# How to find angle something turns as it orbits in orbital mechanics? [closed]

Short version:

• Take a look at this animation really quick. https://commons.wikimedia.org/wiki/File:Orbital_motion.gif

• The direction in which the arrows for $\alpha$ and $\nu$ is adjusted as the sattelite orbits earth in that image. How do I calculate what angle the sattelite must rotate to get the same result if I were to lazily do the same thing?

• Assume I'm the sort of person who did well to understand collegiate trigonometry and first year calculus as you explain this, please. For my own purposes, I'd prefer a 2d explanation, but a 3d one is quite fine and I can work out the 2d one from there on my own.

Long version:

I'm trying to create a basic orbital simulation in 2 dimensions. Despite not writing any code as of yet, I've gotten as far as grasping some of the basic concepts of orbital behavior.

Everyone talks about position in orbit, the velocity of an object, eccentricity, Kepler's three laws, and how to use the universal gravitational constant (or some approximation of it, anyway) to calculate it.

Nobody mentions the fact that, for example, the moon and other such objects apparently rotate as they orbit the Earth in such a way to keep one face towards us at all times, and how to calculate the angle of rotation an apparently stationary object must go through.

How do I ascertain the angle any satellite must rotate through in space to maintain the same face as it orbits like this?

I'd prefer a two dimensional explanation, but traditional 3-dimensional is fine. I can work it down as long as the mathematics don't quickly outstrip collegiate trigonometry and first year calculus.

In one of my responses, I note an image on imgur as part of my question for the person who answered. This is the image in question, so that a copy of it is here on StackExchange.

# For Those Who Get Here Via Search

There are four answers with useful information in them down below, and I would strongly encourage reading all four. While I started to summarize all four of them, it makes far more sense to let them stand on their own, especially as I struggle to take it in myself and work on the sim to make sure things are more-or-less complete.

If you get here via search and the four people help you, please give them the courtesy of upvoting each of them. All four have - together - answered this, at various degrees of required understanding and all with quite some patience and time investment.

Sadly this comes to the point where the checkmark can only have one receiver and one answer pointed to, when I'd like to checkmark all answers. I have given it to what appears on its face to be the generalized solution I initially came looking for.

• Hi and welcome to the Physics SE! The main requirement for this to happen is that the moon spins with the same period around itself and around Earth. But what are your thoughts so far? Please note that you are expected to have thoroughly searched for an answer before asking your question. And it's important to detail where you're stuck and why, in order to attract good answers. But also notice that this question is of the "problem-solving" type, rather than of the (on-topic) conceptual type, so I'm voting to close as off-topic. – stafusa Sep 8 '18 at 9:11
• This would make the second time a stack exchange has all but failed me if asking for help with a physics problem is outside the scope of it >.< [...] As far as concepts, this isn't homework. I'm 27 and teaching now, after all. But that being said, I initially thought to simply fake it; as the same face must face the orbital center, one could just forcefully snap to the line from its centrepoint and some surface point through to the centre of the satellite. ("Earth's core to point on moon surface to moon's core".) This is a classic game-design style solution of faking it until you can make it. – That Guy Sep 8 '18 at 9:35
• Will also add that the physics stack exchange bump reads "Q&A for active researchers, academics and students of physics". I thought I had read it right. Whereas in most SE it seems normal to give information on a problem and try to get a solution, this one is conceptual only? The bump, then, may need some love, or clarification on that point made more blatant. This or I've completely misunderstood stack exchange in general for years (and, oddly, gotten answers based on that misunderstanding). [...] That's not to say you're wrong. I thought I had checked the bump and I had; so we get to here. – That Guy Sep 8 '18 at 9:39
• @stafusa IMHO, the OP isn't necessarily asking for a worked solution. He mostly wants to know what concepts he needs to figure this out. But I agree that the question does look kind of off-topic in its present form. – PM 2Ring Sep 8 '18 at 10:20
• @PM2Ring I mean, it's not like I'm so markedly lazy that I couldn't work it out myself if I just had whatever critical piece I'm missing. [...] Looking at WP: Angular Momentum, I quickly get lost around... Not too quickly, actually, surprising myself. I get some coffee in me and a sheet of paper, I may be able to narrow this question down a bit closer to what's unclear after an hour or two chipping away at it. – That Guy Sep 8 '18 at 10:30

