Effect of spin on gravitational field? If a large sphere of uniform density is spinning around one axis, is its gravitational field uniform? If yes, then what if it's not a sphere or not uniform? For example if


*

*we remove a chunk of the exterior?

*it's made of lead on one side of the equator and aluminum on the other?

*50/50 lead/aluminum, symmetric around the axis?

*it's slightly pancake-shaped?

 A: Without considering frame-dragging (thank you very much, Kerr!) the gravitational field at large R outside an object always acts like the gravitational field from a point mass located at the centre of mass. So regardless of how it is spinning or deformed, it is more or less uniform.
If you consider General Relativity and the effects of frame-dragging, then while that remains more or less true for the magnitude of the force, the geometry of spacetime does some really cool warping and causes the net effect due to gravity (or rather, something with mass) to be less than uniform. For example, a rotating black hole drags spacetime around with it so much that there is a region (the ergosphere) where you physically cannot orbit the black hole against the rotation; all motion has to have a component in the direction of the rotation. But (and I please correct me if I'm wrong), the magnitude of the force of gravity is approximately uniform still.
A: Here comes some partial answers: 
1) Within the Newtonian theory of gravitation: If a large sphere of uniform density is spinning around an axis, its gravitational field is the same as if it wouldn't be spinning at all.
2) However, in Einstein's general relativity theory the spinning causes a non-trivial effect, this goes under the name of Frame-dragging. There is a nice Wikipedia article about this - see this link. 
