[Physics] Gravity on a doughnut-shaped/Möbius planet

Question

How different would the effects of gravity be if the planet we're on is in the shape of a torus (doughnut-shaped)?

For an (approximately) spherical planet, it's slightly clear that objects would tend to be gravitationally attracted towards the center. However a torus would have a hole in its center, and I'm not sure if the attraction towards the center still applies.

In particular, could a person on such a planet walk in the vicinity of the hole without falling off?

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Additional:

Similar question, but now consider a planet in the shape of a Möbius strip. Not only do you have to contend with the hole, but with the "kink" as well. Can a person stand up on the kink?

Answer 1

Gravitational Field from a Ring of Mass

...the force on a unit mass at $P$ from the two masses $M$ is $$F=-\frac{2GMx}{(x^2+a^2)^{3/2}}$$

Now, as long as we look only on the $x$-axis, this identical formula works for a ring of mass $2M$ in the $y$,$z$ plane! It’s just a three-dimensional version of the argument above, and can be visualized by rotating the two-mass diagram above around the $x$ -axis, to give a ring perpendicular to the paper, or by imagining the ring as made up of many beads, and taking the beads in pairs opposite each other. Bottom line: the field from a ring of total mass $M$, radius $a$, at a point $P$ on the axis of the ring distance $x$ from the center of the ring is $$F=-\frac{GMx}{(x^2+a^2)^{3/2}}$$

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Gravity of a Torus

People sometimes think that, perhaps for reasons of symmetry, an object in the interior of a ring of matter would be drawn toward the center, but this is not the case – at least not for objects in the plane of the ring. To see why, 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 $L_1$ and $L_2$. If we rotate this line about the particle through an incremental angle $\mathrm{d}q$, it will sweep out sections of the ring proportional to $L_1\cos(a)\mathrm{d}q$ and $L_2\cos(a)\mathrm{d}q$, 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 $\mathrm{d}q \cos(a) (\frac{1}{L_1} - \frac{1}{L_2}$) in the direction of the $L_1$ intersection point. Hence the net force is in the direction of the closest point on the ring, directly away from the center.

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Gravitation field due to rigid bodies

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Fun additional reading: Ringworld :)