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 $\mathrm{3 R}$ above earth surface. Where $\mathrm{R}$ is Radius of earth (Take $\mathrm{R}=6400 \mathrm{~km}$ ). The depth $\mathrm{d}$ is equal to
Solution
The acceleration due to gravity $g$ at a distance $d$ below the surface of the earth is given by :
<br/><br/>$g_{d}=\frac{G M}{R^{3}}(R-d)$
(depth variation)
<br/><br/>where $G$ is the gravitational constant and $M$ is the mass of the Earth.
<br/><br/>At a height $3R$ above the surface of the Earth, the acceleration due to gravity $g_{h}$ is given by:
<br/><br/>$g_{h}=\frac{G M}{(R+3R)^{2}}$
<br/><br/>Given, $ g_{d}=4 g_{h} \\\\$ , so we can write :
<br/><br/>$$
\begin{aligned}
& \frac{G M}{R^{3}}(R-d)=4 \frac{G M}{(R+3 R)^{2}} \\\\
& \Rightarrow R-d=\frac{R}{4} \\\\
& \Rightarrow d=\frac{3 R}{4} \\\\
& \Rightarrow d=4800 \mathrm{~km}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Gravitation · Topic: Kepler's Laws
This question is part of PrepWiser's free JEE Main question bank. 129 more solved questions on Gravitation are available — start with the harder ones if your accuracy is >70%.