1. Limited time only! Sign up for a free 30min personal 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!

Time average of potential energy in Keplerian orbit

  1. Sep 5, 2009 #1

    acr

    User Avatar

    Hello,

    I'm trying to show that the time average of the potential energy of a 2-body system is equal to the instantaneous potential energy of that system when the two bodies are separated by a distance equal to the semi-major axis, a.

    So I know that

    [tex]U = \frac{-Gm_{1}m_{2}}{r}[/tex]

    and the time average of a function [tex]f(t)[/tex] over a time interval [tex]\tau[/tex] is defined to be

    [tex] <f(t)>=\frac{1}{\tau}\int_0^{\tau} f(t) dt [/tex]

    and

    [tex]r = \frac{a(1-e^{2})}{1+ecos \theta}[/tex]

    Can anyone tell me what's wrong with my attempt? Here it is:

    [tex]<U>=\frac{1}{P} \int_0^P \frac{-Gm_1m_2}{r} dt = \frac{-Gm_{1}m_{2}}{2\pi a(1-e^2)}\int_0^{2\pi}(1+ecos\theta) d\theta
    =\frac{-Gm_1m_2}{a(1-e^2)}
    [/tex],

    but I'm supposed to be verifying that

    [tex]<U>=\frac{-Gm_1m_2}{a}[/tex]

    I can't seem to figure out how to get rid of that factor of (1-e^2) in the denominator. I suppose this can be simplified by just asking why/how

    [tex] <\frac{1}{r}>=\frac{1}{a}[/tex]

    and not

    [tex]\frac{1}{a(1-e^2)}}[/tex]

    Thank you for your time
     
  2. jcsd
  3. Sep 6, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hello acr! Welcome to PF! :smile:

    (nice LaTeX, btw :biggrin:)
    ah, you haven't converted 1/P properly (t = P when θ = 2π doesn't mean P = 2π, does it? :wink:), or dt.

    Hint: use conservation of angular momentum. :wink:
     
  4. Sep 6, 2009 #3

    acr

    User Avatar

    Thanks tiny-tim,

    I'm trying to make sense of how to incorporate the conservation of angular momentum. Would it be correct to assume that

    [tex]
    \textbf{v} \times \textbf{r} = r^2 \frac{d\theta}{dt}=const=\frac{L}{m}
    [/tex]

    is a valid statement? This would indicate that

    [tex]
    dt=\frac{mr^2 }{L} d \theta
    [/tex]

    but that doesn't seem to help much when I evaluate the new integral (or if it does, it's eluding me!).... I'm still not getting something. Any suggestions?
     
  5. Sep 6, 2009 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    That's right. :smile:

    And now use that into the original integral, to substitute for dt. :wink:
     
  6. Sep 6, 2009 #5

    acr

    User Avatar

    Thanks again. Because r is the separation between m1 and m2, L would be m2r2 d theta/dt (choosing to let r represent the position of m2 relative to m1). When I use the result above for dt, the new integral ends up coming out as

    [tex]

    \frac{-Gm_1 m_2^2 a (1-e^2)^{1/2}}{L}

    [/tex]

    Which would indicate that I want L to be equal to

    [tex]

    m_2 a^2 \sqrt{1-e^2} =m_2 a b

    [/tex]

    I don't see how it could be, though. I am still using 2pi in place of P. I'm not sure how I should go about writing P in terms of angular measure, if this isn't correct. Any insights?

    Thank you
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Time average of potential energy in Keplerian orbit
  1. Keplerian Orbits (Replies: 2)

Loading...