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The height ${ }^{\prime} h ^{\prime}$ at which the weight of a body will be the same as that at the same depth $'h'$ from the surface of the earth is (Radius of the earth is $R$ and effect of the rotation of the earth is neglected):
$\frac{\sqrt{5} R - R }{2}$
$\frac{\sqrt{5}}{2} R - R$
$\frac{ R }{2}$
$\frac{\sqrt{3} R - R }{2}$
Solution
$M =$ mass of earth
$M _{1}= mass$ of shaded portion
$R =$ Radius of earth
$M_{1}=\frac{M}{\frac{4}{3} \pi R^{3}} \cdot \frac{4}{3} \pi(R-h)^{3}$
$=\frac{M(R-h)^{3}}{R}$
Weight of body is same at $P$ and $Q$
i.e. $mg _{ P }= mg _{ Q }$
$g _{ P }= g _{ Q }$
$\frac{G M_{1}}{(R-h)^{2}}=\frac{G M}{(R+h)^{2}}$
$\frac{G M(R-h)^{3}}{(R-h)^{2} R^{3}}=\frac{G M}{(R+h)^{2}}$
$( R – h )( R + h )^{2}= R ^{3}$
$R^{3}-h R^{2}-h^{2} R-h^{3}+2 R^{2} h-2 R h^{2}=R^{3}$
$R^{2}-R h^{2}-h^{3}=0$
$R^{2}-R h-h^{2}=0$
$h ^{2}+ Rh – R ^{2}=0 \Rightarrow h =\frac{- R \pm \sqrt{ R ^{2}+4 R ^{2}}}{2}$
ie $h =\frac{- R +\sqrt{5} R }{2}=\left(\frac{\sqrt{5}-1}{2}\right) R$
