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Gravitational acceleration for a rotating torus

  1. May 16, 2008 #1
    gravity of a rotating torus

    While Gm/r^2 seems to apply only for gravitational acceleration towards spheroids, what would be the equation for gravitational acceleration towards a rotating torus? I'm sure the equation would be the same for spheroids and toroids at large distances but what would be the equation in close proximity to the torus (i.e. the outer surface)?
    Last edited: May 16, 2008
  2. jcsd
  3. May 16, 2008 #2
  4. Sep 1, 2008 #3
    Thanks for link Shalaya, while the mathpage is informative and backs up the idea that gravity cancels out at the centre of the torus and that anything that falls inside the torus is pulled away from the centre, is there a more 'rule of thumb' equation on the lines of Gm/r^2?
    Last edited: Sep 1, 2008
  5. Sep 2, 2008 #4
    While it doesn't relate to a rotating torus, I did find an equation that looks at the gravitational field on the polar axis of a torus-


    where x is the distance from the centre of the torus (the torus being on the y,z plane) and a is the radius of the torus.
    source- http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/MoreGravity.htm

    There seems to be a discrepancy with Gm/r^2 over very large distances but it is at least within the same order and it could be argued that over a certain distance, Gm/r^2 is used. Within about 50a, the results are very similar to Gm/r^2 until within close proximity of the torus (~5a) where the gravitational acceleration begins to reduce until it reaches zero at the centre.

    I had previously assumed that gravity on the same plane as the torus would be something like-

    [tex]g=\frac{Gm}{\left(r-r_t \right)^2}[/tex]

    where rt is the radius of the torus. The resulting negative figure for the gravitational field once inside the torus doesn't imply 'negative gravity' as such, just that now as you move to the centre of the torus, gravity is pulling the other way (i.e. to the inside face of the torus). Based on the above, this could be rewritten as-

    [tex]g=\frac{Gm}{\left(r^2-r_t^2 \right)}[/tex]

    providing a very rough rule of thumb equation-

    [tex]g_t=\left|\frac{Gmr}{\left(r^2+r_t^2\right)^{3/2}}sin^2\theta\right|+\left|\frac{Gm}{\left(r^2-r_t^2 \right)}cos^2\theta\right|[/tex]

    so that the first equation dominates at the poles and the second equation dominates at the equator. This produces fairly similar results to Gm/r^2 (up to about 50r from the torus) until about 5r where the results for the pole and the equator begin to diverge. While not perfect, it does produce a fairly simple alternative to Gm/r^2.

    Last edited: Sep 2, 2008
  6. Sep 2, 2008 #5
    Once inside the torus, the gravitational field in the equatorial plane needs to take into account multiple sources and gradually reduce to zero so r is introduced to the top half of the fraction. This also appears to take care of any geometric singularities that might have cropped up at the centre on the equatorial plane.

    Gravity for a torus where r<rt-

    g_t=\left|\frac{Gmr}{\left(r^2+r_t^2\right)^{3/2}}sin^2\theta\right|+\left|\frac{Gmr}{\left(r^2-r_t^2 \right)}cos^2\theta\right|[/tex]

    The way I see it is that this gives an approximation of the gravitational field at any point around the torus, relative to the part of the torus that is most local to the point being looked at.

    Re post #4, the pole equation is actually good to infinity (as is the equatorial equation).
    Last edited: Sep 2, 2008
  7. Sep 8, 2008 #6
    I've looked at this in a bit more detail in the blog linked below, establishing points of reference on the torus which makes it appear as a ring of beads and then calculating the proper distance from each bead-


    Attached below are diagrams that demonstrate the trigonometry involved and an excel spreadsheet based on 8 and 16 points of reference. The only bits of info that need to be punched in are mass, r (radius), radius of torus and plane angle-

    Attached Files:

    Last edited: Sep 8, 2008
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