19 The point is that if we approximate Earth with an oblate ellipsoid, then the surface of Earth is an equipotential surface,$^1$ see e.g. this Phys.SE post. Now, because the polar radius is smaller than the equatorial radius, the density of equipotential surfaces at the poles must be bigger than at the equator.
Assuming spherically symmetric mass distribution within Earth, one can compute gravitational field inside the planet using Gauss' law for gravity. One consequence of the law is that while computing the gravitational field at a distance r < R (with R being the radius of the Earth), one can ignore all the mass outside the radius r from the center \begin {equation} \oint_ {S_r} g_r \cdot dA = -G ...
0 Once an object goes below the surface of Earth the net gravity acting on it is the radius of the mass from the object to the Earth's center. As the object goes deeper the radius gets smaller, and the object has less net gravity pulling it downward. At the Earth's center there would be no net gravity from the Earth.
Hill's sphere is the assumed sphere that a smaller body -rotating around a bigger one- attracts other objects to rotate around it. For example, in the case of Earth, the radius of its Hill's sphere...
So, even if the radius of the universe is the same as its Schwarzschild radius, it does not imply we are inside a black hole with an event horizon at the visible radius of the universe when we use the appropriate metric. Is this simply because to get into the non-observable portion of the universe, you have to go faster than the speed of light?