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Kepler's 3rd law

  1. Feb 23, 2015 #1

    Suraj M

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    While proving Kepler's 3rd law we get the equation
    $$ \frac{ΔA}{Δt} = \frac{2L}{m}$$
    we say L and m are constant,so aerial velocity is constant!
    But consider a body going around a black hole really quickly, then the mass would not be constant, right?
    So is Kepler's 3rd law violated or am i missing something?
     
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  3. Feb 23, 2015 #2

    phyzguy

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    Kepler's laws are a consequence of Newtonian Mechanics. They don't hold in the case of strong gravity or velocities close to the speed of light. For example as you get close to a black hole, there is a radius (clued the ISCO, for Innermost Stable Circular Orbit) within which there are no possible stable orbits. This is still outside of the event horizon.
     
  4. Feb 23, 2015 #3

    Suraj M

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    Oh Okay, is there some defined relation that works in all realistic cases, relating Time period and radius?
     
  5. Feb 23, 2015 #4

    phyzguy

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    Kepler's laws work really well in almost all realistic cases. I suspect it is very difficult to find an astronomical case where the velocity is so high or the gravity so strong that Kepler's laws don't apply. In the relativistic case (velocities close to the speed of light, or very strong gravity), I don't think there is any general rule like Kepler's laws. You have to work out each case.
     
  6. Feb 23, 2015 #5

    Suraj M

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    I guess the object would just get torn apart due to the gravitational force, before we calculate anything!
     
  7. Feb 24, 2015 #6

    A.T.

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    The object could be small which would make tidal forces negligible, but its trajectory could still be highly non-Newtonian.
     
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