<blockquote>
  <p>Is it possible for a planet's axis to precess or wobble in such a way that its season cycle is of lesser duration than its year?</p>
</blockquote>

<p>For this, definitely yes, if you tweak some parameters. On the Wikipedia page for <a href="https://en.wikipedia.org/wiki/Axial_precession#Equations" rel="nofollow noreferrer">Axial Precision</a>, you'll find the set of equations and parameters that allow you to calculate the precessional rate for Earth's axis. Mainly, the Earth's precession is caused by a variety of factors but primarily due to the Moon and Sun (and thus related orbital/body parameters). If you look at those equations, and start tweaking values in them, you'll find you can force the axial precession period to be less than a "year". I'll go through and construct just such a system.</p>

<p>First though, I need to figure out the orbital parameters of the planet such that it has a 100-year long revolution period. This is acheived by using <a href="https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law" rel="nofollow noreferrer">Kepler's Third Law</a>, relating orbital period, orbital distance, and mass of the Sun (and Earth, but that's negligible). The relevant form of the equation, for our purposes is:</p>

<p>$$P^2 = \frac{a^3}{M_{\mathrm{Sun}}}$$</p>

<p>where, $P$ is the orbital period in years, $a$ is the orbital distance in AU (1 AU is the distance between Earth and the Sun), and $M_{\mathrm{Sun}}$ is the mass of the Sun in units of solar masses. If you're dictating that $P=100\ \mathrm{years}$, then I find I can set $a = 10\ \mathrm{AU}$ and $M_{\mathrm{star}} = 0.1\ M_{\odot}$ and make this equation true. What that means is that this new planet orbits a star which is 10 times less massive than our own star (making it a <a href="https://en.wikipedia.org/wiki/Red_dwarf" rel="nofollow noreferrer">red dwarf</a>) and orbits 10 times farther than Earth. My choice of parameters is all serendipitous and you could certainly find other parameters that work as well.</p>

<p>Now, given those parameters above, we can try to calculate the precessional rate for this new planet. I'll assume this new planet has an orbiting moon, just like Earth does. The two important equations for calculating the Earth's precessional rate are to account for the contributions from the Sun and the Moon. The equations are given by:</p>

<p>$$P_{\mathrm{Sun}} = \frac{3}{2}\left(\frac{GM_{\mathrm{Sun}}}{a_{\mathrm{Earth}}(1-e_{\mathrm{Earth}})^{3/2}}\right)\left(\frac{(C-A)}{C}\frac{\cos \epsilon}{\omega}\right)$$</p>

<p>$$P_{\mathrm{Moon}} = \frac{3}{2}\left(\frac{GM_{\mathrm{Moon}}(1-1.5\sin^2 i_{\mathrm{Moon}})}{a_{\mathrm{Moon}}(1-e_{\mathrm{Moon}})^{3/2}}\right)\left(\frac{(C-A)}{C}\frac{\cos \epsilon}{\omega}\right)$$</p>

<p>In these equations, $G$ is the Gravitational Constant, $M$ is mass in kg, $a$ is the orbital distance in meters, $e$ is the orbital eccentricity, $i$ is the inclination of the Moon's orbital plane wrt to the ecliptic, $(C-A)/C = 0.003273763$ is a measurement of the Earth's shape, $\epsilon$ is the Earth's axis' angle, nominally $23.43928^\mathrm{o}$, and finally $\omega$ is the rotation rate in rad/s.</p>

<p>Plugging in all the values for the Earth-Moon-Sun system (which you can find on the wiki page), you find that $P_{\mathrm{Sun}} + P_{\mathrm{Moon}} = 7.789\times 10^{-12}\ \mathrm{rad/s}$ and to convert to a time scale, you can multiply by  $2\times10^{-7}$. Doing so gives you the nominal Earth precessional period of about 26,000 years.</p>

<p>Given all of that, we can start playing with parameters to get it such that the precessional period for our new system is less than 100 years. I already said that $a_{\mathrm{Earth}}$ is 10 times larger and that $M_{\mathrm{Sun}}$ is 10 times smaller so we could get the period of 100 years. I found that I could make the following additional changes.</p>

