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Gravity at very large distance

  1. Mar 18, 2009 #1
    Is gravity at very large distances zero?

    ie g = (k(m1 m2)/r^2) - k1 (you can also use the relativity formule for gravity here if you want)where k1 is some constant that makes gravity zero when (k(m1 m2)/r^2) reaches some threshold value

    so if I had and electron 10^6 billion light years from another electron (ie the opposite ends of the universe or the furtherest distance possible (you choose)).

    Now lets assume electron 1 is doing work, through gravity, on electron 2. I was wondering would the energy associated with the work electron 1 does on electron 2 be smaller than the smallest indivisibe unit of energy, which I think is planks constant?. Of course this assumes energy is quantised.

    If you assume 2 photons can have a gravitational effect on each other then you can replace electron 1 and 2 with photon 1 and 2 at opposite ends of the universe.
     
  2. jcsd
  3. Mar 18, 2009 #2

    madmike159

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    Using the inverse square law on 2 electrons 13.5 billion light years apart you get a force of 3.4x10-123N. I guess this would make the question pointless because by trying to measure the force QM laws would change your results any way. There isn't much point thinking about gravity in that way.
     
  4. Mar 18, 2009 #3
    Those electrons would be via entangled, and gravity is universal.
     
  5. Mar 21, 2009 #4
    Gravity depends of space-time deformation. If you have large size bodies having reasonnable mass, they will deform space - time on large distances but deformations will be low. For atomic particles, deformations are extremely important but they act on short distances. Consequently, interactions between atomic particles only exist at small scales and interactions btween planets and stars exist on large scales. For black holes, you are in the case of strong deformations with large bodies. Then their influence govern univers expansion... and it accelerate !! interesting isn't it ?
     
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