<p>The answer to this is certainly tidal forces, but that doesn't explain the exact mechanism for how tidal forces result in <em>tidal locking</em>, i.e., an orbiting body showing the same face to the central body as it orbits due to the rotation rate and revolution rate being equal. I'll describe this mechanism using the Earth-Moon system so I can be specific, but it equally applies for any system.</p>

<p>To start, tidal forces are a result of differential gravitational forces across the distributed mass of a body. The Moon is not a point mass, it has an extended size. The force of gravity on the Moon by the Earth is dependent upon distance (as is the force of gravity for anything). What this means is that on the Earth-facing side of the Moon, the force of gravity is stronger and as you progress through the Moon to the Earth-opposing side, the force of gravity gets weaker. This means the Earth-facing side of the Moon gets pulled stronger and closer to the Earth while the Earth-opposing side, while still getting pulled towards the Earth, doesn't get pulled as strongly or as close. Ultimately, this results in the Moon being <em>deformed</em> such that it becomes slightly oblate and stretched out in the direction of the Earth. This flexing of the surface is known as tides.</p>

<p>Now let's assume the Moon is not currently tidally locked with the Earth, and in fact spins slightly faster than it orbits. The Earth is causing tides on the Moon and the Moon is spinning on its axis. The tides, caused by the tidal forces, want to remain aligned with the Earth-Moon line since that's the direction the tidal forces are applied. However, it takes time and a lot of energy to deform the Moon. Once the Moon is deformed, it's going to spin along and try to pull that tidal deformity along with it, effectively moving the tidal bulge ahead of the Earth-Moon line. The Earth is still applying the tidal force along that Earth-Moon line to try and pull the tidal bulge back in line. This constant force trying to pull the tidal bulge back (or ahead if the Moon is spinning too slow) allows for a transfer of momentum to slow down the Moon (or again, speed it up if it is too slow). The main point here is that tidal locking is a equilibrium state because if the Moon is spinning too slow or too fast, the Earth trying to pull the tidal bulge into the Earth-Moon line will change the rotation rate of the Moon until it becomes tidally locked. Once it is tidally locked, that tidal bulge will always be along the Earth-Moon line and this force will disappear.</p>

<p>That covers about half of the answer though. When trying to tidal lock the Moon to the Earth, you have to consider two time periods. The first, discussed in the preceding paragraph, is the rotation time of the Moon around its axis. The other is the revolution time of the Moon around the Earth. Both must match. The previous paragraph described how the rotation time of the Moon around its axis could be affected, but there's also a means of affecting the revolution time of the Moon around the Earth. Fortunately, this is by a nearly identical mechanism as above. In effect, the Moon also causes tidal bulges on the Earth and since the Earth is spinning these tidal bulges won't be directly aligned with the Earth-Moon line. This unaligned tidal bulge on Earth acts to transfer energy to the Moon's orbital speed, causing it to speed up or slow down. Incidentally, through conservation of angular momentum, this necessarily causes the Moon to drift away from us at a small but persistent rate.</p>

<p>In summation, tidal forces cause tidal locking, but it happens through intricate and slow forces over a long time period affecting both the Moon's orbital velocity and rotation rate until a equilibrium is found. That equilibrium is tidal locking.</p>

