The value of the acceleration due to gravity | vedclass

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Question

$g_{1}=\frac{G M}{\left(R+\frac{R}{2}\right)^{2}} \ldots(1)$

$g_{2}=\frac{G M(R-d)}{R^{3}} \ldots(2)$

$g_{1}=g_{2}$

$\frac{G M}{\left(\frac{3

Vedclass · Accepted Answer

$\frac{5}{9}$

Vedclass · Answer

$g_{1}=\frac{G M}{\left(R+\frac{R}{2}\right)^{2}} \ldots(1)$

$g_{2}=\frac{G M(R-d)}{R^{3}} \ldots(2)$

$g_{1}=g_{2}$

$\frac{G M}{\left(\frac{3 R}{2}\right)^{2}}=\frac{G M(R-d)}{R^{3}}$

$\Rightarrow \frac{4}{9}=\frac{( R - d )}{ R }$

$4 R=9 R-9 d$

$5 R =9 d \Rightarrow \frac{ d }{ R }=\frac{5}{9}$

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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