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Undefined Gravity

  1. Nov 9, 2015 #1
    Newton's law of universal equation is equal to:

    F= G (m1 * m2)\ r squared

    However if r = 0, then F is undefined. What does it please mean if gravitational force (F) is undefined?
     
  2. jcsd
  3. Nov 9, 2015 #2
    It cant be 0. That would mean that two objects will be in the same place. Also because of strong force.
     
  4. Nov 9, 2015 #3

    Nugatory

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    Staff: Mentor

    It means that the equation you're using doesn't apply under the conditions in which you're trying to use it. When this happens, you have to look at the problem and understand why. In this case there are at least two (mutually consistent) explanations for what's going on:
    1) There are no ideal point masses in the real world, so Newton's law is properly understood as being about the centers of gravity of real objects with non-zero size. In that case the centers of gravity are always separated by some non-zero amount, ##r## is never zero, and the infinity that you're getting when you plug ##r=0## into the equation is natures way of telling you that you're making a mistake.
    2) Quantum-mechanical and general-relativistic effects come into play at extremely small distances, and Newton's law doesn't allow for these.

    There's more real understanding in #1 than in #2, but there's no contradiction between them.
     
  5. Nov 9, 2015 #4

    HallsofIvy

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    Gravitational force is the force between two massive objects. Taking r= 0 means that two massive objects would be occupying the same location which is impossible.
     
  6. Nov 9, 2015 #5
    It is impossible for fermions but not for bosons (not to speak about dark matter). As already mentioned by Nugatory the real problem is the size. There are no point masses in reality.
     
  7. Nov 9, 2015 #6

    mfb

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    2016 Award

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    If you want to add quantum mechanics: the probability to find two particles at exactly the same spot is zero, and the expectation value for the potential energy is zero for every physical continuous distribution for the relative position. This is true for both fermions and bosons, it does not depend on the spin.
     
  8. Nov 9, 2015 #7
    Now is the time to change from B to A.
     
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