The value of the acceleration due to gravity | vedclass

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(D) The acceleration due to gravity at a height $h$ is given by $g_{h} = \frac{GM}{(R+h)^{2}}$.Given $h = \frac{R}{2}$,we have $g_{1} = \frac{GM}{(R +

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$\frac{5}{9}$

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(D) The acceleration due to gravity at a height $h$ is given by $g_{h} = \frac{GM}{(R+h)^{2}}$.Given $h = \frac{R}{2}$,we have $g_{1} = \frac{GM}{(R + R/2)^{2}} = \frac{GM}{(3R/2)^{2}} = \frac{4GM}{9R^{2}} \ldots(1)$The acceleration due to gravity at a depth $d$ is given by $g_{d} = g(1 - \frac{d}{R}) = \frac{GM}{R^{2}}(1 - \frac{d}{R}) = \frac{GM(R-d)}{R^{3}} \ldots(2)$Since $g_{1} = g_{d}$,we equate equations $(1)$ and $(2)$:$\frac{4GM}{9R^{2}} = \frac{GM(R-d)}{R^{3}}$$\frac{4}{9} = \frac{R-d}{R}$$\frac{4}{9} = 1 - \frac{d}{R}$$\frac{d}{R} = 1 - \frac{4}{9} = \frac{5}{9}$

The value of the acceleration due to gravity is $g_{1}$ at a height $h = \frac{R}{2}$ ($R$ = radius of the earth) from the surface of the earth. It is again equal to $g_{1}$ at a depth $d$ below the surface of the earth. The ratio $\left(\frac{d}{R}\right)$ equals

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