# Question #17b7c

Apr 27, 2017

Using Gauss's Law for Gravity:

$\int {\int}_{S} m a t h b f g \cdot d m a t h b f S = - 4 \pi G M$

Where:

• $S$ is any closed surface (the boundary of an arbitrary volume V),
• $d m a t h b f A$ is the usual outward pointing surface area vector,
• $m a t h b f g$ is the gravitational field,
• $G$ is the universal gravitational constant, and
• $M$ is the total mass enclosed within the surface S.

Say we are at radius ${R}_{o}$, inside the earth's radius ${R}_{e}$, with mass as ${M}_{e}$, assuming uniform density and using a concentric Gaussian sphere:

$g \left({R}_{o}\right) \cdot 4 \pi {R}_{o}^{2} = - 4 \pi G \frac{\frac{4}{3} \pi {R}_{o}^{3}}{\frac{4}{3} \pi {R}_{e}^{3}} {M}_{e}$

$\implies g \left({R}_{o}\right) = - \frac{G \setminus {M}_{e}}{{R}_{e}^{3}} \cdot {R}_{o}$

And

${\lim}_{{R}_{o} \to 0} g \left({R}_{o}\right) = 0$

Showing this within a coordinate system is way uglier but do-able.

In practical terms, if you are at the centre of the earth, you are being pulled out radially in every direction, and the force in each direction has an exact but opposite force. These superimpose to leave no force.

Gravity is also zero anywhere inside a hollow shell.