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Hall's solution for Mercury's precession

  1. Jan 10, 2014 #1
    Hall showed that a slight increase in the radius exponent in Newton's equation would cause precession. What would the result be if the exponent was slightly less than 2.
  2. jcsd
  3. Jan 11, 2014 #2

    In which Hall gives a formula, attributed to Bertrand,

    [itex]\theta = {\pi \over {\sqrt{n+3}}}[/itex]

    where [itex]\theta[/itex] is the angle between minimum and maximum radius vector for an orbit of small eccentricity, and [itex]n[/itex] is the exponent to be used in the equations of motion. He says, "If [itex]n = -2[/itex] we have the Newtonian law." So I imagine you can figure it out from there.
    Last edited: Jan 11, 2014
  4. Jan 11, 2014 #3
    Thanks for this. I must be thick, but I'm still a bit puzzled.
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