1. Standard memberXanthosNZ
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    25 Feb '07 03:30
    If you wish to use either the Shell theorem or converting to a point source you must prove that they apply in the case of a torus.
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    25 Feb '07 03:54
    Originally posted by XanthosNZ
    If you wish to use either the Shell theorem or converting to a point source you must prove that they apply in the case of a torus.
    Good point. They clearly apply in the case of a circle (an infinitely thin torus) but, after some thought, I don't think everything cancels out for a torus of finite thickness. Things would be slightly off.
  3. Standard memberXanthosNZ
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    25 Feb '07 06:201 edit
    Originally posted by GregM
    Good point. They clearly apply in the case of a circle (an infinitely thin torus) but, after some thought, I don't think everything cancels out for a torus of finite thickness. Things would be slightly off.
    The math would get very complex certainly because a torus doesn't resolve to a simple figure in spherical coordinates.
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    25 Feb '07 08:24
    Originally posted by XanthosNZ
    The math would get very complex certainly because a torus doesn't resolve to a simple figure in spherical coordinates.
    Why not use torodial coordinates then? It is not very difficult when you know how.
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    25 Feb '07 11:44
    Originally posted by FabianFnas
    Why not use torodial coordinates then? It is not very difficult when you know how.
    Because in that case the formula for the orbit would get rather complicated, or so I suspect.

    Richard
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    25 Feb '07 11:511 edit
    Originally posted by Shallow Blue
    Because in that case the formula for the orbit would get rather complicated, or so I suspect.

    Richard
    So the coordinate system chosing doesn't matter then.

    I would put the torus, as a matrix, into a mathematical program, like MatLab or something, and do the math through numerical methods. Shouldn't be too hard. But this is an approximate method, that doesn't say anything for sure if the orbits are stable or not.

    I did something like it in my university studies as a project and found some interesting result.
  7. Subscribersonhouse
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    25 Feb '07 13:42
    Originally posted by FabianFnas
    So the coordinate system chosing doesn't matter then.

    I would put the torus, as a matrix, into a mathematical program, like MatLab or something, and do the math through numerical methods. Shouldn't be too hard. But this is an approximate method, that doesn't say anything for sure if the orbits are stable or not.

    I did something like it in my university studies as a project and found some interesting result.
    What was the 'something like it' study? Was it your dissertation?
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    25 Feb '07 15:57
    Originally posted by sonhouse
    What was the 'something like it' study? Was it your dissertation?
    It was a time ago, I don't remember the details. But it was a orbit analysis with more than two bodies, with a look ahead of some kind.

    Using the (more or less) same program I could do the calculations about orbits around a torodial shaped body, but I don't think I can find the papers anymore...
  9. Subscribersonhouse
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    25 Feb '07 19:48
    Originally posted by FabianFnas
    It was a time ago, I don't remember the details. But it was a orbit analysis with more than two bodies, with a look ahead of some kind.

    Using the (more or less) same program I could do the calculations about orbits around a torodial shaped body, but I don't think I can find the papers anymore...
    It's clear an orbit aroung the 'equator' of a toroid woult be pretty much like a spheroid planet but it gets tricky when going over the
    'poles' or 'holes' for sure, to say nothing of threading the needle kind of orbit.
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    25 Feb '07 22:271 edit
    Originally posted by sonhouse
    It's clear an orbit aroung the 'equator' of a toroid woult be pretty much like a spheroid planet but it gets tricky when going over the
    'poles' or 'holes' for sure, to say nothing of threading the needle kind of orbit.
    With numerical methods it's much easier. But you never get exact results, only approximated to a certain level.

    Give a particle a location and a velocity, then you let it go and see what happens. Nice and fun to see the graph in 3d of the particles orbit, stable or not.
  11. Standard memberAThousandYoung
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    25 Feb '07 23:26
    It seems to me that there will be unequal gravity from two directions; towards the opposing side of the torus there will be more gravity than the other way. This seems like it should prevent such an orbit.
  12. Subscribersonhouse
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    26 Feb '07 15:17
    Originally posted by AThousandYoung
    It seems to me that there will be unequal gravity from two directions; towards the opposing side of the torus there will be more gravity than the other way. This seems like it should prevent such an orbit.
    Well, another question would be, can you establish a stable orbit around the 'flattened' part of the torous completely external, NOT going through the hole? It's clear an orbit around the toroid 'equator' would be stable because the gravitaional field would be more or less equal around the perimeter of the toroid. Gravity over the "poles" would vary from point to point in a "polar" orbit.
  13. Standard memberXanthosNZ
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    28 Feb '07 06:11
    No one has mentioned the obvious.

    Imagine an orbit that is harmonic motion on a line tangent to the interior wall of the torus and directly through the centre of the hole. On this line all forces not on the line cancel and we are left with an "orbit" that is nothing but back and forth motion passing through the hole twice per period.
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    28 Feb '07 07:27
    Originally posted by XanthosNZ
    No one has mentioned the obvious.

    Imagine an orbit that is harmonic motion on a line tangent to the interior wall of the torus and directly through the centre of the hole. On this line all forces not on the line cancel and we are left with an "orbit" that is nothing but back and forth motion passing through the hole twice per period.
    That's an orbit, but it isn't stable. A comet only has to breathe on this planet and it falls over.

    Richard
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    28 Feb '07 07:59
    Originally posted by Shallow Blue
    That's an orbit, but it isn't stable. A comet only has to breathe on this planet and it falls over.

    Richard
    I think I mentioned it a week ago:

    "Yes, you are right. In the middle of the torus hole, at the torus mass centre, the gravitation is zero. In the rotational axis the gravitation is pointed towards the mass centre. But off the rotational centre the gravitation is pointed towards the nearest surface. Correct me if I'm wrong."

    The "In the rotational axis the gravitation is pointed towards the mass centre." means that you have a back and forth movement.

    ...and I think others has mentioned the samt thing too.
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