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?
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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.
Charlie
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