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Author Topic: What would gravity be like on a doughnut shaped planet?  (Read 15808 times)

Offline Atomic-S

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What would gravity be like on a doughnut shaped planet?
« Reply #25 on: 13/05/2010 06:49:27 »
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Today I believe that the ‘me’ of yesterday was clearly a moron. I now believe you could stand on any point of it’s surface, even within the hole, because the ground which is closest will exert the greatest gravitational pull and would stop you from floating away. But would this make the centre of gravity not a single point but a line running internally through the centre-line of the doughnut? That can’t be right.
Condidering the non-rotating case: If we imagine such a planet to be a very thin ring, like a wire hoop, then it is clear, by considering the attraction as it applies to a long linear mass (which falls off inversely as the distance), that the dominant force will be near the surface, whether inside or outside. On that basis, we may assert that for a planet of proper dimensions, a person would be able to stand on any part of its surface without falling off, even if the planet is not rotating.
 

Offline Geezer

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What would gravity be like on a doughnut shaped planet?
« Reply #26 on: 13/05/2010 06:59:52 »
On that basis, we may assert that for a planet of proper dimensions, a person would be able to stand on any part of its surface without falling off, even if the planet is not rotating.

I'm pretty good at asserting things too, but a proof is much more convincing (unfortunately).
 

Offline imatfaal

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What would gravity be like on a doughnut shaped planet?
« Reply #27 on: 13/05/2010 18:24:03 »
There is more detail of the proof of the thin ring and also of the donut here; it is beyond me to give a precis, so here is a link

http://www.mathpages.com/home/kmath402/kmath402.htm
 

Offline Geezer

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What would gravity be like on a doughnut shaped planet?
« Reply #28 on: 13/05/2010 21:15:49 »
There is more detail of the proof of the thin ring and also of the donut here; it is beyond me to give a precis, so here is a link

http://www.mathpages.com/home/kmath402/kmath402.htm

Thanks Imat!

It looks as if my initial WAG was about right  ;D
 

Offline Ken Hughes

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What would gravity be like on a doughnut shaped planet?
« Reply #29 on: 13/07/2010 12:50:24 »
Gravitational attraction/acceleration, as you know, decays with distance from the surface of any massive object in accordance with the inverse square law. If we imagine this decay from the surface of the doughnut (at any position), let`s say on the internal "equator", then the attraction becomes much smaller very quickly as you take off towards the centre of the system, but the effects from the opposite parts of the doughnut are even less significant at any position, until we get to the centre of the hole where all attractions are equal and opposite. If we go back to the Earth/Moon system where there is a neutral point where the Earth`s field counteracts the Moon`s, then we have the same situation here and the neutral point is a point in the centre of the hole in the equatorial plane. At this position, there is no net effect in any direction as the vector sum of the fields is zero in all directions, and one would simply float at this position. One could, however, walk freely all over the surface of the doughnut without noticing the small decrease in weight on the internal equator and the slight weight gain on the outside equator.
At the outside equatorial plane, we have a thin "disc" of space, spreading towards infinity, where the gravitational effects are the same as if the doughnut where spherical.
As we move up, over and around from this imaginary disc and curl towards the neutral position, then through an axial section cutting the doughnut, we see two effects superimposed. The first is the "normal" gravitation from the immediate part of the doughnut, but superimposed on this is the gradually increasing effect from the opposite section of the doughnut. I ignore the effects to the sides of our considered section as these are equal and opposite and therefore cancel.
The distribution of the gravitational acceleration around this curve is one of a normal spherical planetary attraction, but skewed slightly towards the other side of the doughnut. The equivalent gravitational force will therefore vary slightly, as previously described, with the maximum on the external equator and the minimum on the internal equator.
I trust this helps.
 

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What would gravity be like on a doughnut shaped planet?
« Reply #29 on: 13/07/2010 12:50:24 »

 

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