<p>There's no back and forth motion on the way towards tidal locking.   There's kind of two ways to look this kind of "settling", there's overshooting and correction, basically a back and forth, and there's a very gradual force that grows smaller, basically moving towards zero but not overshooting.    I realize that's just re-writing your question, but it helps to understand the 2nd option.</p>

<p>I'll use the Earth as an example, because the Earth is in the gradual process of becoming tidally locked to the Moon.  It's the same process, but easier to visualize. </p>

<p>Because the Earth rotates faster than the Moon orbits, the tidal bulge moves ahead of the moon.   This is called the tidal bulge offset.</p>

<p><a href="https://i.sstatic.net/XgtZe.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/XgtZe.jpg" alt="http://pages.uoregon.edu/jimbrau/BrauImNew/Chap07/FG07_23.jpg"></a></p>

<p><a href="http://pages.uoregon.edu/jimbrau/astr121-2005/Notes/Chapter7.html" rel="nofollow noreferrer">Source</a>.</p>

<p>It's the size of the bulge and the angle of the offset (and the mass and distance of the Moon) that causes the familiar gravitational effects, the Earth's rotation slowing down and the Moon moving away from the Earth.</p>

<p><a href="https://i.sstatic.net/DbFCu.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/DbFCu.jpg" alt="https://ase.tufts.edu/cosmos/pictures/Explore_figs_8/Chapter5/Fig5_28copy.jpg"></a></p>

<p><a href="https://ase.tufts.edu/cosmos/print_images.asp?id=5" rel="nofollow noreferrer">Source</a></p>

<p>The current angle of offset for the Earth's tidal bulge to the Moon is about 3 degrees, often exaggerated in diagrams so it's easier to see.  (<em>I should find a source for that - I've not been able to find a good one)</em>.   </p>

<p>From the Moon's point of view (and if we assume a circular orbit), the tidal offset stays in the same place.  It's a factor of the relative rotation speed.   As Earth slows down, the angle of offset will move towards zero.    At tidal locking, the angle of offset becomes zero and there's no force to continue in the other direction.    In a simple mathematical model, tidal locking wouldn't ever happen completely, it would just gradually get closer and closer, like an object shot from the Earth at exactly escape velocity.</p>

<p>In reality, because objects aren't perfectly symmetrical and because orbits aren't perfectly circular, the "take forever to get to zero", doesn't happen.  But it's still a gradual approach to zero and no overshooting and returning because the force on the rotation approaches zero as the angle of offset approaches zero and the force needed to change a planet's rotation is enormous and there's essentially zero force to make that happen.</p>

<p>With more eccentric orbits, like Mercury and it's 3:2 rotation to orbit resonance, that might be different.   Not sure how eccentricity might play into this.   That's more complicated.</p>

CC BY-SA 3.0
2017-07-11T00:33:24.530Z
So are you saying the moon, previous to being tidally locked, continues to have complete rotations showing all sides but the rotation rate is just continuously slowing down until it's so slow it just stops and the same face is towards Earth with minimal to no rocking into position?
Brooks Nelson
CC BY-SA 3.0
2017-07-11T03:29:14.537Z
@BrooksNelson  Basically yes.   I don't see any other way that it could work.   The angular momentum changes very slowly (outside of a truly enormous impact).   That's a given and as the moon moves closer to tidal locking the tidal offset angle grows smaller and smaller, leading to smaller and smaller force.     I see no mechanism where it would over-slow-down and then speed up again.   It should just be a gradual approach towards tidal locking.
userLTK
CC BY-SA 3.0
2017-07-11T13:24:16.310Z
Try to avoid "I can't see" responses, since those are subjective.
Carl Witthoft
CC BY-SA 4.0
2020-07-14T07:23:46.793Z
speaking of "back and forth"; see [this comment](https://space.stackexchange.com/questions/45342/how-locked-are-pluto-and-charon-how-much-does-each-librate-as-seen-from-the-o#comment147131_45349) on an answer to *How “locked” are Pluto and Charon? How much does each librate as seen from the other?*
uhoh
CC BY-SA 3.0
2017-07-10T19:05:47.433Z
2017-07-10T19:05:47.433Z
userLTK
<p>For an animated lesson, see Isaac Author's amazing space engineering anthology, including <a href="https://youtu.be/lmc5XqChJpY" rel="nofollow noreferrer">TIDAL LOCKING EXPLAINED</a>. </p>

