<p><strong>The Earth-facing side of the Moon doesn’t change currently due to any of the forces on it by other bodies (excepting librations)</strong>.</p>
<p>From <a href="http://pi.math.cornell.edu/%7Erand/randpdf/gladman.pdf" rel="noreferrer">Gladman et. al.</a>: “<em>tidal dissipation in the satellite drives it to a state where</em> […] <em>the spin axis of a satellite in a generalized Cassini state will appear stationary.</em>”  The <span class="math-container">$S_2$</span> Cassini state is one of the two possible eventual stable states for a general moon system.  But, according to Gladman, our Moon has &quot;<em><span class="math-container">$S_2$</span> as the only possible Cassini state</em>&quot;.</p>
<p>Before achieving a Cassini state, the tidal forces exert an average torque of <span class="math-container">$-SC\cos\theta\sin\theta$</span>, where <span class="math-container">$\theta$</span> is the obliquity of the moon's orbit measured with respect to the orbit normal, <span class="math-container">$C$</span> is the satellite’s moment of inertia about the spin axis, and <span class="math-container">$$S=\frac{3Gm_p}{2r^3}\frac{C-(A+B)/2}{C}$$</span> where <span class="math-container">$m_p$</span> is the mass of the planet, <span class="math-container">$G$</span> is the gravitational constant, <span class="math-container">$r$</span> is the orbital radius of the satellite, and <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are the other two principal moments of inertia.</p>
<p>Note that <span class="math-container">$S$</span> has <span class="math-container">$r^3$</span> in the denominator (also as pointed out by Stuart Robins in a comment above), which means the torque exerted by tidal forces was much higher when the moon was closer to the Earth.  In other words, since the Moon’s face towards the earth is not precessing now, it certainly wouldn’t have in the past when the potential tidal torque was higher.</p>
<p>I think <a href="http://pi.math.cornell.edu/%7Erand/randpdf/gladman.pdf" rel="noreferrer">Gladman et. al.</a> give pretty credible evidence for their claim of tidal lock, in that they account for wobbles, highly eccentric orbits, and oblate spheroids in both the planet and the moon.  They don’t specifically account for the gravitation of other bodies like the Sun.  The above equation also gives us a comparison of relative tidal effects on the Moon by noting <span class="math-container">$m/r^3$</span> is 9.2e7 for the Earth/Moon system, but only 5.9e5 for the Sun/Moon system. Gravitational perturbations from the Sun and other solar system objects affect the Moon's orbital elements, but don't have a significant effect on it's rotation rate.</p>
<p>There is plenty of lunar geological evidence supporting <a href="http://pi.math.cornell.edu/%7Erand/randpdf/gladman.pdf" rel="noreferrer">Gladman et. al.</a>’s claim that the Cassini state is stable, in that we see a fundamental asymmetry between the near and far faces of the Moon.  The near face has a thinner crust and more volcanic maria.  The far side has a thicker crust and more visible asteroid craters:
<a href="https://i.sstatic.net/VweX5.jpg" rel="noreferrer"><img src="https://i.sstatic.net/VweX5.jpg" alt="enter image description here" /></a></p>
<p><a href="https://iopscience.iop.org/article/10.1088/2041-8205/788/2/L42/meta" rel="noreferrer">Roy et. al.</a> claim the asymmetry between the near and far faces is due to heat from the Earth causing the early tidal locked Moon to cool more slowly on the lunar face pointing to the molten cooling Earth. <strong>If their theory is correct, then the Moon's face towards Earth hasn't significantly changed since the Earth/Moon system infancy.</strong></p>
