# Effect of gravity at center of earth

Imagine I was a hypothetical ant man the size of an atom, and I position myself at the exact, down to the atom, center of mass of the earth. A move in any direction will move me out of the center. Would I experience gravity pulling outward on my body in all directions?

• Is there a small hollow chamber for the ant man? Or is he being crushed by hot metal under high pressure? – PM 2Ring Mar 20 at 22:46
• Exactly at the centre of the Earth u won't feel any force. The question about motion is more tricky. For instance, you can consider tunnel on the diameter of the Earth and if you move out of the center, you will oscillate in this tunnel. The question about the direction of force depends on geometry: you should specify your position from the center and then one can calculate the direction of force – Artem Alexandrov Mar 20 at 22:50
• He is in the very very small space in between the metal atoms – spmoose Mar 20 at 22:51
• In that case, all the gravity is balanced, so there is no net force, due to the shell theorem. – PM 2Ring Mar 20 at 23:21

There's actually a very useful way to solve this using the Gauss law for gravity, which is given by:

$$\oint\vec{g}\cdot d\vec{A}=-4\pi GM_{enc}$$

where $$\vec{g}$$ is the gravitational field, $$\vec{A}$$ the area enclosed in the surface of interest, and $$M_{enc}$$ the enclose mass of the object by the gaussian surface.

Assuming the Earth has an uniform volumetric density $$\rho$$, let's consider the two situations proposed:

a) You're at the exact center of the planet: in this case, all forces will cancel each other due to the symmetry of the object, so you will experience zero gravity. From Gauss Law, this is equivalent to having no enclosed mass.

b) You move away a distance $$r$$ from the object: in this case, the enclosed surface will be $$4\pi r$$, and assuming a constant density

$$\rho=3M/4\pi R^3=3M_{enc}/4\pi r^3 \ \ \rightarrow \ \ M_{enc}=4\pi r^3 \rho/3$$

Substituting in the Gauss equation,

$$g(4\pi r^2)=-4\pi G(4\pi r^3 \rho/3)$$

Simplifying,

$$\vec{g}=-\frac{4\pi G\rho}{3}\vec{r}$$

Or in terms of the radius of the Earth,

$$\vec{g}=-GM\left (\frac{r}{R^3}\right )\hat{r}$$

So you can see that only the area enclosed by your Gaussian surface will contribute to the net acceleration you feel towards the center.

Obviously the density of the Earth isn't constant (it's more concentrated on the nucleus than on the surface), so you can have a better approximation using a more empirical model of such density.