Gravity on a torus

[xkcdfan]

Full disclosure: This question was brought on by playing Final Fantasy VI, in which the world map is a torus. If you go leave the map off the east edge, you reappear at the corresponding point on the west edge; the same goes for north and south.

As shown in-game, gravity pulls everyone down to the surface of the planet. Is this plausible? Would this really happen? (I'm just going to ignore the fact that the world appears to be really, really small, as in so small that everyone should go flying every which way whenever they try walking, and there probably shouldn't be an atmosphere... Maybe the oceans are really bigger than they look and they're just rendered smaller on-screen so your airship trips take less time. Sure, let's go with that.)

(Also magic can't explain it because spoilers.)

[idobox]

With gravity as we know it, in "flat" space, it wouldn't be possible.

If gravity was in 1/r3 or more, it could work. Maybe they just have very high iron levels, and the planet is a huge magnet.

Also, maybe a weird space-time topology could allow it.

[gmalivuk]

With gravity as we know it, in "flat" space, it wouldn't be possible.

Are you sure? It wouldn't be constant surface gravity, to be sure, but a torus will still pull toward its surface from every point near the surface.

[idobox]

I haven't done that for a long time, so be indulgent.

If we take a cylinder of infinite height, with a radius smaller than the inner radius of the torus, and aligned with the torus, the mass inside the cylinder will be null.
Gauss theorem tells us the flux integrated on the surface of the torus is null.
By rotational symmetry, we can deduce attraction is null on the surface of the cylinder (which seems very odd, I suspect a gross mistake somewhere)

When the cylinder tangents the torus, we deduce gravity is null at least on the inner equator, or whatever it's called.
When the cylinder is a bit larger, ie when it's surface intersects the torus, there is mass inside it and attraction toward the center of the cylinder.
When the cylinder is larger, you are attracted "normally" to the torus.

Tos sum up, if you were walking on the outer equator, you would be attracted "normally", if you moved away, you would be attracted in the direction of the center of the torus, until you reached the inner equator where you would start to float.

[gmalivuk]

Tos sum up, if you were walking on the outer equator, you would be attracted "normally", if you moved away, you would be attracted in the direction of the center of the torus, until you reached the inner equator where you would start to float.

No, you wouldn't. Imagine the torus has a very small minor radius, so it's more like a thin ring. From any point not on the central axis, attraction is felt toward the nearest point on the ring.

For a detailed discussion, check out this page.

[idobox]

By rotational symmetry, we can deduce attraction is null on the surface of the cylinder

This is where the error lies.

We can only deduce the gravitational field intensity is angle independent, but not height independent. Form far away, the torus is seen as a point mass, and the field is directed toward the center, ie the flux is negative.
To get a null average, we need the flux to be positive, ie going out of the cylinder, somewhere. Near the plane seems a good assumption.
And the math in your link prove it.