JEE MAIN - Physics (2023 - 31st January Morning Shift - No. 8)
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
5260 km
2560 km
640 km
4800 km
Explanation
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:
$$ g_{h}=\frac{G M}{(R+3R)^{2}} $$
Given, $ g_{d}=4 g_{h} \\\\$ , so we can write :
$$ \begin{aligned} & \frac{G M}{R^{3}}(R-d)=4 \frac{G M}{(R+3 R)^{2}} \\\\ & \Rightarrow R-d=\frac{R}{4} \\\\ & \Rightarrow d=\frac{3 R}{4} \\\\ & \Rightarrow d=4800 \mathrm{~km} \end{aligned} $$
$g_{d}=\frac{G M}{R^{3}}(R-d)$ (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:
$$ g_{h}=\frac{G M}{(R+3R)^{2}} $$
Given, $ g_{d}=4 g_{h} \\\\$ , so we can write :
$$ \begin{aligned} & \frac{G M}{R^{3}}(R-d)=4 \frac{G M}{(R+3 R)^{2}} \\\\ & \Rightarrow R-d=\frac{R}{4} \\\\ & \Rightarrow d=\frac{3 R}{4} \\\\ & \Rightarrow d=4800 \mathrm{~km} \end{aligned} $$
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