## 4 Answers

So in the more general case you have a ball of mass $m$ circularly orbiting at distance $R$ with speed $S$ and it also rotates with radius $r$ and surface speed $s$; we would generally describe it with some angular velocities $\Omega=S/R$, $\omega=s/r$. Part of the reason we do this is because we can write it's position as the vector $[R \cos(\Omega t),~R\sin(\Omega t)]$ and then calculus tells us directly that because it is orbiting at distance $R$, with speed $R\Omega$, its acceleration must be $R\Omega^2$. We know that this must be $GM/R^2$ due to the gravitational force, and so we know that at all times $R^3\Omega^2$ is a constant $GM$ that only depends on the mass of the body being orbited, not in the body doing the orbiting. This was Kepler's third law. We would say that $R$ goes proportional to $\Omega^{-2/3}.$

So we have to figure out the moment of inertia for the moon, this boils down to an empirical number $k$ which would max out at $k=2/3$ if all the moon's mass were carried in its crust or $k=0$ if it were carried in its core, the uniform-density solid sphere between these two is $k=2/5$. Whatever it is, the total angular momentum is a constant, assuming both axes align we would have $$L= m R^2\Omega + k m r^2 \omega.$$ Note that to keep $m,r,k,L$ fixed we would have to vary $\omega$ and $\Omega,$ but also $R$ to keep Kepler's third law up.

Now technically our calculus derivation did not work if $\Omega$ is varying in time. But the deviation is hopefully small, so we're talking about tidal braking over very very many orbits, and so we can still use all of this. We would find $a \Omega^{-1/3} + b \omega = 1$ for some constants $a, b$. We could even factor in, if you like, some tiny equatorial bulge, $b=b_0(1+\beta \omega)^2$ or so, and solve the cubic.

But the important lesson here is, none of this is telling you how fast the angular momentum gets shared between $\Omega$ and $\omega$. There is no easy fixed relationship that tells you how to move angular momentum from one of these anglular velocities to the other. That depends on a bunch of complicated stuff, because what's actually happening is that the planet is being deformed a little bit out of a spherical shape, and it is able to do a little bit of that elastically, but it's losing energy to its internal heat as this stretch moves across its surface. And that stretching is going to eventually cause a certain "resonance," a fixed integer ratio of the number of orbits it makes to the number of times it rotates in place. That resonance is not always 1:1, that is just the most common case, where the stretching stays roughly in a static place on the moon.

With that said, it might make sense to introduce a heuristic. If you know you want to get these two to come to equality, you might imagine that this is driven by the difference in their angular velocities, so that as they come closer to 1 to 1, the rate at which the rotation is changing also decreases. That sort of transport process usually ends up with an exponential decay with some time constant $\tau$. So you could model that as say first finding $\Omega_0$ such that $$a \Omega_0^{-1/3} - b \Omega_0 = 1,$$ then you could assume that $$\Omega = \Omega_0 + \alpha e^{-t/\tau}$$ for some constant $\alpha,$ and you could then solve for $\omega(t)$ and the model would probably not be 100% in accordance with exact measurements of any given moon, but it could be realistic enough for a demonstration. Like if you were putting this in a Kerbal Space Program type of gamr, you would probably want to nerd out with a much more detailed simulation, but the exponential decay model is probably good enough for a normal game with some moon-making collisions.

The Moon (and most other celestial bodies) rotates on its axis with an (almost) constant angular velocity. So if the body is locked to its primary, like the Moon is, you can work out its rotation if you know how long it takes to do one orbit.

The core concept here is conservation of angular momentum. As far as their axial rotation goes, celestial bodies are pretty much isolated, so the angular momentum of rotation is constant. However, in most cases, that's just a (very good) approximation. A satellite can exchange angular momentum with its primary due to tidal forces, since neither body is perfectly rigid and perfectly uniform. Thus the Moon was able to achieve tidal locking. Also, due to tidal forces, the Earth's rotation is gradually slowing down and the Moon's orbit is getting larger.

That means the Moon's orbital period is slowly increasing, and due to the tidal lock, so is its rotation period. But the effect is very tiny, and not worth bothering to model in a simple orbit sim which has various other errors that are much larger.

Small bodies close to their primary tend to get tidally locked, but that's not always fhe case. The only planet (currently) locked to the Sun is Mercury, and it's in a 3:2 resonance, not 1:1. Apart from that, rotation periods and rotation axes are fairly random. I discuss that a bit in this answer on the Astronomy SE.

