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(A) Let $M$ be the mass of the earth and $R$ be its radius.At a depth $h$ below the surface,the effective mass $M_1$ attracting the body is the mass o

Vedclass · Accepted Answer

$\frac{\sqrt{5} R - R }{2}$

Vedclass · Answer

(A) Let $M$ be the mass of the earth and $R$ be its radius.At a depth $h$ below the surface,the effective mass $M_1$ attracting the body is the mass of the sphere of radius $(R-h)$.$M_1 = \frac{M}{\frac{4}{3} \pi R^3} \cdot \frac{4}{3} \pi (R-h)^3 = M \frac{(R-h)^3}{R^3}$.The acceleration due to gravity at depth $h$ is $g_P = \frac{G M_1}{(R-h)^2} = \frac{G M (R-h)^3}{R^3 (R-h)^2} = \frac{G M (R-h)}{R^3}$.The acceleration due to gravity at height $h$ above the surface is $g_Q = \frac{G M}{(R+h)^2}$.Given that the weight is the same at both points,$g_P = g_Q$.$\frac{G M (R-h)}{R^3} = \frac{G M}{(R+h)^2}$.$(R-h)(R+h)^2 = R^3$.$(R-h)(R^2 + 2Rh + h^2) = R^3$.$R^3 + 2R^2h + Rh^2 - hR^2 - 2Rh^2 - h^3 = R^3$.$R^2h - Rh^2 - h^3 = 0$.Dividing by $h$ (since $h 
eq 0$): $R^2 - Rh - h^2 = 0$.$h^2 + Rh - R^2 = 0$.Using the quadratic formula $h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$h = \frac{-R \pm \sqrt{R^2 - 4(1)(-R^2)}}{2} = \frac{-R \pm \sqrt{5R^2}}{2}$.Since $h > 0$,we take the positive root: $h = \frac{\sqrt{5}R - R}{2}$.

The height $h$ 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):

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