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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?


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?

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Gravitational Field from a Ring of Mass

enter image description here...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. enter image description hereBottom 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 $$F=-\frac{GMx}{(x^2+a^2)^{3/2}}$$


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.


Gravitation field due to rigid bodies


Fun additional reading: Ringworld :)

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  • $\begingroup$ The Ringworld is the one that doesn't fit here. The concept is irrelevant of its self-gravitation and it would only be a numerical correction of the parameters. Actually, if self-gravitation was significant (and it would be) the idea would probably fail since it would then require compressive strength in the axial direction and tensile strength is easier to manage. $\endgroup$ – Alan Rominger Jan 21 '13 at 20:37
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    $\begingroup$ This very old answer has been flagged as "link-only," which is probably a good flag --- see this guidance for answering. However the links here are particularly nice links, all of which are still alive after eight years. I would rather have someone in the flag review queue take na few minutes to add a little context for one or more of the links than to have this good contribution removed for bending our guidelines. $\endgroup$ – rob Dec 29 '18 at 20:55
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    $\begingroup$ Whoops, I had deleted this from the flag queue. @rob How about we allow a week for someone(s) to enhance this answer with some description; perhaps we can post in chat or meta to bring some attention to it. At the end of that time, if nobody has done so or promised to do so within a short time, I still think it should be deleted; I wouldn't characterize this as "bending our guidelines", it's a flat-out egregious violation of the rule against link-only answers. (Nothing against Pratik, as I know the rules were different/looser back in 2010, I'm just saying it's very clearly not okay now.) $\endgroup$ – David Z Dec 29 '18 at 23:41
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    $\begingroup$ @DavidZ That sounds like a good plan. $\endgroup$ – rob Dec 30 '18 at 4:31
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    $\begingroup$ @Chair Well this answer is a bit of a special case. Ordinarily we'd just delete it, but because of the perceived value of the links we're keeping it around for a little while to give an opportunity to make edits which are necessarily going to be more substantial than is normally allowed. But ideally, the final state after whatever edits do get made should be on par with an independently posted answer, and right now it's not. Our referencing guidelines talk about how answers shouldn't consist entirely of quotes. $\endgroup$ – David Z Dec 30 '18 at 6:43
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I am doing this out of my head, so hopefully it's correct, but please double check.

  1. When you are sufficiently away from the object, the laws will reduce to the classic point mass solution, F = GMmr^-2.

  2. When you are on the plane of the torus hole, furthermore, contributions from the "up" and "down" directions will cancel out. The torus can be considered as a disk with a hole. In this scenario, the gravity will depend only on the amount of mass in the disk defined by the radius from the observer to the centre of of the torus. It will decrease linearly (I think) from the value on the external border (which should be the similar to the one defined at point 1. and the value zero on the inner border).

  3. Inside the torus on the plane of the hole, gravity should be zero.

  4. In other points you actually need to integrate.

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    $\begingroup$ The gravitational potential only depends on the mass inside a certain radius if the mass distribution is spherically symmetric. This theorem does not hold for a disk. This is intuitive if you consider taking the sphere and squashing it down into a disk at the equatorial plane. If you were standing on the equator, the squashing motion brings every bit of mass closer to you (except those already on the equatorial plane), so the squashing motion increases the strength of gravity. If you then put all the mass in the center, gravity would get weaker again. $\endgroup$ – Mark Eichenlaub Nov 8 '10 at 22:23
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On the torus:
Walking towards the inner side of the torus one becomes lighter, because one has gravitational pull under one's feet (which is stronger because it is nearer) and gravitational pull above one's head making you lighter. The outer side of the torus is the side where people are heavier.

On the Möbius strip:
The gravitational pull varies like in the case of the torus, one becomes lighter and then heavier again while making a world trip. Walking from one side to the "other" side [parenthisis because it is the same side] via the thickness of the strip - assuming that is accessible, otherwise one has to fly to the "other" side - the situation is reversed.

As Sklivvz notes:
In the center of the torus or Möbius strip all gravity is cancelled.
At a great distance the form of the object may be neglected, it can be treated as a point mass.

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  • $\begingroup$ "other side" of a Möbius strip? Granted, I didn't specify how thick the strip was but... $\endgroup$ – user172 Nov 8 '10 at 16:28
  • $\begingroup$ The Möbius strip must have a thickness otherwise it couldn't have a mass, so I can make a shortcut during my world trip. Edited. $\endgroup$ – Gerard Nov 8 '10 at 21:34