<p>The near face of the Moon doesn’t appear to be precessing currently, but could it have changed orientation from its original tidal lock configuration? There is a strong suggestion from <a href="https://www.sciencedirect.com/science/article/abs/pii/S0019103583711516" rel="noreferrer">Kadono</a>, that an asteroid impact could have an effect on spin rate.  Unfortunately, the paper is behind a paywall, so lets do a &quot;back of the napkin&quot; calculation to see how much angular momentum a big asteroid impact could have on the moon.</p>
<p>The largest known basin on the Moon is the <a href="https://en.wikipedia.org/wiki/South_Pole%E2%80%93Aitken_basin" rel="noreferrer">South Pole-Aitken basin</a>. The size and speed of the impactor was estimated by <a href="https://web.archive.org/web/20141221162807/http://www.lpi.usra.edu/lpi/potter/publications/Potteretal2012.pdf" rel="noreferrer">Potter et. al.</a> One of their models assumes a chondrite impactor with density <span class="math-container">$\rho$</span> of 3580kg/m3 at 100km radius <span class="math-container">$r$</span> at an impact velocity <span class="math-container">$v$</span> of 10km/s. We can calculate the mass <span class="math-container">$m=\rho 4/3\pi r^3$</span> or 1.5e19kg. The equation for linear momentum is <span class="math-container">$l=mv$</span> or 1.5e20km*kg/s.</p>
<p>The moment of inertia of a sphere is <span class="math-container">$I=2/5MR^2$</span> and we can use <span class="math-container">$M=7.342e22kg$</span> and <span class="math-container">$R=1737.4km$</span> as the mass and radius of the Moon to get <span class="math-container">$I=8.86e28$</span>.  If we set the angular momentum of the Moon <span class="math-container">$I\omega=1.5e20$</span> and solve for <span class="math-container">$\omega$</span> giving us a resulting rotation rate change of 1.69e-9 radians per second or about 3 degrees per year. These calculations assume an impact angle parallel to the lunar surface, which is impossible, but forms an upper bound for a momentum change.  Higher angle impacts still conserve momentum, but more of the momentum goes into changing the orbital parameters rather than rotation rate.</p>
<p>From <a href="https://astronomy.stackexchange.com/questions/1859/when-did-the-moon-become-tidally-locked-to-earth/1862#1862">When did the Moon become tidally locked to Earth?</a>, we know the tidal locking time is <span class="math-container">$$t_{\mbox{lock,Moon}}=7.12753\cdot 10^{-25}wa^6 \frac{\mbox{kg}}{\mbox{Nm}^2 \mbox{km}^3}.$$</span> where <span class="math-container">$w$</span> the spin rate in radians per second, and <span class="math-container">$a$</span> the semi-major axis of the Moon orbit.</p>
<p>I calculated the above equation, supposing the Moon's orbital semi-major axis was less than half it’s current value so long ago. If the moon got struck in the way presented above, it would tidal lock again in less than 2 years, rotating less than 6 degrees in that time.  <strong>So, no, after the Moon was first tidally locked, the same face almost certainly always pointed to the Earth.</strong></p>
<p><strong>Notes:</strong></p>
<ol>
<li><p>It's possible that the asymmetries in the Moon's composition would decrease the time to tidal lock, but it isn't needed for any of the above calculations. Even if the Moon's composition was homogeneous, tidal locking would still occur.</p>
</li>
<li><p>There are many other theories for the lunar face asymmetry, but most of them rely on a steady tidal lock through the early part of the Earth/Moon system.  An exception might be <a href="https://www.nature.com/articles/s41561-020-0559-4" rel="noreferrer">Elardo et. al.</a> who claim the asymmetry is due to a greater abundance of radioactive isotopes on the near side of the Moon, which add heat and lower material melting temperatures.  They call the mix <a href="https://en.wikipedia.org/wiki/KREEP" rel="noreferrer">KREEP</a>.</p>
</li>
<li><p>We can't completely eliminate the possibility that the Moon was hit by an even bigger asteroid then referenced above, with the evidence later being covered by volcanism.  This answer is only based on currently available evidence.</p>
</li>
</ol>

