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I Gravitational Binding Energy of a Torus

  1. Apr 9, 2016 #1
    I was looking at the wikipedia page for the gravitational binding energy of a sphere, but let's say that there was a toroidal planet. What would its gravitational binding energy be?
    I have attempted the solution similar to what they did on wikipedia and obtained:
    [tex]U = -4 G \pi^5 R^2 r^4 \rho^2 [/tex]https://lh3.googleusercontent.com/fPHuQ6I0NONuFyo7tz0OBkPVGWVW6kVChq_TjWU05B-jQRFqvMmnVLfPN7Q9OMCPe3qTMw=s95
    R and r are shown in the diagram to be the total radius of the torus and the radius of the tube of the torus respectively.

    I believe with a substitution for [itex]\rho[/itex] as density, this simplifies to:
    [tex]U = -G M^2 \pi[/tex]

    But this seems to simple. Any ideas?
  2. jcsd
  3. Apr 9, 2016 #2
    A torus can't form or be maintained via gravitational binding. If that were the only "force" (spacetime curvature, to be pedantic) it would have to collapse into a sphere. Even if it were rotating very fast, you wouldn't get a toroidal planet - instead, an oblate spheroid, i.e. disk.

    Strong internal structure of some sort would be required to hold such a shape. Check out Larry Niven's science fiction "Ringworld" for a good example. In such case the tension would produce negative gravity (according to GR); in an extreme case it would actually repel instead of attracting! In a more normal case you could have a small amount of matter, like an asteroid, in any shape imaginable - even a statue of Elvis. But such a small object is maintained by electromagnetic (chemical) binding; gravity would be negligible.

    Bottom line, it can't be analyzed as simply gravitational binding energy. By the way the same would be true for any non-sphere shape: cylinder, sheet, etc.
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