QJEEM23Ph5b43: JEE Main 2023 Physics - Class 11 JEEM23Ph5b43 Solution: At a certain depth "d " below surface of earth, value of…

The acceleration due to gravity $$g$$ at a distance $$d$$ below the surface of the earth is given by :

$g_{d}=\frac{G M}{R^{3}}(R-d)$ (depth variation)

where $$G$$ is the gravitational constant and $$M$$ is the mass of the Earth.

At a height $$3R$$ above the surface of the Earth, the acceleration due to gravity $g_{h}$ is given by:

$$g_h=\frac{GM}{(R+3R{)}^2}$$

Given, $g_{d}=4 g_{h} <span>\</span><span>\</span>$ , so we can write :

$$\begin{array}{rl}& \frac{GM}{R^3}(R-d)=4\frac{GM}{(R+3R{)}^2}\\ \\ & \Rightarrow R-d=\frac{R}{4}\\ \\ & \Rightarrow d=\frac{3R}{4}\\ \\ & \Rightarrow d=4800\mathrm{km}\end{array}$$