While it doesn't relate to a rotating torus, I did find an equation that looks at the gravitational field on the polar axis of a torus-
F=-\frac{Gmx}{\left(x^2+a^2\right)^{3/2}}
where x is the distance from the centre of the torus (the torus being on the y,z plane) and a is the radius of the torus.
source-
http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/MoreGravity.htm
There seems to be a discrepancy with Gm/r^2 over very large distances but it is at least within the same order and it could be argued that over a certain distance, Gm/r^2 is used. Within about 50a, the results are very similar to Gm/r^2 until within close proximity of the torus (~5a) where the gravitational acceleration begins to reduce until it reaches zero at the centre.
I had previously assumed that gravity on the same plane as the torus would be something like-
g=\frac{Gm}{\left(r-r_t \right)^2}
where r
t is the radius of the torus. The resulting negative figure for the gravitational field once inside the torus doesn't imply 'negative gravity' as such, just that now as you move to the centre of the torus, gravity is pulling the other way (i.e. to the inside face of the torus). Based on the above, this could be rewritten as-
g=\frac{Gm}{\left(r^2-r_t^2 \right)}
providing a very rough rule of thumb equation-
g_t=\left|\frac{Gmr}{\left(r^2+r_t^2\right)^{3/2}}sin^2\theta\right|+\left|\frac{Gm}{\left(r^2-r_t^2 \right)}cos^2\theta\right|
so that the first equation dominates at the poles and the second equation dominates at the equator. This produces fairly similar results to Gm/r^2 (up to about 50r from the torus) until about 5r where the results for the pole and the equator begin to diverge. While not perfect, it does produce a fairly simple alternative to Gm/r^2.
Steve