<ul>
<li>The moon is now 5.8 times more massive than our Moon.</li>
<li>The moon is now a quarter of the distance of our Moon.</li>
<li>The planet is spinning twice as fast.</li>
</ul>

<p>With those changes, I find that the precessional period falls down to about 50 years, or half an orbit. You can continue to play with the numbers to get a different result.</p>

<p>An important consideration here is whether or not such a system is likely to even form. I think, given the system I've created, it is not outside the realm of possibilities. To have such a large, almost binary planetary system be so far away from a small star is unlikely, but I don't think impossible. The real difficulty here is that you've set the orbital period to such a long timescale. If you were to relax that a bit to something much shorter (say 10 years) you'd get a more realistic (and likely habitable) system.</p>

<blockquote>
  <p>Earth's precession is very much longer than it's revolution period, but is it possible that these could be reversed and a planet still be habitable?</p>
</blockquote>

<p>Certainly there's nothing to say a planet with a precessional period shorter than the revolution period is uninhabitable. What will make it uninhabitable is being too far from the star. The extreme orbital time you've provided necessitated having a planet that is likely too far from the star to be in the "habitable zone". You just have to play with all the parameters and find something that fits all your needs.</p>

<p>Another thing to consider is that if a planet has such a large and close moon as I've constructed, it will likely cause the planet to slow its rotation rate over short (astronomical) timescales. The planets rotation rate has a significant affect on the precessional rate and slowing it down over short timescales may mean your system doesn't keep the precessional rate under the rotation period for too long.</p>

CC BY-SA 3.0
2017-10-27T13:35:20.673Z
2017-10-27T16:16:40.517Z
2017-10-27T16:16:40.517Z
zephyr
<p>Is it possible for a planet's axis to precess or wobble in such a way that its season cycle is of lesser duration than its year? The gyroscopic precession will thus give seasons to one or the other hemisphere during a possibly hundreds-year-long revolution around its star. Earth's precession is very much longer than it's revolution period, but is it possible that these could be reversed and a planet still be habitable?</p>

CC BY-SA 3.0
2017-10-26T16:24:50.817Z
Here's a related question with some nice answers:   https://worldbuilding.stackexchange.com/questions/91506/can-a-planets-axial-tilt-turn-to-always-be-facing-its-star
userLTK
<planet><precession>
Season cycle < Year?
Bill Diggy
# Season cycle < Year?

(Asked by: username_0 on 2017-10-26)

Is it possible for a planet's axis to precess or wobble in such a way that its season cycle is of lesser duration than its year? The gyroscopic precession will thus give seasons to one or the other hemisphere during a possibly hundreds-year-long revolution around its star. Earth's precession is very much longer than it's revolution period, but is it possible that these could be reversed and a planet still be habitable?

**Tags:** planet, precession

> **Comments:**
> **username_1**: Here's a related question with some nice answers:   https://worldbuilding.stackexchange.com/questions/91506/can-a-planets-axial-tilt-turn-to-always-be-facing-its-star  

---

## 1 answers

### Answer by username_2 on 2017-10-27. Score: 6

> Is it possible for a planet's axis to precess or wobble in such a way that its season cycle is of lesser duration than its year?

For this, definitely yes, if you tweak some parameters. On the Wikipedia page for [Axial Precision](https://en.wikipedia.org/wiki/Axial_precession#Equations), you'll find the set of equations and parameters that allow you to calculate the precessional rate for Earth's axis. Mainly, the Earth's precession is caused by a variety of factors but primarily due to the Moon and Sun (and thus related orbital/body parameters). If you look at those equations, and start tweaking values in them, you'll find you can force the axial precession period to be less than a "year". I'll go through and construct just such a system.

