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Is Gravitational Force conserved at the origin (r=0)?

  1. Sep 1, 2015 #1
    I know gravity is a conservative force field and can be treated as such for all intents and purposes, but I was just thinking that in order to show that a vector field is conservative that vector field must be defined everywhere (gravitational force field is not defined at r=0).

    I was thinking that you may need to examine a curve that encloses the singularity (r=0) like the unit circle, but I'm not sure this is the best way to think about this problem.

    Any additional insight into how to treat/explain the conservative/non-conservative nature of the gravitational force field at the origin would be appreciate.
  2. jcsd
  3. Sep 1, 2015 #2


    Staff: Mentor

    Why not?
  4. Sep 1, 2015 #3


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    Staff: Mentor

    We're talking about the gravitational field of an ideal point mass here, right? If we're dealing with a real planet or other spherically symmetric mass with non-zero size, there won't be a singularity at r=0 and there won't be any problem.

    But with that said, yes, you have the right idea. If there is a singularity at a given point, you can't evaluate a line integral that passes through that point and you must consider only paths that pass arbitrarily close to it. In practice, this is a seldom a concern because we never encounter gravitating point masses in real life; when we model a real planet as a point mass, we're only working with the space outside the surface of the planet so we don't have trajectories that pass through r=0.
  5. Sep 1, 2015 #4
    What does this mean for a radial elliptic trajectory, e.g. for two point masses? Is their position still predictable after they passed the singularity? That shouldn't be possible if the integral can't be evaluated through that point.
  6. Sep 1, 2015 #5
    Would a mass, point or otherwise, pass through the singularity? No impact?
  7. Sep 1, 2015 #6


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    If you want to take it that far, you are into GR - no?
  8. Sep 1, 2015 #7


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    Even worse - a point mass will be a black hole.
  9. Sep 1, 2015 #8


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    Staff: Mentor

    I'm sorry, but I don't understand what you're asking here. The elliptical orbit of two non-trivial masses approximated as point particles doesn't pass through either singularity, so the problem doesn't arise.
  10. Sep 2, 2015 #9
  11. Sep 2, 2015 #10
    I don't know if they pass through the singularity. In the limit of zero angular momentum a classical orbit would result in a degenerate ellipse with a 180° turn in the center. In this case the point masses would bounce off the center. But I do not think it is that easy. The singularity changes the situation.

    That would result in additional complications. Thus we should limit the interaction to gravity only.
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