CC BY-SA 3.0
2017-07-11T13:14:44.837Z
We tend to downvote answers which consist purely of a link, both because links go stale and because they do not actually answer the question.
Carl Witthoft
CC BY-SA 3.0
2017-07-11T14:47:29.870Z
Vote it down if you must for those reasons , but if you vote to delete, and have not cared to view the video or series, then that is dis-service to all. Right, my praise of the authors does not directly answer the questions but damn straight the video with series answers many, so worth the posting by many a moonshot.
Cymatical
CC BY-SA 3.0
2017-07-11T14:49:56.837Z
Besides,  Animated artistic celestial mechanics is such epic learning.
Cymatical
CC BY-SA 3.0
2017-07-11T14:52:11.590Z
Do you know the difference between a downvote and a moderator's right to delete(no vote needed)?
Carl Witthoft
CC BY-SA 3.0
2017-07-11T14:57:20.763Z
I did not.  Do now.  So thanks then I suppose :)
Cymatical
CC BY-SA 3.0
2017-07-11T08:38:08.257Z
2017-12-09T16:12:01.040Z
2017-12-09T16:12:01.040Z
Cymatical
<p>When a moon is in orbit around it's planet but is so close that it will become tidally locked does it experience a large 'wobble' in between the phase when it still has its own rotation and the final phase when it is always showing the same face to the planet?  I can only imagine that it would 'wobble' very widely, back and forth like a pendulum until finally settling into a locked position.  </p>

CC BY-SA 3.0
2017-07-10T14:50:52.497Z
Wow, this is a *really interesting* question! I'll bet there are plenty of simulations, but it would be wonderful to find an animation of the transition in video form. It would be even more interesting to find *an observed example* of a body in this transition period!
uhoh
CC BY-SA 3.0
2017-07-10T14:51:29.823Z
I assume by "wobble" you mean something different than libration, https://en.wikipedia.org/wiki/Libration so, more of a "as it settles into resonance" it first swings back and forth, different than the way it appears to do this as it moves closer and further from the planet due to any eccentricity in the orbit.
userLTK
CC BY-SA 3.0
2017-07-10T14:58:21.053Z
@userLTK it's an important distinction. Assuming a moon had an essentially circular orbit (and both bodies had the necessary physical requirements for tidal locking) there still would be a transition, when viewed from the planet, when it stopped complete apparent rotation and started *not showing* at first a small, then larger fraction of it's back side to the planet, right?
uhoh
CC BY-SA 3.0
2017-07-11T13:17:43.933Z
See also https://astronomy.stackexchange.com/questions/12154/what-causes-objects-to-become-tidally-locked
Carl Witthoft
CC BY-SA 3.0
2017-07-11T13:23:01.563Z
See the reference post in the close vote:  since the moon is always rotating, it has never stopped rotating.  There's no force that would cause it to rotate **slower** than tidal lock.
Carl Witthoft
CC BY-SA 3.0
2017-07-11T13:25:11.130Z
See further  the discussion in Physics.SE  https://physics.stackexchange.com/questions/156845/physics-of-tidal-locking-can-the-process-be-underdamped
Carl Witthoft
CC BY-SA 3.0
2017-07-11T18:08:18.963Z
The "duplicate question" doesn't answer this, but the physics question that @CarlWitthoft linked above, does.
userLTK
<tidal-locking>
Tidal locking of a moon
Brooks Nelson
# Tidal locking of a moon

(Asked by: username_0 on 2017-07-10)

When a moon is in orbit around it's planet but is so close that it will become tidally locked does it experience a large 'wobble' in between the phase when it still has its own rotation and the final phase when it is always showing the same face to the planet? I can only imagine that it would 'wobble' very widely, back and forth like a pendulum until finally settling into a locked position.

**Tags:** tidal-locking

> **Comments:**
> **username_1**: Wow, this is a *really interesting* question! I'll bet there are plenty of simulations, but it would be wonderful to find an animation of the transition in video form. It would be even more interesting to find *an observed example* of a body in this transition period!  
> **username_2**: I assume by "wobble" you mean something different than libration, https://en.wikipedia.org/wiki/Libration so, more of a "as it settles into resonance" it first swings back and forth, different than the way it appears to do this as it moves closer and further from the planet due to any eccentricity in the orbit.  
> **username_1**: @username_2 it's an important distinction. Assuming a moon had an essentially circular orbit (and both bodies had the necessary physical requirements for tidal locking) there still would be a transition, when viewed from the planet, when it stopped complete apparent rotation and started *not showing* at first a small, then larger fraction of it's back side to the planet, right?  
> **username_3**: See also https://astronomy.stackexchange.com/questions/12154/what-causes-objects-to-become-tidally-locked  
> **username_3**: See the reference post in the close vote:  since the moon is always rotating, it has never stopped rotating.  There's no force that would cause it to rotate **slower** than tidal lock.  
> **username_3**: See further  the discussion in Physics.SE  https://physics.stackexchange.com/questions/156845/physics-of-tidal-locking-can-the-process-be-underdamped  
> **username_2**: The "duplicate question" doesn't answer this, but the physics question that @CarlWitthoft linked above, does.  

