# Can a revolving body self-gravitate?

If a body is revolving around a point at radius R with tangential velocity V, does General Relativity predict that at some tangential speed, the body will revolve around the point without any external forces acting on it? Since General Relativity models gravity as a curvature of spacetime, I was thinking that an observer at the center of rotation of the revolving body would measure some curvature of space as a result of the revolution. This is because the body would be contracted in the tangential direction relative to the observer, such that C<2*PI*R where 'C' is the circumference of the revolving body measured by the observer (the revolving body is thought of as a ring of particles). This implies a curvature of space, which would be equivalent to a gravitational force.

## 2 Answers

I'm not sure about axially symmetric solutions, but for spherically symmetric configurations, curvature inside a certain radius is unaffected by the matter distribution outside (as in Newtonian gravity). I suspect the same thing happens in your case. I suspect that if the ring were thick enough, there would be an effect, but that is not exactly the case you are considering.

Oliver Heaviside showed at the end of the nineteenth century, that if a field is moving at a finite speed, then there is a co-field, which is caused by, and acts on, moving charges. Since in this scope, gravity is a field, its charge is weight (ie what is incorrectly called mass: Mass is a measure you find with rulers.).

In any case, the nature of the co-field is similar to magnetism (which is the co-electric field), something demonstrated by OH. It's called variously co-gravitation [Jeffimenko] or gravitomagnetism [wikipedia].

More recent analysis changes some of the numerical constants by a factor of 2.

So if your planet is spinning at something like very fast, one has to take to account the nature of the co-field into account.