM}{R^2}$$\frac{1}{(R+h)^2} = \frac{1}{9 R^2}$Taking the square root on both sides:$\frac{1}{R+h} = \frac{1}{3 R}$$R + h = 3 R$$h = 2 R$Therefore,the height is $2 R$.

The height at which the weight of a body becomes $\left(\frac{1}{9}\right)^{\text{th}}$ of its weight on the surface of the Earth is $(R = \text{radius of Earth})$: (in $R$)

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