<p>Connor Garcia has provided a <a href="https://astronomy.stackexchange.com/a/49105/7982">wonderful answer</a> to this question. I would like to dwell on the final point in his post, that not all pseudo-synchronous rotation states are stable.</p>
<p>When a rapidly rotating secondary is undergoing tidal deceleration, it may experience either one or two torques from its primary.</p>
<ul>
<li>If the secondary is a fluid body (say, a Juipter, or a stellar companion, or hot and molten terrestrial planet), it experiences only the tidal torque -- which produces a tidal bulge running over the secondary's circumference.</li>
</ul>
<p>As Valeri V. Makarov demonstrated in <a href="https://iopscience.iop.org/article/10.1088/0004-637X/810/1/12" rel="nofollow noreferrer">his wonderful work</a>, the tidal torque will make the fluid secondary end up in the pseudosynchronous state, <em>provided the mean viscosity of this body is lower than some critical value</em>.</p>
<ul>
<li>If the secondary posesses a rigid crust and can sustain a permanent triaxiality, it is experiencing an additional torque from the star. A proper name for it would be &quot;the permanent-triaxiality-generated torque&quot;, but for brevity we call it simply the &quot;triaxial torque.&quot; It is then the interplay of the tidal and triaxial torques that defines the rotational dynamics of the secondary and its end state.</li>
</ul>
<p>When the body is away from spin-orbit resonances, the triaxial torque averages out and plays no role. Things become interesting when the rotator has to transcend a spin-orbit resonance (or to get trapped in it). There the triaxial torque comes into play, and the passing of -- or capture in -- a higher spin-orbit state becomes in many situations probabilistic, i.e. dependent on the phase difference between the two torues. It can be demonstrated that the outcome depends on the orientation of the rotator in the pericentre at the moment of crossing of the resonance. (Why exactly in the pericentre -- that is explained in Eric Jensen's comment to Connor Garcia's post: because in the pericentre the tidal forces are the strongest.) For more on this, see <a href="https://iopscience.iop.org/article/10.1088/0004-637X/752/1/73" rel="nofollow noreferrer">another paper by Valeri V. Makarov</a>.</p>
<p>Now, suppose that a triaxial body has successfully crossed all the higher spin-orbit resonances, and is approaching pseudosynchronism. Will it stay there, or will it cross it, to end up in the exact synchronism? Were the tidal torque dominant, pseudosynchronism would be the end state. In real life, however, the elevation due to the permanent figure will normally be higher than the tidal bulge. Consequently, the triaxial torque will be larger. So the body will be captured in the exact synchronism, with its figure tilted very slightly relative to the direction to the perturber (so that the emerging triaxial torque could compensate for the tidal torque). This is what happened to our Moon when it crossed the pseudosynchronous state without stopping there, and went on to despin towards synchronism. See <a href="https://iopscience.iop.org/article/10.1088/0004-637X/764/1/27" rel="nofollow noreferrer">this work</a> for details.</p>