• Due to not being able to attach images in response comments, I've posted the image for my question for your clarification here: imgur.com/Pt6itde [...] Due to stack exchange's overall policy to keep all relevant data in the site, I'm not exactly sure where to put that image to keep it here and properly contextualized. – That Guy Sep 8 '18 at 10:51
• @ThatGuy The main point of my answer is that the rate of spin of a celestial body is essentially constant. There are minute fluctuations, and very long term changes, but a simple orbit sim can safely ignore them. And if a satellite is locked with one face towards its primary its rotation period is identical to its orbital period. So the angular speed of rotation equals its mean angular orbital speed. – PM 2Ring Sep 8 '18 at 11:40
• "a simple orbit sim can safely ignore them" ... This is my fault, not yours, but long term hopes for this if I can get it working right would see it working to simulate approximately, if memory serves, 4.6 billion years or so. I suppose what I consider "simple" is a matter of scale; it'd still only be one component of many things I'd have to research and have sufficient understanding of. [...] But again, that's not your fault. That's mine for mistaking a "normal" person's sense of scale on the matter and mine doubly for being unclear about my intentions longer term. – That Guy Sep 8 '18 at 11:49
• @ThatGuy Your question talks about a basic 2D orbit sim, so that's what my answer is aimed at. FWIW, doing an orbit sim that's accurate over a billion years is extremely difficult. And if your intention is to model our solar system over that time span, forget it. :) You can't even get orbital parameter data of sufficient accuracy for that, what to speak of dealing with all the non-linear stuff that arises when you want to model real planets & not idealized ones. – PM 2Ring Sep 8 '18 at 12:19
• We have very good orbit data for the Moon, due to decades of laser ranging measurements, so we can model its orbit accurate to a few centimetres. But even that's not good enough for ultra long-term work. We can tell you where the Moon's orbit will be in 26 million years, but we cannot tell you where on that orbit the Moon will actually be. I read that figure of 26 million about a decade or so ago. Things may have improved a little since then, but it will still be in that ballpark. – PM 2Ring Sep 8 '18 at 12:24

Given that an orbit is (at least in a simple two-body setup) an ellipse, a fairly regular path for which the angle (of the vector from the planet to the moon relative to some suitable starting point) increases continually over time, then it is not hard to see that it must rotate a full angle (360 degrees, $2\pi$ radians), for every orbit to keep the same face as best as such a path will permit, since it must return to that face upon return to its starting point and moreover must not complete any more cycles in the interim.

As a free body, the rotation of the Moon itself is more or less just usual rotation (i.e. ignoring other perturbing forces at this level of detail), which is created by its possession of rotational angular momentum. With no external torques, the angular momentum of an object does not change. Angular momentum is related to rotation speed by

$$L = I\omega$$

where $\omega$ is the angular velocity (rad/s). (The rotational period is $T = \frac{2\pi}{\omega}$, derived by considering that a full angle is $2\pi$ rad.). With no change in angular momentum $L$, the rotational speed does not change either. It therefore executes this motion, with such uniform rotational speed, while traversing the orbit ellipse, independently of the fact of that elliptical traversal. (The whole reason it rotates at the same rate as it orbits is that it was effectively "torqued down" a long time ago, thus now that torque has ceased.)

Thus for your simple game, you can model the Moon as a circular object which rotates once per sidereal (true orbital) month (2.36 Ms, or about 27 days, which you will probably not be running as literally that period of course :) ), that is, you just define its angle by

$$\theta_\mathrm{orientation}(t) := 2\pi \frac{t}{T}$$

in radians, where $T$ is the period (which you can take as whatever value you need for your simulation) and $t$ the current simulation time. You can use whatever favorite rotation method you want to use to rotate the graphic by this angle (e.g. matrices, quaternions, etc.).

For the actual movement of the Moon along the orbit, however, this is a bit more complicated. In general, the value of orbital angle ("true anomaly", i.e. the angle I just mentioned formed by the Moon's position vector, where the zero point is taken as the periapse) over time is not a simple, even elementary function: in theory, you can write it using a sort of "generalized Lambert W function", but it is best computed numerically.

