Gozza wrote:Imagine that the earth is solid rock and that there is a room sized space at the centre, what would happen if you were in that room?
Hi, Gozza. The gravitational force inside a sphere corresponds to all the mass that lies closer to the center than you. That is: the force comming from the shell that lies above you cancels out. If you are at the center, the forces from all the sphere cancel out, and the acceleration is zero. Note that inside a hollow sphere, no matter how thick is the shell, no matter where you are in the hole, the acceleration is zero.
The reason is the following. Suppose that you are somewhere between the outer surface and the center. The parts of the shell that are close to you contribute with a force that pulls you towards the surface. But the parts of the shell that are "on the other side" of the planet pull you to the other side, by exactly the same amount. Those parts are farther away, and pull you more weakly, but there are more of them, so one thing exactly compensates the other, and the forces cancel out. Your "room at the center of the Earth" is an extreme case of this: a very thick hollow sphere. Ergo, no force.
This phenomenon is a consequence of the 1/r^2 form of the gravitational force and the geometry of 3D space. In other geometries, or with other forces, the behavior could be different. If you know some calculus, you can find an easy to follow demonstration of this in many elementary Physics books. I will let Fridger explain what is different in General Relativity.
Decades ago there was a fashion for a crazy theory of a "hollow Earth", with openings at the poles, such that people could stand on the inside surface. Such situation is a falacy.
Regards,
Guillermo
PS: All this is true in a radially simmetric distribution of mass. If you have more mass on one side of the planet than on others, you end up being pulled that way, obviously.