# Moon's orbit precession

1. Nov 28, 2014

### chuligan

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

By definition k = w^2; and we know already: w = 2pi/18.6y; because the period of precession is 18.6 years.
Therefore: k = (2pi/18.6)^2 = 0.114 1/y^2 = 1.146e-16 1/s^2
But what is this number?

2. Nov 28, 2014

### ShayanJ

You're assuming a linear oscillation but that's not how it works!

3. Dec 8, 2014

### chuligan

This should be rather linear, because the z changes exactly as: z = z_max sin(wt), under a precession.

But You can assume I don't know anything, and simply derive the orbital precession. :)

4. Dec 8, 2014

### ShayanJ

Take a look at here! Its not even an oscillation!
Anyway, the point is, the moon is constantly rotating so, whether what you want is an oscillation or not, your z is constantly changing with no simple pattern. The parameters you should considers are parameters of showing the inclination of moon's orbital plane.
Just consider earth's equator. Now extend it to infinity. This gives you an infinite plane. Then consider the line joining the centre of the earth to centre of the moon. As the moon rotates, this sweeps a plane. That plane is the moon's orbital plane around the earth. The angle between these planes is what you should be considering in you calculations.
I don't know what you're doing. If you want to know how this angle changes, then you should study about deviations the earth-moon system from a Keplerian system. But if you want to assume a particular dependence of time for that angle and see how that affects moon's motion, then you should translate that change in that angle to a dependence of time of what you call z, which even if that angles had a simple oscillation, I don't think z would!

Last edited by a moderator: May 7, 2017
5. Dec 9, 2014

### chuligan

This is a linear motion almost.
You should look at the z only, not at the apse, nor other motion in the x-y plane.

6. Jan 4, 2015

### lpetrich

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.

Last edited: Jan 4, 2015
7. Jan 4, 2015

### lpetrich

To solve the equations for ξ and η, I will guess the likely form of the solution and then solve for that guessed form's parameters. Something sometimes called an Ansatz. I will also impose ω' << ω, so as to get lowest-order perturbations only. The guessed form:
$$\xi = \left( \frac{\omega'}{\omega} \right)^2 ( \xi_0 + \xi_1 \cos (2(L-L')) ) ,\ \eta = \left( \frac{\omega'}{\omega} \right)^2 ( \eta_1 \sin (2(L-L')) )$$
Plugging it in gives
$$\xi_0 = - \frac16 ,\ -7 \xi_1 - 4 \eta_1 = \frac32 ,\ - 4 \xi_1 - 4 \eta_1 = - \frac32 ,\ \xi_1 = -1 ,\ \eta_1 = \frac{11}{8}$$
So for the equation for z, we get
$$k(t) = \omega^2 + \omega'^2 \left( \frac32 + 3 \cos (2(L-L')) \right)$$

To solve for z, we get a hint from Floquet theory - Wikipedia. The z solution has the form
$$z = \sum_k z_k \sin(H + 2k(L-L')) ,\ H = H_0 + \omega_z t$$
We start with ωz being close to ω. Working out the lowest-order perturbed solution, that gives us
$$\omega_z = \omega + \frac34 \frac{\omega'^2}{\omega}$$
giving a lowest-order precession period of (4/3)*((TE)2/TM).

This gives a period of about 17.8 years, close to the actual one of 18.60 years.

Doing a lowest-order calculation of the precession of the Moon's perigee gives the same period, but the actual one is about 8.85 years. This twice-as-fast precession had made Isaac Newton's head ache, and around 1750, some of his successors doubted the inverse-square nature of gravity. But around then, Alexis Clairaut did the calculation to higher order, and he found that most of the discrepancy could be accounted for by the next few terms in the series expansion.

What I've just described is part of the Hill-Brown lunar theory. "Theory" here being a solution of the equations of motion.