Answer to Question #132906 in Mechanics | Relativity for Viccy

Question

Explain how varies gravitational acceleration with depth.

Answer 1

Acceleration due to gravity on earth's surface is given by:

g=$\dfrac{GM}{R^2}$ .................equation (a)

considering $\sigma$ to be density of material of the earth

mass=volume*density

M =$\frac{4}{3}$ $\pi$ R³ *$\sigma$

substituting in equation (a)

g= $\frac{4}{3}$ $\pi$ R³*$\sigma$ *$\frac{G}{R^2}$ ....................equation(b)

=$\frac{4}{3}$ $\pi$GR $\sigma$

considering that the body is to be taken below the surface of earth.

Then,the acceleration due to gravity at the depth below the surface would be;

g_d=$\frac{4}{3}\pi$ (R-d)$\sigma$ .......................equation (c)

hence,deviding equation (c) and (b)

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

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

This expression of acceleration due to gravity at the depth below the earth surface.

the equation is of decreasing function as we move down to radius of earth(R).

At the center the earth d=R reducing the bracket function to zero and acceleration equating to zero.

from this, it indicates that acceleration due to gravity at the center of earth is 0.