Question

Why does the value of 'g' decrease as we go deep inside a mine (or the Earth)?

Answer 1

According to Law of Gravitation:
$F\left(r\right)=G\frac{Mm}{r^2}$
Now, if Earth is considered uniformly dense of density ${\rho }_0$ then mass of Earth within the radius,r is given by:
$M\left(r\right)=\frac{4}{3}\pi r^3{\rho }_0$ (Mass = Volume $\times$ Density)
Now, $F\left(r\right)=G\frac{Mm}{r^2}=mg$
So, acceleration due to gravity will be, $g=\frac{F\left(r\right)}{m}$
$g=\frac{G}{r^2}\frac{4}{3}\pi r^3{\rho }_0=\left(\frac{4}{3}G\pi {\rho }_0\right)r$
This shows that the value of ${}^{\prime }{g}^{\prime }$ depends linearly on $r$. It decreases as $r$ decreases (deep inside a mine).