One final point I want to point out is that not all "other such objects" have this synchronous rotation characteristic. It is the case for the Moon, but moons of other planets may have other rotation patterns, even entirely decoupled from their orbit. The trick is whether or not there are strong enough gravitational interactions and of the right type - here due to the tides - to create some kind of orbital position-dependent torque that will then bring the body into both a rotation and orbit pattern that are related in some fashion. If you look at a planet like Jupiter, you will find the four famous "Galilean" moons have a complex "resonance" pattern caused by the fact that they essentially apply bursts of force to each other during their orbits as they get close. The outer moons of Jupiter, which are very small, effectively large asteroids, and thus neither exert any strong forces on each other nor have the spatial extent to have significant tides raised on them by Jupiter, have effectively decoupled rotational and orbital periods, as is the case with Earth and the Sun (e.g. the moon Himalia orbits at about 11 500 Mm from Jupiter, is about 150 km in size, orbits with a period of 21.6 Ms, about 2/3 of a year, but rotates far faster, very roughly a thousand times, with a period of only 28 ks.).

ADD: It seems that another desired piece of information here is related to finding just how exactly the Moon itself slows down to reach synchronous rotation. This is a bit more involved to explain.

Basically, what happens is that, thanks to Newton's law of universal gravitation, i.e.

$$F_G = G\frac{m_1m_2}{r^2}$$

the gravitational force from a planet, like the Earth, is not uniform with increasing distance, but rather falls off. (This is fortunate, if it weren't, then the whole Universe would have never been able to expand out of the Big Bang! For us and more practically, it also is what allows us to "escape" a gravity well with finite energy - if gravity kept tugging as strong as ever no matter how far away you were, you would eventually stop and fall back down even if projected out very fast, and you would have to continually be burning engines to climb higher and higher.)

What that also means, however, is that if we are given something like the Moon, because it is not a point mass but rather has some spatial extent - in particular, about 3474 km from front to back, it will experience more force ($r$ is smaller) on the side facing the Earth than on the side facing away. This results in effectively a force gradient across the Moon, which has the effect of stretching it like a piece of clay. As a result, it becomes slightly oblong, not a perfect sphere. (The same thing happens to the Earth as a result of the Moon, and this is why we have the tides, and moreover, why this is called a "tidal" force.) From the Moon's point of view, the apparent forces change (similar to how centrifugal force works) and it sees this as though it has a force being applied to its front and back which are pulling in opposite directions, stretching it out, while another force is applied around the "waist" between these two points, trying to squeeze it in. [If you've read anything ahead about black holes, you may have come across the idea of 'spaghettification'. This is a far more extreme example of this same thing, and conversely, this can be considered the effect in an extremely incipient and immature form.]

As the Moon rotates around the Earth in its orbit, since it's rotating at the same speed (ignoring also the slight, but ultimately netting zero due to equal reversal each period, irregularity due to the slight ellipticity of the Moon's orbit) as it orbits (i.e. the periods are equal), this bulge points toward the Earth at all times.

However, were the Moon in rotation, this would not quite be the case: in particular, if I were to apply a small impulse to the Moon to give it a "kick" that spins it slightly ahead, the "bulge" won't simply ripple through it right away and remain aligned because the Moon's material takes time to deform under the realignment of forces. In particular, this is as a result of both inertial (the natural property of matter to resist changing its state of motion in response to a force as in Newton's first and second laws which btw are not the same thing), and also mechanical (i.e. the natural stiffness of the material) effects. Thus, what will happen is the bulge will end up at an awkward angle, with part facing slightly behind the Moon in its orbit and part facing ahead.

However, both parts will be being pulled toward the center of the Earth, or by the Moon's view they will be being tugged apart. Effectively, the force situation looks like this:

If you were to think of these forces as being provided by two people engaged in a tug-of-war with respect to the bar in the middle, you would easily expect it would straighten out, right? And indeed, that's what's going on here. The two forces constitute a couple, and the result is to produce a net torque. Torque (also called "moment"), after all, is defined by

$$\mathbf{\tau} = \mathbf{r} \times \mathbf{F}$$

where $\mathbf{r}$ is the vector to the point of force application from the center (here along the bar), and $\mathbf{F}$ is the applied force. Thanks to the right-hand rule for the cross product, two such equal and contrary forces will combine their torques and thus produce a net torque that will tend to rotate the object.

Moreover, since the Moon is always orbiting, it actually can't stop being angled so long as it rotates faster, and as a result, it continues to experience torque. The torque thus acts to retard the rotation until it finally no longer makes an angle with its bulge toward the gravitator (Earth). If it had initially no rotation (a scenario that would allow us to actually see the back side of the Moon), the same effect would do the opposite and torque it up, speeding it up until it reached synchronous rotation "from below".

