# How to calculate rate of axial precession?

I want to start with the fact that I am by no means an astronomer, or even a hobbyist in the field. I am attempting to build a fantasy world which still reflects accepted physical laws.

My problem comes from the fact that I have no idea how to go about figuring out the period of axial precession for a fantasy earth-like planet. For the Earth, I believe the period is roughly ~26,000 years. I've spent a few days trying to find resources, but in all honesty I can't wrap my head around half of the equations I've seen presented.

How would I go about doing this calculation for a given planet, and which of its properties would I have to specify in order to do this calculation? I've got 2 moons around my earth like planet, one of which is large enough to hold a stable atmosphere and be inhabited (which may make it more of a binary planet system than a planet-moon relationship).

I've asked this same question in the astronomy stack exchange, but it seems that the commentators over there believe you folks' may be able to help also...

• Depends very very very much on the exact specific planet in question. If the planet is perfectly spherical and has a perfectly uniform mass distribution then it won't precess. For a general object with no symmetry, such as a planet, you must do a numerical simulation. So you need to give the exact physical characteristics of the planet, down to actual shape (planets are not perfect spheres or ellipsoids) and the actual mass distribution. Then hope that somebody knows enough physics, has access to the right computer programs, and has the time to work the numerical simulation for you. – AlexP May 4 at 19:30
• ... On the other hand, just pick a number and be done with it. Any period of precession between 5,000 to 50,000 years is perfectly believable. – AlexP May 4 at 19:33
• (a) You're worried about very small details if you're worried about axial precession. Does your story really need that level of detail? If axial precession doesn't play a part in your plot, then I'd simply skip it. (b) Axial precession is a very complicated function of solar system creation. It involves the formation of the star and planets (pretty much all of them). It involves matter passing through the system (asteroids). And it involves bazillions of years to stabilize. Worst of all, it's about as predictable as spinning a top (watch the axis!) @AlexP's pick-a-number answer is good. – JBH May 4 at 23:36
• Not knowing what axial precession is (aside from the explanations here), I'm reluctant to add tags. But would orbital-mechanics be a good choice? – Cyn says make Monica whole May 5 at 1:34
• @Cyn: No, it has nothing to do with orbits. Axial precession is the slow rotation of the axis of rotation of a rotating body. In the case of Earth, the axis of rotation rotates slowly (describing a cone, obviously) with a period of about 26,000 years. You can observe axial precession easily if you take a top (a children's toy) and spin it. Here is a YouTube video showing such a toy spinning; notice that while the top spins at high speed, its axis of rotation also rotates along the surface of a cone. – AlexP May 5 at 9:20

Wikipedia's page on Axial Precession has a good deal of mathematics on it and, unless you have absolutely got to have precise numbers for some obscure story reason (which is probably what I'd call the tail wagging the dog - change the story to avoid that problem) then making up the numbers or just forgetting all about Axial Precession is the way to go. The periods involved will be very long by the standards of any story you're likely to write, so why burden yourself with something you probably don;t need.

The maths that follows only gives a rough approximation anyway, and you don't want to even contemplate the kind of things you need to do to get a better one : it's not worth it.

That said let's have a look at the very basic theory result that Wikipedia gives :

There are two component to axial precession that matter (for Earth) : the one due to the Moon and the one due to the Sun. The Moon's effect is actually larger, but these numbers are very sensitive to the values you use.

The Solar Contribution

$$\frac{d\psi}{dt}=\left[ \frac {GM_s}{a_s^3\left(1-e_s^2\right)^\frac 3 2} \right] \left[ \frac {C-A}C \frac {cos\epsilon}\omega \right]$$

Lots of symbols so what do they mean ?

• $$G$$ - the Universal Gravitational Constant also famous from $$F = \frac {GM_1M_2} {r^2}$$ Newtons law for gravitation.
• $$M_s$$ - The Sun's mass - in your case you need the mass of your planet's star, of course.
• $$a_s$$ - The semi-major axis of the orbit of the planet around it's star
• $$e_s$$ - The eccentricity of the planet's orbit around it's star.

