Orbiting unusually-shaped celestial objects

Izack

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Say you have an enormous and highly dense torus-shaped object in space. Correct me if I'm wrong, but I learned in high-school that objects orbit each other's center of mass (which for a torus/donut with even density throughout would be in the 'hole' in the center). By this reasoning, is it not possible to orbit inside the circumference of the ring?
Similarly, would you be able to orbit an empty Dyson Sphere from within its 'wall?'
Speaking purely theoretically, duh.

Slap me with a physics textbook if this is too silly.
 
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tblaxland

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Definitely you can't orbit inside a Dyson sphere. Not sure about a torus though. Intuitively you would think that you couldn't orbit inside the ring in the same plane as the torus, but perhaps out of plane you could?

BTW, the "orbit each other's center of mass" assumption only applies for spherical objects or at great distances. The ellipsoid shape of the Earth creates perturbations on satellites that are proportional to the altitude.
 

Izack

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Hmm, I had always assumed that the perturbations in satellites' orbits were due to Earth's center of mass not lining up with its center of volume, which would result (under my apparently flawed reasoning) in unpredicted orbital radius measurements if those measurements were taken based on the Earth's volumetric center, and not its center of mass. (Attachment)

So, for the Dyson sphere, is it impossible because the gravity of the near side would be more effective on the orbiting object than that of the far side? What effect would this have on an object orbiting outside the sphere at a close proximity?
 

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statickid

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I don't have too much 'OFFICIAL' knowledge on this matter, but I would believe that the gravity is based specifically on the mass. THEREFORE if you were to make you're way to the center of the torus, you would fall towards the side that you were baaaaaarrely closer too, because you no matter how precise you were, it would be almost 100% impossible to get yourself EXACTLY in the middle. In effect, I believe there is NO reason to be gravitationally attracted to what is, in effect, nothing. If there were a very large, very dense, very rigid torus that could not collapse into a spheroid, would you not be able to walk around on it? Perhaps you could balance the pulls and be stationary at the center, but no "orbit" around nothing in the middle.
 

Hielor

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The "objects orbit each others' centers of mass" assumption is only valid when the objects are fully separate--that is, as long as one object can't be considered to be "inside" the other. An object inside a spherical shell (like a Dyson sphere) will feel no gravitational effects from the sphere, and this is presumably applicable to a Ringworld situation as well.

See the "Inside a Shell" section:
[ame="http://en.wikipedia.org/wiki/Shell_theorem"]Shell theorem - Wikipedia, the free encyclopedia[/ame]
 

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Larry Niven tackled this problem in detail in his Ringworld and The Ringworld Engineers novels, great reading material. The structure had to be stabilized using E.B.A. thrusters.
 

Eagle

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Objects only appear to orbit a body's center of mass when you're far enough away relative to the body's mass distribution. But there is gravitational attraction between an object and a body is between each particle of each. This is why you can have cool things like using gravity gradients to stabilize a satellite.

If you think back to second semester calculus or calculus based physics the equation involves taking a double or triple(for 2d or 3d cases) integral to get the gravitational attraction for each infinitesimal section of the body.

@Izak The motion of an object inside a dyson sphere depends upon if there is a large sun or other bodies inside the sphere. If there is a massive central star, you could probably orbit normally unless you got close to the inside surface of the sphere, at which point it would attract you. Also you could orbit small planets if you were close enough to them. Otherwise, if its empty you're just going to drift towards a wall of the sphere and the direction of your gravitational acceleration will be towards the nearest wall (not the center).
 
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Moach

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by my feeble experience programming orbit simulators (something i do as a hello-world test whenever i get a new physics engine)... i reckon it might be possible to orbit a torus shape in a "figure-eight" path crossing through the middle...

but it is indeed a really bizarre concept....

quite a bit of food for thought!
cheerz!
 

tori

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i reckon it might be possible to orbit a torus shape in a "figure-eight" path crossing through the middle

It should be, but the stability would be low - comparable to a lagrange point. Circular orbits inside of a torus, or any kind of an inertial orbit in the plane of the torus, will not.
 

eveningsky339

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Wouldn't the orbiting object be pulled to one side when passing through the inside of a torus?
 

