Gravity of a Torus

What is the gravitational field of a torus?  In which direction does
the net force of gravity point if you live on the surface of a planet
shaped like a doughnut?  An object in the hole of the "doughnut" will 
not necessarily be drawn toward the center.  Consider a very thin ring
of mass treated as a circle of radius R in the plane, and a particle 
inside this ring at a distance r from the center.  Construct an 
arbitrary line passing through this particle, striking the ring in 
two opposite directions at distances u and v.  If we rotate this line
about the particle through an incremental angle h, it will sweep out 
sections of the ring proportional to u*h*cos(a) and v*h*cos(a), where
"a" is the angle the chord makes with the normals to the circle at 
the points of intersection.

The net gravitational force exerted by these two opposing sections of 
the ring is proportional to the masses in these small sections divided 
by the squares of the distances, i.e., the force is proportional to
h cos(a) (1/u - 1/v) in the direction of the u intersection point.  

Of course if the particle was inside a spherical shell instead of a 
circular ring, the conical region surrounding each ray through the 
particle would sweep out a section of the shell proportional to the 
SQUARE of the distances u and v, which would then cancel out with 
the inverse squares in the force expressions, leaving a net zero 
force on the particle.  However, since our mass is essentially just 
a one-dimensional ring, the distances u and v appear only to the first 
power in the mass terms, and so they cancel only one of the powers in 
the inverse squares.

As a result, I think the net force along this ray (and every ray) is 
positive in the direction of the closer point on the ring, which 
suggests that a particle in the equatorial plane inside the ring would 
move outward toward the ring (along a radial line).  There would be 
a stationary point in the center of the ring, but it wouldn't be
stable.  I think the potential energy would be a local maximum at 
the center.

Working out the actual formulas for this "ring planet" of radius R 
and total mass M, the force of gravity on an interior point particle of 
mass m in the plane of the ring at a distance r < R from the center 
of the ring would be
                     _                                           _
           G M m r  |         9   / r \2      75  / r \4          |
  F(r)  =  -------  |  1  +  --- ( --- )  +   -- ( --- )  +  ...  |
             R^3    |_       16   \ R /       64  \ R /          _|

in the outward radial direction.  Thus at r=0 the force is zero, but 
at any non-zero value of r there will be a net outward force, initially 
proportional to r, and then increasing sharply as r approaches R.  
(The ring is assumed to have infinite density so the outward force 
goes to infinity at r=R.)

On the other hand, an exterior particle (r > R) will see an inward 
radial force of magnitude

                    _                                           _
           G M m   |         3   / R \2      45  / R \4          |
  F(r)  =  ------  |  1  +  --- ( --- )  +   -- ( --- )  +  ...  |
            r^2    |_        4   \ r /       64  \ r /          _|

So this starts out infinite on the outer surface, drops off steeply in 
the near field, and then approach an inverse-square relation in the
far field.

It would be interesting to know how closely the gravitational field
of a general torus is to that of the same mass distributed on a thin
ring.  Obviously if we arrange a set of spheres in a circle, their
field is the same as if their masses were located at points on the
ring.  We can also arrange a set of overlapping spheres with most of
the overlap removed by internal spheres, all of which are centered
on the ring, so in this way we can approximate a general torus with
just positive and negative spheres that can be treated as points on
the central ring.