CC BY-SA 4.0
2023-01-22T03:33:20.663Z
2023-01-24T00:32:41.967Z
2023-01-24T00:32:41.967Z
Michael_1812
<p>According to <a href="https://ui.adsabs.harvard.edu/abs/1981A%26A....99..126H/abstract" rel="nofollow noreferrer">Hut [1981]</a>, for eccentric orbits, pseudo-synchronous states are states of</p>
<blockquote>
<p>near synchronization of revolution and rotation at periastron</p>
</blockquote>
<p>Kepler's 2nd law says that a planet sweeps out equal area in equal time from the star.  Periastron is where the planet is closest to the star and thus moving the fastest, with the fastest angular speed around the star.  That means that other than at periastron, the orbital angular speed is less than the rotation rate of the planet.</p>
<p>The planet Mercury is in such a state.  Also from <a href="https://ui.adsabs.harvard.edu/abs/1981A%26A....99..126H/abstract" rel="nofollow noreferrer">Hut [1981]</a>:</p>
<p><a href="https://i.sstatic.net/dx864.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/dx864.png" alt="enter image description here" /></a></p>
<p>We can contrast this with the Moon, which is in synchronous rotation with the Earth.  Hence its rotation is less than revolution at perigee, but greater at apogee.</p>
<p>Pseudo-synchronization does not have to be at resonant frequencies.  See Greg Laughlin's explanation <a href="https://oklo.org/2006/12/11/pseudo-synchronization/" rel="nofollow noreferrer">here</a>, with an equation that calculates the spin period of a pseudo-synchronous orbit as a function of the orbital period and eccentricity:</p>
<p><span class="math-container">$$P_\text{SPIN}=\frac{(1+3e^2+\frac{3}{8}e^4)(1-e^2)^{3/2}}{1+\frac{15}{2}e^2+\frac{45}{8}e^4+\frac{5}{16}e^6} P_\text{ORBIT}$$</span></p>
<p><strong>Note</strong>:</p>
<ol>
<li>Our Moon presumably passed through a pseudo-synchronous rotation state on its way to tidal lock with the Earth, demonstrating that not all pseudo-synchronous rotation states are stable.</li>
</ol>

CC BY-SA 4.0
2022-04-16T01:31:44.383Z
This happens in binary star systems as well, for eccentric orbits.  The reason the rotation rates end up matched at periastron is that that is where the tidal forces are the strongest.
Eric Jensen
CC BY-SA 4.0
2023-01-22T02:53:37.050Z
Dear Connor Garcia, thank you for this detailed explanation. I think, you made a slip of the pen in the closing line. You probably wanted to say not "pseudo-synchronous orbits" but "pseudosynchronous rotation states."
Michael_1812
CC BY-SA 4.0
2022-04-14T17:49:15.450Z
2023-01-23T17:29:41.643Z
2023-01-23T17:29:41.643Z
Connor Garcia
<p>I understand the term synchronous rotation - when the plant spin frequency is the same as the orbital frequency. However, I've seen the term pseudo-synchronization coming up in a few places and I'm not sure I understand what is the physical meaning behind it. Does it relate to eccentric orbits? Where the angular velocity changes across different phases of the orbit?</p>

CC BY-SA 4.0
2022-04-14T16:46:26.353Z
My guess would be a spin-orbit resonance (like Mercury's 3:2)
WarpPrime
<orbital-mechanics>
What is pseudo-synchronisation when a planet is orbiting a host star?
ValientProcess
# What is pseudo-synchronisation when a planet is orbiting a host star?

(Asked by: username_0 on 2022-04-14)

I understand the term synchronous rotation - when the plant spin frequency is the same as the orbital frequency. However, I've seen the term pseudo-synchronization coming up in a few places and I'm not sure I understand what is the physical meaning behind it. Does it relate to eccentric orbits? Where the angular velocity changes across different phases of the orbit?

**Tags:** orbital-mechanics

> **Comments:**
> **username_1**: My guess would be a spin-orbit resonance (like Mercury's 3:2)  

---

## 2 answers

### Answer by username_3 on 2022-04-14. Score: 9 | Accepted answer

According to [Hut [1981]](https://ui.adsabs.harvard.edu/abs/1981A%26A....99..126H/abstract), for eccentric orbits, pseudo-synchronous states are states of

> near synchronization of revolution and rotation at periastron

Kepler's 2nd law says that a planet sweeps out equal area in equal time from the star. Periastron is where the planet is closest to the star and thus moving the fastest, with the fastest angular speed around the star. That means that other than at periastron, the orbital angular speed is less than the rotation rate of the planet.

The planet Mercury is in such a state. Also from [Hut [1981]](https://ui.adsabs.harvard.edu/abs/1981A%26A....99..126H/abstract):

[![enter image description here](https://i.sstatic.net/dx864.png)](https://i.sstatic.net/dx864.png)

We can contrast this with the Moon, which is in synchronous rotation with the Earth. Hence its rotation is less than revolution at perigee, but greater at apogee.

