1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Precession in a central potential

  1. Aug 23, 2007 #1
    1. The problem statement, all variables and given/known data

    Consider the following parametrization of an orbit in polar form,

    [itex] \ell u = 1 + e \cos[(\phi -\phi_0)\Gamma][/itex]

    where u = 1/r.

    I'm trying to find the shift in the angular position of the periapsis after one complete orbit.

    3. The attempt at a solution

    Choose axes so that the point of first closest approach is [itex]\phi_0[/itex].

    [itex] u'(\phi) = - \Gamma e \sin[(\phi -\phi_0)\Gamma][/itex]

    Setting [itex]u'(0) =0[/itex] we obtain

    [itex] (\phi -\phi_0)\Gamma = n \pi[/itex] where n is an integer.

    So after one complete orbit I guess the shift is [itex]\Delta \phi = \phi - \phi_0 = \frac{\pi}{\Gamma}[/itex], or should that be [itex]\frac{2pi}{\Gamma}[/itex]?

    Thanks.
     
  2. jcsd
  3. Aug 23, 2007 #2
    I preffer this:

    [itex]\frac{2pi}{\Gamma}-\frac{2pi}{1}[/itex]

    To solve this problem, you even don't need to consider the exact trajectory.
    You simply need to check how much (an angle) the trajectory deviates from periodicity.
    For [itex]\Gamma = 1[/itex] the trajectory is periodic and there is no shift.
     
  4. Aug 23, 2007 #3
    How did you obtain 2Pi/Gamma - 2Pi?
     
  5. Aug 23, 2007 #4
    On reflection I get [itex]2\pi/\Gamma[/itex]. I think the first perihelion occurs at 0, the first aphelion occurs at [itex]\pi/\Gamma[/itex] and the second perihelion at [itex]2\pi/\Gamma[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Precession in a central potential
Loading...