The height at which the weight of a body beco | vedclass

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Question

$g^{\prime}=\frac{g}{\left(1+\frac{h}{R}\right)^{2}}, g^{\prime}=\frac{g}{9}$

$\Rightarrow \frac{g}{9}=\frac{g}{\left(1+\frac{h}{R}\right)^{2}} \Ri

Vedclass · Accepted Answer

$h = 2R$

Vedclass · Answer

$g^{\prime}=\frac{g}{\left(1+\frac{h}{R}\right)^{2}}, g^{\prime}=\frac{g}{9}$

$\Rightarrow \frac{g}{9}=\frac{g}{\left(1+\frac{h}{R}\right)^{2}} \Rightarrow 1+\frac{h}{R}=3 \Rightarrow h=2 R$

The height at which the weight of a body becomes $\frac{1}{9} ^{th}$ its weight on the surface of earth (radius of earth is $R$)

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