Pseudo-synchronization does not have to be at resonant frequencies. See Greg Laughlin's explanation [here](https://oklo.org/2006/12/11/pseudo-synchronization/), with an equation that calculates the spin period of a pseudo-synchronous orbit as a function of the orbital period and eccentricity:

$$P\_\text{SPIN}=\frac{(1+3e^2+\frac{3}{8}e^4)(1-e^2)^{3/2}}{1+\frac{15}{2}e^2+\frac{45}{8}e^4+\frac{5}{16}e^6} P\_\text{ORBIT}$$

**Note**:

1. Our Moon presumably passed through a pseudo-synchronous rotation state on its way to tidal lock with the Earth, demonstrating that not all pseudo-synchronous rotation states are stable.

> **Comments:**
> **username_4**: This happens in binary star systems as well, for eccentric orbits.  The reason the rotation rates end up matched at periastron is that that is where the tidal forces are the strongest.  
> **username_2**: Dear username_3, thank you for this detailed explanation. I think, you made a slip of the pen in the closing line. You probably wanted to say not "pseudo-synchronous orbits" but "pseudosynchronous rotation states."  

---

### Answer by username_2 on 2023-01-22. Score: 2

username_3 has provided a [wonderful answer](https://astronomy.stackexchange.com/a/49105/7982) to this question. I would like to dwell on the final point in his post, that not all pseudo-synchronous rotation states are stable.

When a rapidly rotating secondary is undergoing tidal deceleration, it may experience either one or two torques from its primary.

* If the secondary is a fluid body (say, a Juipter, or a stellar companion, or hot and molten terrestrial planet), it experiences only the tidal torque -- which produces a tidal bulge running over the secondary's circumference.

As Valeri V. Makarov demonstrated in [his wonderful work](https://iopscience.iop.org/article/10.1088/0004-637X/810/1/12), the tidal torque will make the fluid secondary end up in the pseudosynchronous state, *provided the mean viscosity of this body is lower than some critical value*.

* If the secondary posesses a rigid crust and can sustain a permanent triaxiality, it is experiencing an additional torque from the star. A proper name for it would be "the permanent-triaxiality-generated torque", but for brevity we call it simply the "triaxial torque." It is then the interplay of the tidal and triaxial torques that defines the rotational dynamics of the secondary and its end state.

When the body is away from spin-orbit resonances, the triaxial torque averages out and plays no role. Things become interesting when the rotator has to transcend a spin-orbit resonance (or to get trapped in it). There the triaxial torque comes into play, and the passing of -- or capture in -- a higher spin-orbit state becomes in many situations probabilistic, i.e. dependent on the phase difference between the two torues. It can be demonstrated that the outcome depends on the orientation of the rotator in the pericentre at the moment of crossing of the resonance. (Why exactly in the pericentre -- that is explained in username_4's comment to username_3's post: because in the pericentre the tidal forces are the strongest.) For more on this, see [another paper by Valeri V. Makarov](https://iopscience.iop.org/article/10.1088/0004-637X/752/1/73).

Now, suppose that a triaxial body has successfully crossed all the higher spin-orbit resonances, and is approaching pseudosynchronism. Will it stay there, or will it cross it, to end up in the exact synchronism? Were the tidal torque dominant, pseudosynchronism would be the end state. In real life, however, the elevation due to the permanent figure will normally be higher than the tidal bulge. Consequently, the triaxial torque will be larger. So the body will be captured in the exact synchronism, with its figure tilted very slightly relative to the direction to the perturber (so that the emerging triaxial torque could compensate for the tidal torque). This is what happened to our Moon when it crossed the pseudosynchronous state without stopping there, and went on to despin towards synchronism. See [this work](https://iopscience.iop.org/article/10.1088/0004-637X/764/1/27) for details.


---

