Imagine earth to be a solid sphece of mass M | vedclass

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Question

$\Rightarrow g(h)=\frac{g}{\left(1+\frac{h}{R}ight)^2}$

$	herefore \frac{g}{4}=\frac{g}{\left(1+\frac{h}{R}ight)^2}$

$	herefore 4=\left(1+

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

$\Rightarrow g(h)=\frac{g}{\left(1+\frac{h}{R}ight)^2}$

$	herefore \frac{g}{4}=\frac{g}{\left(1+\frac{h}{R}ight)^2}$

$	herefore 4=\left(1+\frac{h}{R}ight)^2$

$	herefore 2=1+\frac{h}{R}$

$	herefore \frac{h}{R}=1 \quad \ldots(1)$

$\Rightarrow g(d)=g\left(1-\frac{d}{R}ight)^2$

$	herefore \frac{g}{4}=g\left(1-\frac{d}{R}ight)^2$

$	herefore \frac{1}{4}=1-\frac{d}{R}$

$	herefore \frac{d}{R}=1-\frac{1}{4}$

$	herefore \frac{d}{R}=\frac{3}{4} \quad \ldots(2)$

$	herefore$ Taking ratio of equ.$(1)$ and $(2)$

$\frac{h}{d}=\frac{4}{3}$

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