Question

The value of ‘g’ at the depth from the ground goes on __________ .
A) increasing
B) fluctuating
C) decreasing
D) varying

Answer 1

First we need to know what does the acceleration due to gravity on the surface depend upon. Hence Further we need to obtain a relationship of acceleration due to gravity at a distance d inside the Earth’s core itself. Hence further comparing with respect to the radius of the Earth, will enable us to determine the correct answer.

Formula used:
${{g}_{d}}=g\left( 1-\dfrac{d}{R} \right)$

Complete step-by-step answer:
Consider the Earth to be a sphere of mass ‘M’ and Radius R and center O. The acceleration due to gravity ‘g’ on the surface of the Earth any point of the Earth will be,
$g=\dfrac{GM}{{{R}^{2}}}$ where ‘G’ is the gravitational constant.
Assuming the Earth to be a homogenous sphere of volume ‘V’ and average density $\rho$, them its total mass will be,
$\begin{aligned} & M=\rho V \\\ & \Rightarrow M=\dfrac{4}{3}\pi {{R}^{3}}\rho \\\ \end{aligned}$
Hence the expression for acceleration due to gravity now becomes,
$\begin{aligned} & g=\dfrac{GM}{{{R}^{2}}} \\\ & \Rightarrow g=\dfrac{G\dfrac{4}{3}\pi {{R}^{3}}\rho }{{{R}^{2}}} \\\ & \therefore g=\dfrac{4G\pi R\rho }{3}......(1) \\\ \end{aligned}$
Let us say we take a point A which is at a distance ‘d’ from the surface of the Earth. Hence the mass of the Earth (m) enclosed below the distance ‘d’ from the surface is equal to,
$m=\dfrac{4}{3}\pi {{(R-d)}^{3}}\rho$
Hence the acceleration due to gravity ${{g}_{d}}$ at the point A is equal to,
$\begin{aligned} & {{g}_{d}}=\dfrac{Gm}{{{(R-d)}^{2}}} \\\ & \Rightarrow {{g}_{d}}=\dfrac{G\dfrac{4}{3}\pi {{(R-d)}^{3}}\rho }{{{(R-d)}^{2}}}=G\dfrac{4}{3}\pi (R-d)\rho \\\ & \therefore {{g}_{d}}=\dfrac{4}{3}G\pi (R-d)\rho .....(2) \\\ \end{aligned}$
Taking the ratio of equation 1 and 2 we get,
$\begin{aligned} & \dfrac{g}{{{g}_{d}}}=\dfrac{\dfrac{4G\pi R\rho }{3}}{\dfrac{4}{3}G\pi (R-d)\rho }=\dfrac{R}{R-d} \\\ & \Rightarrow {{g}_{d}}=g\left( \dfrac{R-d}{R} \right) \\\ & \therefore {{g}_{d}}=g\left( 1-\dfrac{d}{R} \right) \\\ \end{aligned}$
From the above equation we can conclude that as the value of ‘d’ increases, the acceleration due to gravity also decreases.

So, the correct answer is “Option C”.

Note: From the above expression we can clearly see that the acceleration due to gravity decreases with the increase in depth. This is why the acceleration due to gravity is less in mines than that of Earth’s surface. It is also to be noted that the acceleration due to gravity with altitude as well.