Time average of potential energy in Keplerian orbit

  • Thread starter acr
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  • #1
acr
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0
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
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
Welcome to PF!

Hello acr! Welcome to PF! :smile:

(nice LaTeX, btw :biggrin:)
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]
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:
 
  • #3
acr
6
0
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?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
[tex]
dt=\frac{mr^2 }{L} d \theta
[/tex]
That's right. :smile:

And now use that into the original integral, to substitute for dt. :wink:
 
  • #5
acr
6
0
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
 

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