Time average of potential energy in Keplerian orbit

[acr]

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

$$U = \frac{-Gm_{1}m_{2}}{r}$$

and the time average of a function $$f(t)$$ over a time interval $$\tau$$ is defined to be

$$<f(t)>=\frac{1}{\tau}\int_0^{\tau} f(t) dt$$

and

$$r = \frac{a(1-e^{2})}{1+ecos \theta}$$

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

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

but I'm supposed to be verifying that

$$<U>=\frac{-Gm_1m_2}{a}$$

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

$$<\frac{1}{r}>=\frac{1}{a}$$

and not

$$\frac{1}{a(1-e^2)}}$$

Thank you for your time

 

[tiny-tim]

Welcome to PF!

Hello acr! Welcome to PF!

(nice LaTeX, btw )

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

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

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

Hint: use conservation of angular momentum.

 

[acr]

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

$$\textbf{v} \times \textbf{r} = r^2 \frac{d\theta}{dt}=const=\frac{L}{m}$$

is a valid statement? This would indicate that

$$dt=\frac{mr^2 }{L} d \theta$$

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?

 

[tiny-tim]

$$dt=\frac{mr^2 }{L} d \theta$$

That's right.

And now use that into the original integral, to substitute for dt.

 

[acr]

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

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

Which would indicate that I want L to be equal to

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

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