Now that second term in square bracket, which is also in the expression for the Lunar contribution. This one is trickier.

• $$C$$ - moment of inertia (of Earth) around the axis of rotation
• $$A$$ - moment of inertia around the equator
• $$\epsilon$$ - the angle between the equatorial plane and the ecliptic plane (see below)
• $$\omega$$ - Earth's angular velocity (due to it's rotation, not it's orbit)

Now this expression is really poorly dealt with in Wikipedia because of two problems.

$$epsilon$$ in Wikipedia is assumed to be the same for both the Solar and Lunar contributions. This is not (AFAIK) correct. The angle should be the angle between axis of rotation of the body and the plane of the orbit of the other body (which means it's different for the Sun and Moon).

The $$C-A$$ and $$A$$ terms are really hard to deal with for mere mortals (and frankly just messy for anyone else). For your purposes I would propose the following compromise term instead. It's based on modeling the planetary bulge as an ellipsoid of constant density compared with the $$A$$ value for an ideal sphere - both objects have the same mass and density, which I'm taking as constant . I'll spare you the derivation :

$$\frac {C-A} C \approx 1 - \frac{R^2} {a^2}$$

where in this case $$R$$ is the average radius of the planet and $$a$$ is the equatorial radius of the planet.

The Lunar Contribution

$$\frac{d\psi}{dt}=\left[ \frac {GM_l}{a_l^3\left(1-e_l^2\right)^\frac 3 2} \left(1-\frac 3 2 sin^2i\right) \right] \left[ \frac {C-A}C \frac {cos\epsilon}\omega \right]$$

Not much change here expect that the masses and so on refer to the Moon and not the Sun (hence the different subscripts). There is one additional term which is the factor :

$$\left(1-\frac 3 2 sin^2i\right)$$

This corrects for the effect that the angle of inclination of Moon's orbit to the ecliptic is not zero. The ecliptic being the plane with the Sun and the Earth's orbit in it.

You have to decide these numbers for yourself.

The total effect :

The total effect is simply the sum of the two other effects so :

$$\frac {d\psi}{dt} = \frac {d\psi_l}{dt} + \frac {d\psi_s}{dt}$$

If you had multiple moons you would need multiple lunar correction terms.

Just for clarification that $$\frac {d\psi}{dt}$$ means the rate of change of the angle $$\psi$$ with respect to time $$t$$. To get how big an angle you'd move through in a century you do this :

$$\Delta \psi \approx \frac {d\psi}{dt} \Delta t$$

• I don't really need an exact, I was more or less looking for a rough estimate within a couple centuries or so of accurate seeing as the rate of axial precession changes over time to. To many variables to get an exact value. – Tristan Schott May 6 at 23:59
• Trust me - this is not exact. :-) I should probably say that a rough estimate of zero is probably fine. These really are small effects normally. – StephenG May 7 at 0:17
• I'm doing the calculations now, I actually already had (almost) all of the other numbers required. – Tristan Schott May 7 at 1:31

First, find tonight's pole star. As the planet spins on its axis, over the course a day/night, all the stars in the sky will move in a circular path, with the exception of the pole star. You can use a fixed sight-line (e.g. the ends of two sticks stuck in the ground) to determine if a star has moved after just a few minutes. You should then be able to narrow down the pole star within an hour. Keep in mind- most worlds do not have an "exact" pole star, even our Polaris moves in a tiny circle.

Now that you have a pole star (or pole star circular-path), you can set up some much more permanent apparatus to fix its position for a long time. If the planet has an axial procession, then the pole star will eventually move out of this "permanently-fixed" position. The RATE at which it moves out of this position would determine the rate of axial precession. The time it takes for the star to return to its original position would define a full precession cycle. It maximum displacement during this period would define the precessions angular maximum.

I suspect, with some of that crazy math you mentioned, they can figure out how long a full cycle will be, and the angular maximums of the precession, before a full cycle has completed; but I'm not really sure how.

Edit: just to clarify- these stars are NOT moving, rather your planet is spinning, and precessing, which makes the stars appear to move. In fact, the above is based on the assumption that they do NOT move.