Discuss the variation of acceleration due to gravity with depth. What happens to g at the center of the Earth.

Hint: Acceleration due to gravity depends on distance from the center of the Earth. Replace R by (R-d) as distance from center decreases as we move towards center.

Complete step by step answer:
Acceleration due to gravity is acceleration gained by an object while moving under the influence of gravitational force.
On the Earth’s surface it is given as
$g=\dfrac{GM}{{{R}^{2}}}$
Where, G is universal constant of gravitation
     M is Mass of earth
     R is radius of earth
If $\rho$is the density of earth then
$\begin{align}
  & M=Volume\times \rho \\
 & M=\dfrac{4}{3}\pi {{R}^{3}}\rho \\
\end{align}$
Therefore,
$g=\dfrac{G}{{{R}^{2}}}\times \dfrac{4}{3}\pi {{R}^{3}}\rho $
$g=\dfrac{4}{3}\pi R\rho G$…...….……………………………….(1)
Now if a body is taken to depth d then
${{g}_{d}}=\dfrac{4}{3}\pi (R-d)\rho G$……………………………….(2)
Dividing equation (2) by equation (1)
\[\begin{align}
  & \dfrac{{{g}_{d}}}{g}=\dfrac{R-d}{R} \\
 & {{g}_{d}}=g\dfrac{(R-d)}{R} \\
 & {{g}_{d}}=g(1-\dfrac{d}{R}) \\
\end{align}\]
This an expression for acceleration due to gravity at depth d.
From equation (2) we can say that as depth increases acceleration due to gravity decreases and at the center as d=R, acceleration due gravity becomes zero.
Therefore, at center of Earth as d=R
\[{{g}_{d}}=0\]
Acceleration due to gravity decreases linearly with increase in depth.

Additional Information:
Acceleration due to gravity also decreases as we move above Earth’s surface. Thus, acceleration due to gravity is maximum at the Earth’s surface and is zero at the center of the Earth and at infinite distance above the Earth’s surface.

Note: No need to remember the whole derivation, just remember to replace R by (R-d) where d is depth. Further calculation is just a simple mathematical manipulation. Once you have obtained expression for acceleration due to gravity at depth d then compare it with acceleration due to gravity at earth’s surface.