At a certain depth d below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height 3 R above earth surface. Where R is Radius of earth (Take R = 6400 km

Question

At a certain depth d below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height 3 R above earth surface. Where R is Radius of earth (Take R = 6400 km ). The depth d is equal to

TestHub · Accepted Answer

The magnitude of acceleration due to gravity $g_{1}$ at a depth of $d$ below the surface of Earth is

$g_{1} = G \frac{M}{R^{3}} (R - d) = G \frac{M}{R^{2}} (1 - \frac{d}{R})$ ....(1)

The magnitude of acceleration due to gravity $g_{2}$ at a height $d$ above the surface of Earth is

$g_{2} = G \frac{M}{{(R + d)}^{2}}$ ....(2)

Given here, $g_{1} = 4 g_{2}$

$\frac{G M}{R^{2}} (1 - \frac{d}{R}) = \frac{4 G M}{{(R + 3 R)}^{2}}$

$\Rightarrow \frac{G M}{R^{2}} [1 - \frac{d}{R}] = \frac{4 G M}{{(4 R)}^{2}}$

$\Rightarrow 1 - \frac{d}{R} = \frac{1}{4} \Rightarrow \frac{d}{R} = \frac{3}{4} \Rightarrow d = \frac{3}{4} R$

$\Rightarrow d = 4800 km$

Physics - Gravitation Question with Solution | TestHub

Options:

Doubts & Discussion