<p>Wikipedia's page on <a href="https://en.wikipedia.org/wiki/Axial_precession" rel="nofollow noreferrer">Axial Precession</a> 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.</p>

<p>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.</p>

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

<p>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.</p>

<p><strong>The Solar Contribution</strong></p>

<p><span class="math-container">$$\frac{d\psi}{dt}=\left[ \frac {GM_s}{a_s^3\left(1-e_s^2\right)^\frac 3 2} \right] \left[ \frac {C-A}C \frac {cos\epsilon}\omega \right]$$</span></p>

<p>Lots of symbols so what do they mean ?</p>

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

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

<ul>
<li><span class="math-container">$C$</span> - moment of inertia (of Earth) around the axis of rotation</li>
<li><span class="math-container">$A$</span> - moment of inertia around the equator</li>
<li><span class="math-container">$\epsilon$</span> - the angle between the equatorial plane and the ecliptic plane (see below)</li>
<li><span class="math-container">$\omega$</span> - Earth's angular velocity (due to it's rotation, not it's orbit)</li>
</ul>

<p>Now this expression is really poorly dealt with in Wikipedia because of two problems.</p>

<p><span class="math-container">$epsilon$</span> 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).</p>

<p>The <span class="math-container">$C-A$</span> and <span class="math-container">$A$</span> 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 <span class="math-container">$A$</span> 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 :</p>

<p><span class="math-container">$$\frac {C-A} C \approx 1 - \frac{R^2} {a^2}$$</span></p>

<p>where in this case <span class="math-container">$R$</span> is the <em>average</em> radius of the planet and <span class="math-container">$a$</span> is the equatorial radius of the planet.</p>

<p><strong>The Lunar Contribution</strong></p>

<p><span class="math-container">$$\frac{d\psi}{dt}=\left[ \frac {GM_l}{a_l^3\left(1-e_l^2\right)^\frac 3 2} \left(1-\frac 3 2 sin^2i\right) \right] \left[ \frac {C-A}C \frac {cos\epsilon}\omega \right]$$</span></p>

<p>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 :</p>

<p><span class="math-container">$$\left(1-\frac 3 2 sin^2i\right)$$</span></p>

<p>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.</p>

<p>You have to decide these numbers for yourself.</p>

<p><strong>The total effect :</strong></p>

<p>The total effect is simply the sum of the two other effects so :</p>

<p><span class="math-container">$$\frac {d\psi}{dt} = \frac {d\psi_l}{dt} + \frac {d\psi_s}{dt}$$</span></p>

<p>If you had multiple moons you would need multiple lunar correction terms.</p>

<p>Just for clarification that <span class="math-container">$\frac {d\psi}{dt}$</span> means the rate of change of the angle <span class="math-container">$\psi$</span> with respect to time <span class="math-container">$t$</span>.  To get how big an angle you'd move through in a century you do this :</p>

<p><span class="math-container">$$\Delta \psi \approx \frac {d\psi}{dt} \Delta t$$</span></p>

CC BY-SA 4.0
2019-05-04T23:14:05.773Z
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.
StephenG - Help Ukraine
CC BY-SA 4.0
2019-05-04T23:09:55.403Z
2019-05-04T23:09:55.403Z
StephenG - Help Ukraine
<p>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.</p>

<p>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.</p>

<p>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?</p>

<p>Edit <em>Important? Details</em>:
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).</p>

CC BY-SA 4.0
2019-05-04T05:25:46.527Z
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
CC BY-SA 4.0
2019-05-04T06:19:24.607Z
I suspect this would be more appropriate for a hard-science question [Worldbuilding SE](https://worldbuilding.stackexchange.com/) 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.
StephenG - Help Ukraine
CC BY-SA 4.0
2019-05-04T06:32:42.097Z
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.
StephenG - Help Ukraine
CC BY-SA 4.0
2019-05-04T18:20:06.677Z
I've posted the same question in worldbuilding just now with the hard science tag, hopefully that'll help me.
Tristan Schott
<earth-like-planet><precession>
How would one figure out the rate of axial precession of a planet?
Tristan Schott
# How would one figure out the rate of axial precession of a planet?

(Asked by: username_0 on 2019-05-04)

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).

**Tags:** earth-like-planet, precession

> **Comments:**
> **username_1**: 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.  
> **username_2**: I suspect this would be more appropriate for a hard-science question [Worldbuilding SE](https://worldbuilding.stackexchange.com/) 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.  
> **username_2**: 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.  
> **username_0**: I've posted the same question in worldbuilding just now with the hard science tag, hopefully that'll help me.  

---

## 1 answers

### Answer by username_2 on 2019-05-04. Score: 1 | Accepted answer

Wikipedia's page on [Axial Precession](https://en.wikipedia.org/wiki/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 :

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

> **Comments:**
> **username_2**: 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.  

---