CC BY-SA 4.0
2020-12-13T01:57:50.067Z
2020-12-13T02:05:34.007Z
2020-12-13T02:05:34.007Z
Connor Garcia
<p>Title. Does the Earth-facing side of the Moon slowly precess due to perturbations and torques exerted by other bodies? Or is the side of the Moon we see today the same as when it first got tidally locked?</p>
<p>P.S. I know we don't see the exact same side of the Moon   everytime due to libration. That's not what I'm asking.</p>

CC BY-SA 4.0
2020-12-09T23:21:50.190Z
Interesting question! I have a hunch that there may have been a period of time when capture was happening when the motion might have been a little chaotic, then some damped oscillations before settling down, but I could be totally wrong. ([libration animations](https://space.stackexchange.com/q/19936/12102) (for those who haven't seen them))
uhoh
CC BY-SA 4.0
2020-12-10T00:49:54.173Z
I'm fairly sure it's pretty much the same orientation (i.e., no long term precession in the sense you mean, although the Moon does undergo several types of precession), but not certain enough to post an actual answer. I suspect that the lunar mascons are an important factor in locking the orientation.
PM 2Ring
CC BY-SA 4.0
2020-12-10T06:13:05.507Z
As I understand it, the Earth-Moon after formation were very hot, basically liquid rock, possibly a synestia or vaporized rock.   As the bodies cooled the Earth-side of the Moon remained hot longer, being heated by the Earth, which was slower to cool.  The Moon also formed under strong tides and as a result it formed lumpy and probably slightly elongated at the time and it was denser on the Earth side.   It's unlikely to have changed faces since it's formation, at least that's my understanding, but a professional and sourced answer would probably be better than my general understanding.
userLTK
CC BY-SA 4.0
2020-12-12T18:53:06.617Z
Tidal force goes as the inverse-cube of the distance, as opposed to gravity which is inverse-square. That means that tidal force falls away VERY quickly as you move away from an object. So, there's nothing out there that is close and large enough to budge the moon out of the tidally locked state that it's in, even the 2012 doomsday Planet X idiocy wouldn't've been able to do that. It's a kinda straightforward calculation, but it's been way too long since I've done stuff like that so can't really post an answer, but it could also be why you can't find a research paper on it.
Stuart Robbins
CC BY-SA 4.0
2020-12-12T21:23:08.773Z
I would understand “tidally locked” to mean that the Moon *always* points the same side to the Earth, but that’s semantics. I know, though, that the aspect difference between the two sides of the Moon is usually explained by the fact that Earth’s gravity, stronger on one side, “pulled” magma out of the Moon to form maria, whereas the other side was more exposed to meteoritic bombardment without Earth acting as a shield. This would imply that the Moon has (basically) *always* pointed the same side towards Earth and does *not* precess.
Pierre Paquette
<the-moon><tidal-locking>
Has the Earth-facing side of the Moon that we see today always faced us ever since the Moon got tidally locked? Or does it precess?
user177107
# Has the Earth-facing side of the Moon that we see today always faced us ever since the Moon got tidally locked? Or does it precess?

(Asked by: username_0 on 2020-12-09)

Title. Does the Earth-facing side of the Moon slowly precess due to perturbations and torques exerted by other bodies? Or is the side of the Moon we see today the same as when it first got tidally locked?

P.S. I know we don't see the exact same side of the Moon everytime due to libration. That's not what I'm asking.

