# Gravity on a doughnut-shaped/Möbius planet

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?

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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Consider Super Mario Galaxy as a case study :D – BBischof Nov 8 '10 at 14:47
I think Mobius strip is still like a donut so not much will change. – Pratik Deoghare Nov 8 '10 at 15:16
A Möbius strip is special as a 2D surface, but gravity works in 3D space... – kennytm Nov 8 '10 at 19:09
Maybe you should consider a Klein's bottle – adhalanay Nov 8 '10 at 20:31
@adhalanay, maybe. :) – user172 Nov 8 '10 at 20:33

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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. – Alan Rominger Jan 21 '13 at 20:37

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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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. – Mark Eichenlaub Nov 8 '10 at 22:23

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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"other side" of a Möbius strip? Granted, I didn't specify how thick the strip was but... – user172 Nov 8 '10 at 16:28
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. – Gerard Nov 8 '10 at 21:34

## protected by Qmechanic♦Feb 18 '14 at 16:57

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