First slide

Acceleration due to gravity and its variation

Question

The height ‘h’ at which the weight of a body will be the same as that of 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):

Moderate

A

$\frac{\sqrt{3} R - R}{2}$
B

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

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

$\frac{R}{2}$

Solution

g at height $h = \frac{G M_{e}}{{(R_{e} + h)}^{2}}$
g at depth $d = \frac{G M_{e}}{R_{e}^{2}} (1 - \frac{d}{R_{e}})$
For Weight to be equal $\frac{G M_{e}}{R_{e}^{2} {(1 + \frac{h}{R_{e}})}^{2}} = \frac{G M_{e}}{R_{e}^{2}} (1 - \frac{d}{R_{e}})$
Given d = h
$\begin{array}{l} \Rightarrow 1 = {(1 + \frac{h}{R_{e}})}^{2} (1 - \frac{h}{R_{e}}) \\ \Rightarrow h^{2} + h R - R^{2} = 0 \\ c o m p a r i n g w i t h a x^{2} + b x + c = 0 \\ a = 1, b = R a n d c = - R^{2} \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ h = \frac{- R \pm \sqrt{R^{2} - 4 (1) (- R^{2})}}{2 (1)} \\ h = \frac{- R \pm \sqrt{R^{2} + 4 R^{2}}}{2} \\ \Rightarrow h = \frac{\sqrt{5} R - R}{2} \end{array}$

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