**Tags:** the-moon, tidal-locking

> **Comments:**
> **username_1**: Interesting question! I have a hunch that there may have been a period of time when capture was happening when the motion might have been a little chaotic, then some damped oscillations before settling down, but I could be totally wrong. ([libration animations](https://space.stackexchange.com/q/19936/12102) (for those who haven't seen them))  
> **username_2**: I'm fairly sure it's pretty much the same orientation (i.e., no long term precession in the sense you mean, although the Moon does undergo several types of precession), but not certain enough to post an actual answer. I suspect that the lunar mascons are an important factor in locking the orientation.  
> **username_3**: As I understand it, the Earth-Moon after formation were very hot, basically liquid rock, possibly a synestia or vaporized rock.   As the bodies cooled the Earth-side of the Moon remained hot longer, being heated by the Earth, which was slower to cool.  The Moon also formed under strong tides and as a result it formed lumpy and probably slightly elongated at the time and it was denser on the Earth side.   It's unlikely to have changed faces since it's formation, at least that's my understanding, but a professional and sourced answer would probably be better than my general understanding.  
> **username_4**: Tidal force goes as the inverse-cube of the distance, as opposed to gravity which is inverse-square. That means that tidal force falls away VERY quickly as you move away from an object. So, there's nothing out there that is close and large enough to budge the moon out of the tidally locked state that it's in, even the 2012 doomsday Planet X idiocy wouldn't've been able to do that. It's a kinda straightforward calculation, but it's been way too long since I've done stuff like that so can't really post an answer, but it could also be why you can't find a research paper on it.  
> **username_5**: I would understand “tidally locked” to mean that the Moon *always* points the same side to the Earth, but that’s semantics. I know, though, that the aspect difference between the two sides of the Moon is usually explained by the fact that Earth’s gravity, stronger on one side, “pulled” magma out of the Moon to form maria, whereas the other side was more exposed to meteoritic bombardment without Earth acting as a shield. This would imply that the Moon has (basically) *always* pointed the same side towards Earth and does *not* precess.  

---

## 1 answers

### Answer by username_6 on 2020-12-13. Score: 9 | Accepted answer

**The Earth-facing side of the Moon doesn’t change currently due to any of the forces on it by other bodies (excepting librations)**.

From [Gladman et. al.](http://pi.math.cornell.edu/%7Erand/randpdf/gladman.pdf): “*tidal dissipation in the satellite drives it to a state where* […] *the spin axis of a satellite in a generalized Cassini state will appear stationary.*” The $S\_2$ Cassini state is one of the two possible eventual stable states for a general moon system. But, according to Gladman, our Moon has "*$S\_2$ as the only possible Cassini state*".

Before achieving a Cassini state, the tidal forces exert an average torque of $-SC\cos\theta\sin\theta$, where $\theta$ is the obliquity of the moon's orbit measured with respect to the orbit normal, $C$ is the satellite’s moment of inertia about the spin axis, and $$S=\frac{3Gm\_p}{2r^3}\frac{C-(A+B)/2}{C}$$ where $m\_p$ is the mass of the planet, $G$ is the gravitational constant, $r$ is the orbital radius of the satellite, and $A$ and $B$ are the other two principal moments of inertia.

Note that $S$ has $r^3$ in the denominator (also as pointed out by Stuart Robins in a comment above), which means the torque exerted by tidal forces was much higher when the moon was closer to the Earth. In other words, since the Moon’s face towards the earth is not precessing now, it certainly wouldn’t have in the past when the potential tidal torque was higher.

I think [Gladman et. al.](http://pi.math.cornell.edu/%7Erand/randpdf/gladman.pdf) give pretty credible evidence for their claim of tidal lock, in that they account for wobbles, highly eccentric orbits, and oblate spheroids in both the planet and the moon. They don’t specifically account for the gravitation of other bodies like the Sun. The above equation also gives us a comparison of relative tidal effects on the Moon by noting $m/r^3$ is 9.2e7 for the Earth/Moon system, but only 5.9e5 for the Sun/Moon system. Gravitational perturbations from the Sun and other solar system objects affect the Moon's orbital elements, but don't have a significant effect on it's rotation rate.

