Okay, that program is taking an extra minute or two to work out since I forgot I would need another way to add up the total mass, and I know it will be in proportion to sqrt(r^2-x^2), but I'll still have some work to do on it, so I went ahead and worked out the integration for the gravity on a particle on the apex of a cone of uniform density. This time, I made the apex the origin, and the cone lies along the x axis, so it should go like this . . .

Int G D x dx dy dz / d^3, where d = sqrt(x^2 + y^2 + z^2), (x/h)^2 r^2 = y^2 + z^2 at any point along the surface at the edge of the cone, h is the height of the cone, and r is the radius of the base.

= Int 2 G D x dx dy sqrt[(x/h)^2 r^2 - y^2] / (x^2 + y^2) / sqrt[x^2 + y^2 + (x/h)^2 r^2 - y^2] for z = +/- sqrt[(x/h)^2 r^2 - y^2]

= Int 2 G D dx dy sqrt[(r/h)^2 x^2 - y^2] / (x^2 + y^2) / sqrt[1 + (r/h)^2]

= Int 4 G D dx / sqrt[1 + (r/h)^2] * [sqrt[x^2 + (xr/h)^2] atan[sqrt(1 + (r/h)^2)(xr/h) / sqrt[(xr/h)^2 - (xr/h)^2]] / x - atan[(xr/h) / sqrt[(xr/h)^2 - (xr/h)^2]]] for y = +/- (xr/h), where the denominator in the atans then become zero, so we get atan(infinity) for each of them, but don't worry, these infinities work out to atan(infinity) = pi/2, so we get

= Int 4 G D dx (pi / 2) / sqrt[1 + (r/h)^2] * [sqrt[x^2 + (xr/h)^2] / x -1], for x = 0 to h

= 2 pi G D h [1 - 1 / sqrt[1 + (r/h)^2]], where D = M / V and V = pi r^2 h / 3, so

= 2 pi G M h [3 / (pi r^2 h)] [1 - 1 / sqrt[1 + (r/h)^2]]

= (G M / r^2) * 6 [1 - 1 / sqrt[1 + (r/h)^2]]

It looks like if we were to set a particle on the apex of a cone with a very small height, many times smaller than the radius of the base, then the gravity would strengthen toward a limit of 6 times greater than that of a sphere with the same mass and at a distance equal to the radius of the base.

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Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."