It is a bit involved to compute the effects of the torque in this situation, however the relevant equation is

$$\frac{d \omega}{dt} = \frac{45}{8} k G \frac{M_E^2 A^3}{M_M r^6} \sin(2\alpha)$$

where $M_E$ is the mass of the Earth, $M_M$ the mass of the Moon, $r$ the orbital distance, $\alpha$ the angle the bulge forms with the line between Earth and Moon, and $k$ is something known as the "tidal Love number" which takes account of the composition of the object. $A$ is the Moon's cross-sectional area.

Now the angle $\alpha$ itself depends on the rotation speed $\omega$ - in particular, we could imagine $\alpha = C (\omega_\mathrm{orbit} - \omega)$ where $C$ is some inertial constant (the harder the Moon would rotate, the further "back" it can keep its "bulge"), and thus technically the equation has an $\omega$ in the sine, i.e.

$$\frac{d \omega}{dt} = \frac{45}{8} k G \frac{M_E^2 A^3}{M_M r^6} \sin(C [\omega_\mathrm{orbit} - \omega])$$

and thus is nonlinear. If however $\alpha$ is suitably small (under about 500 mrad), we can approximate the sine as $\alpha$ itself and so if the Moon isn't rotating too fast (the inaccuracy should not matter for a game), we get

$$\frac{d \omega}{dt} = \frac{45}{8} k G \frac{M_E^2 A^3}{M_M r^6} \cdot C (\omega_\mathrm{orbit} - \omega)$$

Which has the generic form

$$\frac{d\omega}{dt} = K \omega + J$$

and we recognize this as an exponential equation for $\omega$. $K$ will be negative if the rotation is slowing, so over the long term, $\omega$ will decay exponentially - providing justification for @CR Drost's answer.

Note, however, that this also depends very crucially on the orbital radius $r$, and in fact that is not constant. In particular, the loss of angular momentum ($L = I\omega$ so $\frac{dL}{dt} = I \frac{d\omega}{dt}$) to rotation ends up as a gain in angular momentum by the rest of the system - both the Earth and the Moon's orbit. Thus technically speaking we actually have a coupled system of differential equations, and it will not be possible to solve exactly, but it can be numerically integrated.

That said, I'd suspect that CR Drost's suggestion to just use the exponential will suffice for simplistic modeling in game physics. It will though be rather poor if you are considering the Moon all the way back to the point of formation. If you want to get nitpicky, one possible way to improve on it is to fake (still approximate - see a trend here? This is physics! This is why that there's the long-standing joke about the "spherical cow", you try and toss away what you can and when you can't you still try to toss away as much as you can and still get away with it! If the cow can't be perfectly spherical, it better be an ellipsoid. If it can't be that, make it triaxial. Etc.!) it with an assumed circular orbit, i.e. take for the Moon that

$$G \frac{M_E M_M}{r^2} = M_M r \omega_\mathrm{orbit}^2$$

and since for the orbital angular momentum we have that $L_\mathrm{orbit} = (M_M r^2) \omega_\mathrm{orbit}$ that

$$G \frac{M_E M_M}{r^2} = \frac{L_\mathrm{orbit}^2}{M_M r}$$

you find how $L_\mathrm{orbit}$ affects $r$, the orbital distance. If we then also use another assumption that all lost angular momentum goes into the orbit, then $\frac{dL_\mathrm{orbit}}{dt} = -\frac{dL}{dt}$ where $L$ is the Moon's rotational angular momentum from before, and also that the rotation speed of the Moon $\omega$ is related to $L$ by $L = I\omega$, we can combine these to get a differential equation for $\omega$ taking into account the orbital outspiral. This will not be exponential in its early stages but contain a rapidly dropping-off power term, I believe (at least - our favorite word - approximately!). You can numerically integrate this (e.g. Runge-Kutta) for your simulation. Though personally, I'd say just stick to the exponential :)