There is plenty of lunar geological evidence supporting [Gladman et. al.](http://pi.math.cornell.edu/%7Erand/randpdf/gladman.pdf)’s claim that the Cassini state is stable, in that we see a fundamental asymmetry between the near and far faces of the Moon. The near face has a thinner crust and more volcanic maria. The far side has a thicker crust and more visible asteroid craters:
[![enter image description here](https://i.sstatic.net/VweX5.jpg)](https://i.sstatic.net/VweX5.jpg)

[Roy et. al.](https://iopscience.iop.org/article/10.1088/2041-8205/788/2/L42/meta) claim the asymmetry between the near and far faces is due to heat from the Earth causing the early tidal locked Moon to cool more slowly on the lunar face pointing to the molten cooling Earth. **If their theory is correct, then the Moon's face towards Earth hasn't significantly changed since the Earth/Moon system infancy.**

The near face of the Moon doesn’t appear to be precessing currently, but could it have changed orientation from its original tidal lock configuration? There is a strong suggestion from [Kadono](https://www.sciencedirect.com/science/article/abs/pii/S0019103583711516), that an asteroid impact could have an effect on spin rate. Unfortunately, the paper is behind a paywall, so lets do a "back of the napkin" calculation to see how much angular momentum a big asteroid impact could have on the moon.

The largest known basin on the Moon is the [South Pole-Aitken basin](https://en.wikipedia.org/wiki/South_Pole%E2%80%93Aitken_basin). The size and speed of the impactor was estimated by [Potter et. al.](https://web.archive.org/web/20141221162807/http://www.lpi.usra.edu/lpi/potter/publications/Potteretal2012.pdf) One of their models assumes a chondrite impactor with density $\rho$ of 3580kg/m3 at 100km radius $r$ at an impact velocity $v$ of 10km/s. We can calculate the mass $m=\rho 4/3\pi r^3$ or 1.5e19kg. The equation for linear momentum is $l=mv$ or 1.5e20km\*kg/s.

The moment of inertia of a sphere is $I=2/5MR^2$ and we can use $M=7.342e22kg$ and $R=1737.4km$ as the mass and radius of the Moon to get $I=8.86e28$. If we set the angular momentum of the Moon $I\omega=1.5e20$ and solve for $\omega$ giving us a resulting rotation rate change of 1.69e-9 radians per second or about 3 degrees per year. These calculations assume an impact angle parallel to the lunar surface, which is impossible, but forms an upper bound for a momentum change. Higher angle impacts still conserve momentum, but more of the momentum goes into changing the orbital parameters rather than rotation rate.

From [When did the Moon become tidally locked to Earth?](https://astronomy.stackexchange.com/questions/1859/when-did-the-moon-become-tidally-locked-to-earth/1862#1862), we know the tidal locking time is $$t\_{\mbox{lock,Moon}}=7.12753\cdot 10^{-25}wa^6 \frac{\mbox{kg}}{\mbox{Nm}^2 \mbox{km}^3}.$$ where $w$ the spin rate in radians per second, and $a$ the semi-major axis of the Moon orbit.

I calculated the above equation, supposing the Moon's orbital semi-major axis was less than half it’s current value so long ago. If the moon got struck in the way presented above, it would tidal lock again in less than 2 years, rotating less than 6 degrees in that time. **So, no, after the Moon was first tidally locked, the same face almost certainly always pointed to the Earth.**

**Notes:**

1. It's possible that the asymmetries in the Moon's composition would decrease the time to tidal lock, but it isn't needed for any of the above calculations. Even if the Moon's composition was homogeneous, tidal locking would still occur.
2. There are many other theories for the lunar face asymmetry, but most of them rely on a steady tidal lock through the early part of the Earth/Moon system. An exception might be [Elardo et. al.](https://www.nature.com/articles/s41561-020-0559-4) who claim the asymmetry is due to a greater abundance of radioactive isotopes on the near side of the Moon, which add heat and lower material melting temperatures. They call the mix [KREEP](https://en.wikipedia.org/wiki/KREEP).
3. We can't completely eliminate the possibility that the Moon was hit by an even bigger asteroid then referenced above, with the evidence later being covered by volcanism. This answer is only based on currently available evidence.


---