First though, I need to figure out the orbital parameters of the planet such that it has a 100-year long revolution period. This is acheived by using [Kepler's Third Law](https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law), relating orbital period, orbital distance, and mass of the Sun (and Earth, but that's negligible). The relevant form of the equation, for our purposes is:

$$P^2 = \frac{a^3}{M\_{\mathrm{Sun}}}$$

where, $P$ is the orbital period in years, $a$ is the orbital distance in AU (1 AU is the distance between Earth and the Sun), and $M\_{\mathrm{Sun}}$ is the mass of the Sun in units of solar masses. If you're dictating that $P=100\ \mathrm{years}$, then I find I can set $a = 10\ \mathrm{AU}$ and $M\_{\mathrm{star}} = 0.1\ M\_{\odot}$ and make this equation true. What that means is that this new planet orbits a star which is 10 times less massive than our own star (making it a [red dwarf](https://en.wikipedia.org/wiki/Red_dwarf)) and orbits 10 times farther than Earth. My choice of parameters is all serendipitous and you could certainly find other parameters that work as well.

Now, given those parameters above, we can try to calculate the precessional rate for this new planet. I'll assume this new planet has an orbiting moon, just like Earth does. The two important equations for calculating the Earth's precessional rate are to account for the contributions from the Sun and the Moon. The equations are given by:

$$P\_{\mathrm{Sun}} = \frac{3}{2}\left(\frac{GM\_{\mathrm{Sun}}}{a\_{\mathrm{Earth}}(1-e\_{\mathrm{Earth}})^{3/2}}\right)\left(\frac{(C-A)}{C}\frac{\cos \epsilon}{\omega}\right)$$

$$P\_{\mathrm{Moon}} = \frac{3}{2}\left(\frac{GM\_{\mathrm{Moon}}(1-1.5\sin^2 i\_{\mathrm{Moon}})}{a\_{\mathrm{Moon}}(1-e\_{\mathrm{Moon}})^{3/2}}\right)\left(\frac{(C-A)}{C}\frac{\cos \epsilon}{\omega}\right)$$

In these equations, $G$ is the Gravitational Constant, $M$ is mass in kg, $a$ is the orbital distance in meters, $e$ is the orbital eccentricity, $i$ is the inclination of the Moon's orbital plane wrt to the ecliptic, $(C-A)/C = 0.003273763$ is a measurement of the Earth's shape, $\epsilon$ is the Earth's axis' angle, nominally $23.43928^\mathrm{o}$, and finally $\omega$ is the rotation rate in rad/s.

Plugging in all the values for the Earth-Moon-Sun system (which you can find on the wiki page), you find that $P\_{\mathrm{Sun}} + P\_{\mathrm{Moon}} = 7.789\times 10^{-12}\ \mathrm{rad/s}$ and to convert to a time scale, you can multiply by $2\times10^{-7}$. Doing so gives you the nominal Earth precessional period of about 26,000 years.

Given all of that, we can start playing with parameters to get it such that the precessional period for our new system is less than 100 years. I already said that $a\_{\mathrm{Earth}}$ is 10 times larger and that $M\_{\mathrm{Sun}}$ is 10 times smaller so we could get the period of 100 years. I found that I could make the following additional changes.

* The moon is now 5.8 times more massive than our Moon.
* The moon is now a quarter of the distance of our Moon.
* The planet is spinning twice as fast.

With those changes, I find that the precessional period falls down to about 50 years, or half an orbit. You can continue to play with the numbers to get a different result.

An important consideration here is whether or not such a system is likely to even form. I think, given the system I've created, it is not outside the realm of possibilities. To have such a large, almost binary planetary system be so far away from a small star is unlikely, but I don't think impossible. The real difficulty here is that you've set the orbital period to such a long timescale. If you were to relax that a bit to something much shorter (say 10 years) you'd get a more realistic (and likely habitable) system.

> Earth's precession is very much longer than it's revolution period, but is it possible that these could be reversed and a planet still be habitable?

Certainly there's nothing to say a planet with a precessional period shorter than the revolution period is uninhabitable. What will make it uninhabitable is being too far from the star. The extreme orbital time you've provided necessitated having a planet that is likely too far from the star to be in the "habitable zone". You just have to play with all the parameters and find something that fits all your needs.

Another thing to consider is that if a planet has such a large and close moon as I've constructed, it will likely cause the planet to slow its rotation rate over short (astronomical) timescales. The planets rotation rate has a significant affect on the precessional rate and slowing it down over short timescales may mean your system doesn't keep the precessional rate under the rotation period for too long.


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