---

## 2 answers

### Answer by username_2 on 2017-07-10. Score: 4

There's no back and forth motion on the way towards tidal locking. There's kind of two ways to look this kind of "settling", there's overshooting and correction, basically a back and forth, and there's a very gradual force that grows smaller, basically moving towards zero but not overshooting. I realize that's just re-writing your question, but it helps to understand the 2nd option.

I'll use the Earth as an example, because the Earth is in the gradual process of becoming tidally locked to the Moon. It's the same process, but easier to visualize.

Because the Earth rotates faster than the Moon orbits, the tidal bulge moves ahead of the moon. This is called the tidal bulge offset.

[![http://pages.uoregon.edu/jimbrau/BrauImNew/Chap07/FG07_23.jpg](https://i.sstatic.net/XgtZe.jpg)](https://i.sstatic.net/XgtZe.jpg)

[Source](http://pages.uoregon.edu/jimbrau/astr121-2005/Notes/Chapter7.html).

It's the size of the bulge and the angle of the offset (and the mass and distance of the Moon) that causes the familiar gravitational effects, the Earth's rotation slowing down and the Moon moving away from the Earth.

[![https://ase.tufts.edu/cosmos/pictures/Explore_figs_8/Chapter5/Fig5_28copy.jpg](https://i.sstatic.net/DbFCu.jpg)](https://i.sstatic.net/DbFCu.jpg)

[Source](https://ase.tufts.edu/cosmos/print_images.asp?id=5)

The current angle of offset for the Earth's tidal bulge to the Moon is about 3 degrees, often exaggerated in diagrams so it's easier to see. (*I should find a source for that - I've not been able to find a good one)*.

From the Moon's point of view (and if we assume a circular orbit), the tidal offset stays in the same place. It's a factor of the relative rotation speed. As Earth slows down, the angle of offset will move towards zero. At tidal locking, the angle of offset becomes zero and there's no force to continue in the other direction. In a simple mathematical model, tidal locking wouldn't ever happen completely, it would just gradually get closer and closer, like an object shot from the Earth at exactly escape velocity.

In reality, because objects aren't perfectly symmetrical and because orbits aren't perfectly circular, the "take forever to get to zero", doesn't happen. But it's still a gradual approach to zero and no overshooting and returning because the force on the rotation approaches zero as the angle of offset approaches zero and the force needed to change a planet's rotation is enormous and there's essentially zero force to make that happen.

With more eccentric orbits, like Mercury and it's 3:2 rotation to orbit resonance, that might be different. Not sure how eccentricity might play into this. That's more complicated.

> **Comments:**
> **username_0**: So are you saying the moon, previous to being tidally locked, continues to have complete rotations showing all sides but the rotation rate is just continuously slowing down until it's so slow it just stops and the same face is towards Earth with minimal to no rocking into position?  
> **username_2**: @BrooksNelson  Basically yes.   I don't see any other way that it could work.   The angular momentum changes very slowly (outside of a truly enormous impact).   That's a given and as the moon moves closer to tidal locking the tidal offset angle grows smaller and smaller, leading to smaller and smaller force.     I see no mechanism where it would over-slow-down and then speed up again.   It should just be a gradual approach towards tidal locking.  
> **username_3**: Try to avoid "I can't see" responses, since those are subjective.  
> **username_1**: speaking of "back and forth"; see [this comment](https://space.stackexchange.com/questions/45342/how-locked-are-pluto-and-charon-how-much-does-each-librate-as-seen-from-the-o#comment147131_45349) on an answer to *How “locked” are Pluto and Charon? How much does each librate as seen from the other?*  

---

### Answer by username_4 on 2017-07-11. Score: -1

For an animated lesson, see Isaac Author's amazing space engineering anthology, including [TIDAL LOCKING EXPLAINED](https://youtu.be/lmc5XqChJpY).

> **Comments:**
> **username_3**: We tend to downvote answers which consist purely of a link, both because links go stale and because they do not actually answer the question.  
> **username_4**: Vote it down if you must for those reasons , but if you vote to delete, and have not cared to view the video or series, then that is dis-service to all. Right, my praise of the authors does not directly answer the questions but damn straight the video with series answers many, so worth the posting by many a moonshot.  
> **username_4**: Besides,  Animated artistic celestial mechanics is such epic learning.  
> **username_3**: Do you know the difference between a downvote and a moderator's right to delete(no vote needed)?  
> **username_4**: I did not.  Do now.  So thanks then I suppose :)  

---

