chuligan said:
I want to compute the precession of the Moon's orbital plane, sometimes called as the precession of nodes.
I assume simple harmonic motion;
the Moon orbit lies in a plane X-Y, so it oscillates along z:
z'' = -kz
So you wish to find k. One can do that in the limit of a small oscillation out of the Earth's orbit plane.
$$ {\ddot z} = - \frac{\partial V}{\partial z} = - \left( \frac{\partial^2 V}{\partial z^2} \right)_{z=0} z + \cdots $$
to lowest order, where V is the potential per unit mass. I'll find V in the limit where m(Moon) << m(Earth), and the Sun is much farther away from the Earth than the Moon is. One can easily extend this discussion to the case of m(Moon) ~ m(Earth). The potential is
$$ V = - \frac{GM_E}{\sqrt{x^2 + y^2 + z^2}} - \left( \frac{GM_S}{\sqrt{(x' - x)^2 + (y' - y)^2 + z^2}} - \frac{GM_S}{\sqrt{x'^2 + y'^2}} \right) $$
Thus,
$$ \left( \frac{\partial^2 V}{\partial z^2} \right)_{z=0} = \frac{GM_E}{r^3} + \frac{GM_S}{r'^3} $$
where radii ## r = \sqrt{x^2 + y^2} ## and ## r' = \sqrt{x'^2 + y'^2} ##.
It seems like we've got it made, but for one little thing. The Sun's perturbations on r have a relative size of (second potential term) / (first potential term), so we must include those terms also. Thus, k must be a function of time: k(t). So we must now calculate them. First, the equations of motion:
$$ \frac{ d^2 {\mathbf x}}{dt^2} = - \frac{GM_E {\mathbf x}}{r^3} + \frac{GM_S}{r'^3} (3 ({\mathbf x} . {\mathbf {\hat x'}}) {\mathbf {\hat x'}} - {\mathbf x}) $$
Let's use polar coordinates:
$$ \frac {d^2 r}{dt^2} - r \left( \frac{d\theta}{dt} \right)^2 = - \frac{GM_E}{r^2} + \frac{GM_S r}{r'^3} (3 ({\hat r} \cdot {\hat r'})^2 - 1) $$
$$ r \frac {d^2 \theta}{dt^2} +2 \frac{dr}{dt} \frac{d\theta}{dt} = \frac{GM_S r}{r'^3} (3 ({\hat r} \cdot {\hat r'}) ({\hat \theta} \cdot {\hat r'}) ) $$
Now put in perturbations of circular orbits:
$$ L = L_0 + \omega t ,\ r = a (1 + \xi) ,\ \theta = L + \eta ,\ {\hat r} = \{ \cos \theta , \sin \theta \} ,\ {\hat \theta} = \{ - \sin \theta , \cos \theta \} ,\ \omega^2 = \frac{GM}{a^3} $$
Substituting in,
$$ k(t) = \omega^2 (1 - 3\xi) + \omega'^2 $$
$$ {\ddot \xi} - \omega^2 \xi - 2 \omega {\dot \eta} = 2 \omega^2 \xi + \frac12 \omega'^2 (3 \cos(2(L-L')) + 1) $$
$$ {\ddot \eta} + 2 \omega {\dot \xi} = - \frac12 \omega'^2 (3\sin (2(L-L'))) $$
I'll continue in my next post here.
