Variation of g with depth

Let's suppose that our earth is spherical.

Somehow, if anyone could reach the center of the spherical earth, he might feel that his weight has become zero.

Since, the acceleration due to gravity is zero at the center of earth.

Let's understand how the acceleration due to gravity becomes zero at the center of earth.

So, we need to derive the variation of acceleration due to gravity with depth of the earth.

Let the shape of earth be spherical having mass $M_e$ and density $\rho$.

Let's also take the center of earth at $O$ and radius as $R$.

If a small body of mass $m$ is placed at the surface of earth, then, acceleration due to gravity is given by,

We will write this expression in the form of density of the earth $\rho$. So, the mass of earth can be written as,

Thus, $g$ can also be given by,

Now, let's assume that the body is taken to a depth of $d$ from the surface of the earth.

So, the effective mass of the earth which will pull the body towards itself, will be because of mass of sphere of radius $R-d$ only.

In this case, let the acceleration due to gravity becomes $g^{\prime }$ which can be given by,

Just as we calculated the mass of whole earth, we will calculate the mass of this sphere also, which can be given by,

Thus, the acceleration due to gravity $g^{\prime }$ can be given by,

Now, we will compare the value of $g$ when the body is at the surface of the earth and when it is taken to a depth $d$.

So, we can write,

Hence, we can see from the expression that when the depth increases, the acceleration due to gravity decreases.

If the body is taken to the center of the earth, acceleration due to gravity becomes zero.

Revision

The variation of acceleration due to gravity with the depth of earth is given by,

It decreases with increasing depth of the earth.

The acceleration due to gravity reduces to zero at the center of spherical earth.

The End