At a place, value of $ g $ is less by $ 1 \% $ than its value on the surface of the Earth (Radius of Earth $ =6400 \mathrm{~km} $ ). The place is

1. $ 64 \mathrm{~km} $ below the surface of the Earth
2. $ 64 \mathrm{~km} $ above the surface of the Earth
3. $ 30 \mathrm{~km} $ above the surface of the Earth
4. $ 32 \mathrm{~km} $ below the surface of the Earth.

Here, $g$ is the gravitational acceleration on the surface and let $g_{d}$ be the $1$ % less gravity at the distance $d$ from the surface.

Radius of the earth $R=6400\ km$

$g_d=g(1-\frac{d}{R})$

Or $0.99g=g(1-\frac{d}{R})$

Or $0.99=1-\frac{d}{R}$

Or $\frac{d}{R}=1-0.99$

Or $\frac{d}{R}=0.01$

Or $d=0.01R$

Or $d=0.01\times 6400\ km$

Or $d=64\ km$

Therefore, at $64\ km$ below the surface of the earth $g$ is $1$ % less.

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Updated on: 10-Oct-2022

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