I haven't done this calculation, so I'm not going to immediately say you're wrong, but the example you use (a hollow sphere) doesn't generalize to other shapes.
Gravity is zero inside a hollow sphere or inside a hollow infinite cylinder. Here's the conceptual reason why. Imagine that you're near the left side of the sphere or cylinder. Then the mass to your left is closer to you, and therefore every kg of it exerts a strong gravitational pull (as force is proportional to 1/r^2). But although the mass to your right is farther away and thus exerts less gravitational force per kg, there's a lot more of it: roughly speaking, the amount of mass in a given direction is proportional to r^2 (times the solid angle it subtends). This factor of r^2 precisely cancels out the effects of the 1/r^2 weaker force per kg. (And yes, a careful derivation using calculus or Gauss's law would be more rigorous.)
None of that applies in the case of a torus. If I'm in the hole of the torus close to the left side, there's still a 1/r^2 difference in gravitational force per kg favoring the pull toward the left. But now, roughly speaking, the extra mass to my right is only greater by a factor of r (coming from the circumference of a circle rather than the surface area of a sphere). So the far side of the torus becomes less and less significant the father away it gets.
On top of all that, my impression from the article was that this sort of planet would require a really fast rate of rotation. I get the sense that the virtual "centrifugal" force plays a major role in the physics here. (Or if you prefer, that it's not reliable to ignore the effects of being in an accelerated reference frame.)
Dammit! You're right, I'm (very) wrong. (The stackexchange answer I used as a "check" was wrong too. That'll show me.) I was lazily using Gauss's law and ignoring the fact that a torus isn't spherically symmetric.
Could you please point out where a stackexchange answer was wrong on that question? I'm a heavy user on this site and have written (and continue to write) a lot on this subject there. Thanks.
I guess an intuitive arguement for why the gravity should be nonzero is to consider the limit as the major radius gets bigger and bigger. When it is very big, for a person on the surface it seems as if you are standing on the surface of a cylinder (the rest of the torus is far away), so you feel some gravity. So it can not be identically zero.
Note, though, that the analogous argument for gravity inside a sphere doesn't work: make the sphere very big and stand at some point on the inside. You'd think that you'd feel the same gravity as you would due to an infinite plane (constant g toward the ground), but in fact it would come out to zero because there's so much distant mass in other directions (the r^2/r^2 argument I made earlier). So it pays to be careful!
Yes. But standing inside the sphere it is kindof intuitive that the far-away parts of the sphere are not negligible---they fill up the entire night sky!
Gravity is zero inside a hollow sphere or inside a hollow infinite cylinder. Here's the conceptual reason why. Imagine that you're near the left side of the sphere or cylinder. Then the mass to your left is closer to you, and therefore every kg of it exerts a strong gravitational pull (as force is proportional to 1/r^2). But although the mass to your right is farther away and thus exerts less gravitational force per kg, there's a lot more of it: roughly speaking, the amount of mass in a given direction is proportional to r^2 (times the solid angle it subtends). This factor of r^2 precisely cancels out the effects of the 1/r^2 weaker force per kg. (And yes, a careful derivation using calculus or Gauss's law would be more rigorous.)
None of that applies in the case of a torus. If I'm in the hole of the torus close to the left side, there's still a 1/r^2 difference in gravitational force per kg favoring the pull toward the left. But now, roughly speaking, the extra mass to my right is only greater by a factor of r (coming from the circumference of a circle rather than the surface area of a sphere). So the far side of the torus becomes less and less significant the father away it gets.
On top of all that, my impression from the article was that this sort of planet would require a really fast rate of rotation. I get the sense that the virtual "centrifugal" force plays a major role in the physics here. (Or if you prefer, that it's not reliable to ignore the effects of being in an accelerated reference frame.)