Izack

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Thanks for the responses.
Assuming that gravity was an action between objects instead of particles is what made me think this could work. In hindsight I should have known better.

The idea of the figure-eight orbit is interesting, though. Can anyone elaborate?
 

tori

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It looks like a fig. 8 wouldn't work either. Try doing a figure 8 orbit between two equally massive stars (e.g. here) - that's a torus simplified into 2D. It won't work, the closest side of the torus will pull the orbiter in. The only orbit inside of a torus that could work is a kind of pogo orbit lying inside of the torus' axis, going above and below its center of gravity.
 

Eagle

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It looks like a fig. 8 wouldn't work either. Try doing a figure 8 orbit between two equally massive stars (e.g. here) - that's a torus simplified into 2D. It won't work, the closest side of the torus will pull the orbiter in. The only orbit inside of a torus that could work is a kind of pogo orbit lying inside of the torus' axis, going above and below its center of gravity.
You can orbit 2 stars, its called a double free-return. And yes you can figure 8. The problem is these are not stable orbits and perturbations will eventually force you into a more stable orbit.

Stable as in how the L4 and L5 are stable and if you're near enough to it you'll stay there forever, compared to the unstable L1, L2 and L3 that you will eventually fall out of. There are objects at Earth's and Jupiter's L4 and L5, but nothing stays at their L1, L2 or L3 for very long.
 

tori

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I ment a figure 8 going like this:
8v.png


simplified to a 2D problem that is easier to visualize in an applet:
8s.png


A double-free return (which is what I've been doing in the Earth-Moon system for the past couple days by the way :) would look like this if I understand you correctly:
8d.png


The "pogo" orbit that I ballpark to be stable-ish like an L1 point I imagine to look something like this:
1p.png

- from every point on that green line all parts of the homogenous torus are equidistant, and therefore the gravity vector points always either straight "up", or straight "down". It's not really an orbit, but what are you gonna do.
 

Hielor

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I ment a figure 8 going like this:
8v.png
Won't work--when you're at the "top left" part of the orbit, you won't get pulled back to the "right" since you're closer to the "left" part of the torus.
 

Linguofreak

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@Izak The motion of an object inside a dyson sphere depends upon if there is a large sun or other bodies inside the sphere. If there is a massive central star, you could probably orbit normally unless you got close to the inside surface of the sphere, at which point it would attract you. Also you could orbit small planets if you were close enough to them. Otherwise, if its empty you're just going to drift towards a wall of the sphere and the direction of your gravitational acceleration will be towards the nearest wall (not the center).

Wrong: assuming that the sphere is of uniform density and thickness, the gravitational field on the inside will be the exact same as if it wasn't there at all. So if there's nothing in the center, the inside of the sphere will be flat space. If there's a star in the center, everything inside will orbit the star with the same orbital parameters as if the star weren't there.

Basically, if you're off center, the nearest point on the sphere attracts you more than the farthest, but more of the sphere is in the general direction of the farthest point, and if the sphere is uniform, these two exactly cancel out. See the following link, or Google for "gravitational field of a uniform hollow sphere": http://www.grc.nasa.gov/WWW/K-12/Numbers/Math/Mathematical_Thinking/grvtysp.htm

For a torus, see here: http://www.mathpages.com/home/kmath402/kmath402.htm

Also, the case of a torus is *not* analogous to two orbiting stars, because the stars are orbiting each other, whereas the two ends of the torus are just sitting there (plus, there may be gravitational effects from other parts of the torus.
 

tori

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Linguofreak, the two-star analogy assumes those stars just hanging in space, just like in that gravity sim. Imagine those two points (star masses) as an intersection of a plane and a circle. It behaves exactly as described in [2] (thanks for the links, good stuff), only in 2D.
 
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