# How would one figure out the rate of axial precession of a planet?

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?

Edit Important? Details: My Earthlike planet has a mass 1.2597x the mass of our earth, and two fairly large moons (actually, one of them may be large enough to classify it as a binary planet but I'm not sure).

• It's a great question, welcome to Stack Exchange! I've adjusted the wording a little bit to better suit the format here. Have a look and feel free to edit further.
– uhoh
May 4, 2019 at 5:25
• I suspect this would be more appropriate for a hard-science question Worldbuilding SE where such a question is a very good fit IMO. ("hard-science" is a tag on WB SE which means that answers must be backed by real science, maths, etc., not just wordy opinions or wordy logic) Fantasy world questions aren't a good fit for Astronomy SE, again IMO. May 4, 2019 at 6:19
• Note the presence or absence of a large moon (like Earth has) is very important to know. Earth's moon is the largest contribution to axial precession, and the Sun next. You would frankly do as well to ignore the effect entirely unless this is absolutely critical to your fantasy world. May 4, 2019 at 6:32
• I've posted the same question in worldbuilding just now with the hard science tag, hopefully that'll help me. May 4, 2019 at 18:20

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 :

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

• This answer is posted on Worldbuilding SE (by me) and I figure posting it here as a community wiki would (a) let people make useful contributions to it (and fix any errors :-) ) and (b) avoid me getting votes in two SEs for the same answer, which would not be in the spirit of SE generally. I do think the question is useful enough to warrant an answer so either fix mine ( :-) ) or make a better one. Thanks. May 4, 2019 at 23:14