Question 109: Gravitation

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 $\mathrm{3 R}$ above earth surface. Where $\mathrm{R}$ is Radius of earth (Take $\mathrm{R}=6400 \mathrm{~km}$ ). The depth $\mathrm{d}$ is equal to

Accepted Answer

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}
$

Answer

5260 km

Answer

2560 km

Answer

640 km

Answer

4800 km

Gravitation

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

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