Question

At what depth from the surface of the earth (in terms of radius of earth) the acceleration due to gravity will be $\dfrac{2g}{5}$ ?
A. $\dfrac{2}{5}R$
B. $\dfrac{3}{5}R$
C. $\dfrac{4}{25}R$
D. $\dfrac{9}{25}R$

Answer 1

Acceleration due to gravity or g does not remain constant above and below the surface of Earth instead it changes.
As the depth $d$ increases, the Earth can be thought of as being composed of a smaller sphere of radius $\left( {{R}_{E}}-d \right)$ with a spherical shell of thickness $d$ and since Mass of a sphere is proportional to the cube of its Radius, so consequently, mass of the Earth decreases.
Thus the Force on the point mass decreases and so does the acceleration due to gravity at depth $d$ which can be formulated as $g\left( d \right)=g\left( 1-\dfrac{d}{{{R}_{E}}} \right)$ where ${{R}_{E}}$ is the Radius of Earth.
Hence, the acceleration due to gravity at depth $d$ can be said to decrease by a factor of $\left( 1-\dfrac{d}{{{R}_{E}}} \right)$ .

Complete step-by-step answer:
Let the depth from the surface of Earth at which the acceleration due to gravity is $\dfrac{2g}{5}$ be $d$ .
We know, acceleration due to gravity at depth $d$ ,
$g\left( d \right)=g\left( 1-\dfrac{d}{{{R}_{E}}} \right)$
Putting the value of $g\left( d \right)=\dfrac{2g}{5}$ , we get
$\Rightarrow \dfrac{2}{5}=\left( 1-\dfrac{d}{{{R}_{E}}} \right)$
$\Rightarrow \dfrac{d}{{{R}_{E}}}=\left( 1-\dfrac{2}{5} \right)$
$\Rightarrow \dfrac{d}{{{R}_{E}}}=\dfrac{3}{5}$
$\Rightarrow d=\dfrac{3}{5}{{R}_{E}}$
Therefore, the depth $d$ at which the acceleration due to gravity is $\dfrac{2g}{5}$ is equal to$\dfrac{3}{5}{{R}_{E}}$.
Hence, the correct option would be (B) $\dfrac{3}{5}{{R}_{{}}}$

So, the correct answer is “Option B”.

Note: One is advised to must remember that just like the acceleration due to gravity decreases by a factor of $\left( 1-\dfrac{d}{{{R}_{E}}} \right)$ at depth $d$ , it also decreases by a factor of $\left( 1-\dfrac{2h}{{{R}_{E}}} \right)$ at a height h above the Surface of Earth and these changes are only subjected to Earth and not any other planet. And at the surface of Earth, $g$ has a stable value of $9.8m{{s}^{-2}}$ .