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I have been working through some lecture notes and am quite confused on something. I am trying to understand how to average a quantity over an orbit (Keplerian) but I am struggling to get a clear idea on this. The notes I am using is: http://www.sns.ias.edu/sites/default/files/isima1.pdf

So I am trying to do the exercise on page 8, but have no idea how to get the solutions shown.

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The time average of a periodic quantity $Q(t)$ is, by definition,

$$\langle Q \rangle=\frac{1}{T}\int_0^TQ(t)\,dt$$

where $T$ is the period.

For Keplerian orbits, it is usually easiest to change the integration variable to the angular coordinate $\psi$ and express all quantities being integrated in terms of $\psi$.

The exercise thus consists of doing a variety of angular integrals. There is a typo in one of the results.

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  • $\begingroup$ thanks for this. I have tried to use this in solving the integral but I yield answers. Could you please go through an example question (question 1 for example)? I am also using r = a(1-e^2)/( 1 + e cos(phi) ) $\endgroup$
    – Warrenmovic
    Commented May 13, 2020 at 8:54
  • $\begingroup$ Complete solutions to homework-like exercises aren’t allowed on this site. What problem are you running into? You have the correct $r$. Do you know how to change the variable of integration from $t$ to $\psi$? You need to make use of (12). $\endgroup$
    – G. Smith
    Commented May 13, 2020 at 16:38
  • $\begingroup$ I am actually having an issue with the resulting integral. Take question 2 as an example, I end up with an integral that I am struggling to solve. $\endgroup$
    – Warrenmovic
    Commented May 13, 2020 at 17:58
  • $\begingroup$ The integrals with powers of $1+e\cos\psi$ in the denominator are difficult, although Mathematica can do them. I know of two ways to do them without a computer. One is to turn the integral into a contour integral around the unit circle in the complex plane, and use the residue theorem. The other is to expand in powers of $e$, integrate term-by-term, and then try to recognize what function of $e$ the resulting infinite series is. If you are only having a problem doing integrals, ask on Math SE. $\endgroup$
    – G. Smith
    Commented May 13, 2020 at 18:07
  • $\begingroup$ I used Mathematica to check all the results. The first answer should have a + rather than a -. $\endgroup$
    – G. Smith
    Commented May 13, 2020 at 18:11

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