Secular precession of pericenter

Now we compute the orbital average of perturbations, i.e., the orbital average of $\dot{{\bf L}}$ which will give its secular perturbations. The derivative of the first component ($L_1$), from (29) is

$$\dot{L_1} = -\frac{3\epsilon\mu A}{r^5}y$$

The real work is to compute the orbital average

$$\frac{1}{2\pi}\int_{0}^{2\pi} \frac{y}{r^5} \' dM,$$

where $M$ is mean anomaly. But since the semi-major axis of the ellipse was aligned with the x-axis, parts of the integral with $y>0$ and $y<0$ will cancel each other, yielding zero for the integral. Therefore,

$$\dot{L_1} = 0$$

We have to do actual computation for $\dot{L_2}$, i.e., we need

$$\frac{1}{2\pi}\int_{0}^{2\pi} \frac{x}{r^5} \rd M,$$

We change the variable of integration from $M$ to $E$ (eccentric anomaly). Both change from $0$ to $2\pi$ for a full orbit. We have Kepler's equation

$$M = E - e \sin E$$

which gives

$$\,d M = \,d E -e \cos E \,d E = (1-e\cos E)\,d E$$

We also have

$$x = a(\cos E - e), \;\;\; r = a(1-e \cos E)$$

Then

$$\frac{1}{2\pi}\int_{0}^{2\pi} \frac{x}{r^5} \,d M = \frac{... ...i a^4}\int_{0}^{2\pi} \frac{\cos E - e}{(1-e \cos E)^4}\,d E$$

This integral can be computed analytically. In principle, we can compute $\dot{L_2}$ for any value of $e$. This computation will be left as an exercise.

We, however, take the easy road and use an expansion in the eccentricity $e$. The Taylor series gives us

$$\frac{1}{(1-e\cos E)^4} = 1 + 4 e \cos E + O(e^2).$$

Therefore we have

$$\frac{\cos E - e}{(1-e \cos E)^4} \approx (\cos E -e )(1+4 e \cos E) = \cos E -e + 4 e \cos^2 E - O(e^2).$$

From Trigonometry we recall that

$$\cos^2 E = \frac{1}{2}(\cos 2 E + 1),$$

and we get

$$\cos E -e + 4 e \cos^2 E = \cos E -e + 2 e \cos 2 E + 2 e.$$

Substituing this in the integral we get

$$\frac{1}{2\pi a^4}\int_{0}^{2\pi} \frac{\cos E - e}{(1-e \c... ...\cos E -e + 2 e \cos 2 E + 2 e \right)\,d E = \frac{e}{a^4}.$$

With this, we have

$$\dot{L_2} = \frac{3\epsilon\mu A}{a^4} e,$$

where we (somewhat sloppily) replaced the $\approx$ with $=$.