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What proof is there that the center of gravity of a disc is at its center when the gravitational field is not uniform, such as objects close to the edge of the disc, or lying within the body of the disc?

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Symmetry. It couldn't be anywhere else. –  Andrew Jan 5 '12 at 18:55
    
I think the question need to be clarified. @Andrew has the right answer to the question as asked, but I suspect that john meant to ask about the direction of gravitational force, which is a different game entirely (i.e. for test particles much nearer the surface than to the nearest edge the force is effective downward along the local normal, and not toward the geometric center). –  dmckee Jan 5 '12 at 22:04
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2 Answers 2

I assume here a general relativistic standpoint, as this is the most general case, and because it has interesting observational consequences.

The idea about center-of-mass is usually applied to calculate the gravitational field outside spherically symmetric matter distributions. This does not assume that the field is uniform. In particular it may decrease as a function of distance form the center of the disk or sphere.

In the theory of general relativity Birkhoff's theorem states (in simple terms) that the gravitational field outside a spherically symmetrical matter distribution is only a function of the total mass, and not the extend of the matter or its distribution as a function of distance from the center, or even as a function of time!

Note that this assumes a spherically symmetric matter distribution. Namely it has to be isotropic, and because of the symmetry it will thus have a uniquely defined center, which is completely analogous to the center of mass. Also it assumes that the test particle is outside the matter distribution. Inside there are higher order corrections to the gravitational field.

The physical consequence is that no measurement of only the gravitational field can be made to distinguish between a homogeneous sphere of mass M, a point (all located at the center of the sphere) gravitational source of the same mass (eg. a black hole) or a radially oscillating star.

This all assumes that the observer, or test particle, is in the same plane as the gravitating object. In the case of a 2D disk then, this assumes that the test particle is in the plane of the disk. Outside this plane, the theorem will not hold true. Surely enough, a particle initially traveling towards the center, may well end up intersecting the disk away from the center, because of the anisotropic gravitational field.

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Actually, Center Of Mass is the only critical issue, symmetry is irrelevant. At the COM the gravitational field is flat, i.e. zero G, so COM is coincident with COG. Pretty much by definition.

The question is about G at the center of symmetry of a disc when there are other objects in the system. Since these will move the COM, the 0g point will move away from the center of the disc itself. So there is no such proof. There is only the integral of Gm1m2/r^2 of the delta masses over their radii from every point in the system, solved for the point where the integral is 0. This is only coincident with the geometrical center of a disc or sphere (or any symmetrical shape for that matter) when there are no other massy objects involved.

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