Secular precession of the node

The motion of the node is described by changes in the angular momentum vector ${\bf A}$. We derived

$$\dot{{\bf A}} = {\bf r}\times {\bf F}$$

Only the noncentral part of ${\bf F}$ needs to be computed. This means that we need only

$$U = \frac{3 \mu J_2 R^2}{2r^5} z^2$$

when we compute the perturbations in ${\bf A}$. The corresponding force is

$${\bf F}= -\frac{3 \mu J_2 R^2}{2} (-5x z^2/r^7, -5y z^2/r^7, -5z^3/r^7+2 z/r^5).$$

Observe that we only have to keep the noncentral part of this, i.e.,

$${\bf F}= -\frac{3 \mu J_2 R^2}{2} (0, 0, 2 z/r^5),$$

which gives

$${\bf r}\times {\bf F}= -\frac{3 \mu J_2 R^2}{2} (2yz/r^5, -2xz/r^5, 0)$$

In particular, as expected, the $z$ component of ${\bf A}$ is conserved.

For the sake of simplicity, we compute the secular change in the node only for circular orbits, but for arbitrary inclination. In this case, we can choose a coordinate system in which the ascending node is on the positive $x$ axis. We then have

$${\bf A}= (0, -A \sin i, A \cos i)$$

where $A=\sqrt{\mu a (1-e^2)}$.

Consider a coordinate system $(x_1, y_1, z_1)$ attached to the orbital plane. In this,

$$x_1 = a \cos M, \;\;\; y_1 = a \sin M, \;\;\; z_1 = 0$$

where we can measure the mean anomaly $M$ from any point. We transform the orbit into the original frame by rotation around the $x$ axis and get

$$x = x_1, \;\;\; y = y_1 \cos i, \;\;\; z = y_1 \sin i.$$

Then, we have

$$yz = y_1^2 \cos i \sin i = a^2 \sin^2 M \cos i \sin i$$

and

$$xz = x_1 y_1 \sin i = a^2 \cos M \sin M \sin i.$$

Since for a circular orbit $r = a$, we have to compute, in essence, only

$$\frac{1}{2\pi}\int_{0}^{2\pi} \sin^2 M \, dM = \frac{1}{2}$$

and

$$\frac{1}{2\pi}\int_{0}^{2\pi} \cos M \sin M \, dM = 0.$$

We collect our results to get

$$\dot{A_1} = -\frac{3 \mu J_2 R^2}{2a^3} \cos i \sin i,$$

and

$$\dot{A_2} = 0, \;\;\; \dot{A_3} = 0.$$

Hence, the derivative of ${\bf A}$ is perpendicular to ${\bf A}$ and it is in the $(x,y)$ plane. The radius of the circle traversed by ${\bf A}$ (as the node changes) is $A \sin i$. Hence, the secular change in the node is

$$\Omega = \Omega_0 - \frac{3 \mu J_2 R^2}{2a^3A} (\cos i) \;t.$$

As before, we can use the mean motion to write this as

$$\Omega = \Omega_0 - \frac{3 n J_2 R^2}{2a^2(1-e^2)} (\cos i) \;t.$$

Of course, since we took $e=0$ as our reference orbit, this is valid only for $e=0$.