= Solution
The orbital plane has normal parallel to the conserved <angular momentum>. If its inclination to the equatorial plane is $\psi$, then $L_z=L\cos\psi$. Therefore
$$
p_\theta^2=L^2-\frac{L_z^2}{\sin^2\theta}
=\frac{L^2}{\sin^2\theta}(\sin^2\theta-\cos^2\psi),
$$
so
$$
p_\theta(\theta)=\pm\frac{L}{\sin\theta}
\sqrt{\sin^2\theta-\cos^2\psi}.
$$
It vanishes when $\sin\theta=|\cos\psi|$, equivalently at the greatest northern and southern latitudes reached by the inclined orbit. These are the turning points of the polar motion.
Back to article page