That said, I should also point out what was mentioned in the comments that this also affects the Earth as well, and in fact the Earth is slowing down. More accurately, it is slowing right now (mind the $\frac{1}{r^6}$) at a rate of about 0.54 ns/s, equiv. 0.54 ms/Gs (milliseconds per Gigasecond). As a result, while we usually think of a day as 86 400 s - or equivalently, most commonly as 24 hours - and use this interval to define it in clocks, it is actually slightly longer: it is now 86 400.002 s: or about two milliseconds longer. This is because the current definition of a second is a fixed time unit based on the atomic clock, and it was established to refer to 1/86 400 of the day of the year 1900. There have now been 108 years, or 3.4 gigaseconds, since then, and so the day has grown by 0.54 x 3.4 = 1.8 ms, which I rounded to 2 ms (it's actually a bit more irregular than that due to other, more complex, factors.). Over your lifetime - typically 2.5 Gs (79 years) for the rich world, 2.2 Gs for the global average - you can expect it will lose another 1.3-1.4 ms or so. A whole second will then be had by around 2000 Gs from now (over 60,000 years), and the day will be 86 401 s long. The Earth's rotation will never get a chance to slow to the ~2.4 Ms length of the Moon's orbit: by this formula (we will actually need the full exponential to be more accurate, and that actually means, since $r$ is increasing, it will be longer, so this is a lower bound) it would take - using that 0.54 ms/Gs = 0.54 Ms/Es (know your SI prefixes well - three up from milli is mega, three up from giga is exa!) 4.4 Es for that to happen. But one exasecond is over twice the elapsed lifetime of the Universe, and the Sun will die long before then. In its death throe, the Earth and Moon will be dragged into the greatly expanded stellar atmosphere and be vaporized.

Contrary to what H.G. Wells imagined in his old novel "The Time Machine" (where his time traveler traveled far into the future and reported the slowing down of the Earth's rotation), there would not be time for this to occur in reality. (At his time, much less was understood about astrophysics and especially stellar dynamics than now.)

(Keep in mind this is the mean solar day, not the sidereal day: the difference is the latter is the true rotation period, the former is the time between two noons or midnights. The two are not equal due to the effects of the motion of the Earth in its orbit around the Sun.)

• So, for starters, you can have an upvote for actually explaining some of the core orbital mechanics. My last question left is, in the case where torque has not been (functionally) exhausted, how does one calculate these things? Torque I'm assuming observes conservation of angular momentum (which would tie multiple people here together nicely), but in what way is that calculated and reflected? (In other words, "how does one solve for the general case?") – That Guy Sep 8 '18 at 11:36
• I am not quite clear what you are asking for. Are you asking to find how you'd actually compute the torque in the case of a moon that is undergoing tidal deceleration but has not yet settled into a synchronous rotation pattern? – The_Sympathizer Sep 8 '18 at 11:39
• More or less. The chief interests here being how one calculates the torque and how that affects the rotational speed of the satellite (slash "moon"). – That Guy Sep 8 '18 at 11:44
• @ThatGuy: 'the torque' due to tidal effects is slowing the Earth's day (it used to be eighteen hours...) and causing Luna to move away. It isn't an orbital effect, but a redistribution of mass (tidal bulges). This is why we have leap seconds. – Whit3rd Sep 9 '18 at 6:02
• @That Guy: Added a crapload of details! – The_Sympathizer Sep 9 '18 at 6:58

Luna is not exactly presenting the same face toward Earth; it is in a slightly elliptical orbit, but has simple rotation (no significant angular acceleration) that has become locked to the average orbital rotation by tidal effects.

So, there is slight libration of the face presented to Earth, and we can see more than half of the lunar surface (59%).

While Luna's rotation is not variant during a month, its angular motion across the sky is slightly faster at perigee, and slower at apogee, and that means our Earthly viewing angle accelerates/decelerates through the orbital phases, just as Kepler's third law dictates. If the orbit were exactly circular, there would be no libration, and the lunar rotation and orbital angle would be synchronous.

• I fail to see how this answers the question. – Kyle Kanos Sep 8 '18 at 10:28
• I feel like there's a forest/trees problem here... So the moon is ostensibly (and probably actually) not a very good example of what I'm trying to understand. This nevertheless does nothing to address what I am actually trying to understand. – That Guy Sep 8 '18 at 10:29
• What you say is correct. But you failed to mention that the Moon's axial motion is almost perfectly uniform. And that's the main thing the OP is asking about. – PM 2Ring Sep 8 '18 at 10:33
• Actually, I'm asking for the mathematical model in physics that explains this behavior in the general case, with prefernce for 2d over 3d. I'll have an image for you here in a moment in regards to your answer and what you seem to be driving at to get towards the right answer. – That Guy Sep 8 '18 at 10:48
• @ThatGuy : How is the moon 'not a very good example'? The rotation of any moon or planet is CONSTANT, its angle (with respect to the fixed stars) is changing at 360 degrees per siderial "day"; only over megayear scales do any variations become measurable. – Whit3rd Sep 12 '18 at 0:01