CC BY-SA 3.0
2017-07-07T05:43:01.493Z
If this goes for any system, does that mean Earth will eventually be tidal locked to the Sun?
Liren
CC BY-SA 3.0
2017-07-07T05:52:46.177Z
@Liren Yes, although it will take a lot longer. In fact, the Earth will become tidally locked to the Moon first.
Mark H
CC BY-SA 3.0
2017-07-07T08:23:37.170Z
I wonder which countries will get to see the moon... what a tourist attraction! :)
NikoNyrh
CC BY-SA 3.0
2017-07-07T17:35:09.353Z
@Liren Technically it could happen if there was only the Earth/Sun. The problem is we have lots of other planets like Jupiter which have noticeable gravitational perturbations on the Earth that prevent us from ever reaching that equilibrium of tidal locking. But, we definitely do see planet's tidally locking to their stars, e.g. Mercury (albeit at a 2:3 resonance rather than a 1:1 like the Moon) or the planets in the recently announced TRAPPIST-1 system.
zephyr
CC BY-SA 3.0
2017-07-07T17:36:21.510Z
@NikoNyrh Probably none of them. It's predicted that the Earth will take upwards of 50 billion years to tidally lock to the Moon (the Moon's affects on the Earth are just too small to do it sooner) and we know the Solar System will be long gone by then. The Sun will go Red Giant in about 5 or so billion years.
zephyr
CC BY-SA 3.0
2017-07-11T04:15:38.003Z
As already mentioned in zephyr's excellent answer above this is about gravity having a weaker pull on the far side of an orbiting object than the near side. If you think about it, this is why tidal locking is more about the distance from the "mother object" than the mass of the objects. This enlightened me as to why tidally locked habitable zone planets around M dwarfs are more likely than around sun-like stars. Remember, the force of gravity is proportional to the mass of the objects, but inversely proportional to the square of the distance.
Jack R. Woods
CC BY-SA 3.0
2017-07-11T04:29:29.563Z
A note about the comment by Mark H. This is obvious if you think about ocean tides. The sun effects tides, too, but not nearly as much as the moon. The length of a day increased 1.7 milliseconds over the last century so I think it will be a little while before we become tidally locked to the moon. (I haven't done the math, but I'm guessing it will be well after the Earth becomes uninhabitable.)
Jack R. Woods
CC BY-SA 3.0
2017-07-11T04:54:28.417Z
According to this guy https://www.universetoday.com/128350/will-earth-lock-moon/ .. or whomever did the math, we're looking at 50 billion years (a little longer than I thought).
Jack R. Woods
CC BY-SA 4.0
2023-06-16T10:57:48.717Z
I don't think it's right to say if a moon has equal rotational and orbital periods it is tidally locked to its host planet. For example, if the Moon had an orbital and rotational period of 24 hours would it be called tidally locked again?
Snack Exchange
CC BY-SA 3.0
2017-07-06T16:15:15.757Z
2017-07-06T16:15:15.757Z
zephyr
<p>The simple answer is: tidal forces, which are a secondary effect of gravity. In the same way that the Moon causes low and high tides of the oceans here on Earth, the Earth also has a similar effect on the Moon. </p>

<p>The force is of the same origin, however far stronger due to the mass of the Earth. These tidal forces cause a torque on the rotation on the Moon and this is why it only shows the same face to the Earth. </p>

<p>Fun fact, Pluto and it's only moon Charon show only one face during their orbit due to the tidal force. Also the tidal forces are slowing down the rotation of the Earth too. </p>

CC BY-SA 3.0
2017-07-06T14:51:57.467Z
Do tidal forces depend on asymmetry?   IOW, if the moon were more perfectly symmetrical all around, would tidal forces still exist or have the same effect?
0tyranny0poverty
CC BY-SA 3.0
2017-07-06T15:02:15.407Z
They depend upon mass and the  non constant nature of the gravitational force. That is, the gravitational attraction between two bodies is more strong at the sides facing each other. Hope that helps.
Rumplestillskin
CC BY-SA 3.0
2017-07-06T15:57:12.017Z
@0tyranny0poverty Sort of. If you could make the Moon's mass distribution perfectly symmetrical (the lunar mass distribution is [notoriously uneven](https://en.wikipedia.org/wiki/Mass_concentration_%28astronomy%29)) it would reduce the rate of change of the Moon's rotational period, but tidal forces themselves create an asymmetry in the surface: the tidal bulge. To eliminate the effect of tidal locking you'd need to have a perfectly rigid body, and that's physically impossible.
PM 2Ring
CC BY-SA 3.0
2017-07-06T15:57:40.120Z
FWIW, the Moon also creates a tidal bulge in the solid matter of Earth, but it's not as noticeable as the tidal bulge it creates in the oceans.
PM 2Ring
CC BY-SA 3.0
2017-07-06T17:48:46.657Z
@0tyranny0poverty  Symmetry doesn't matter as long as the object is flexible.   Tidal forces will bend a sphere into an oval shape and the oval is asymmetric.    The Moon is very asymmetrical and that may have sped up it's tidal locking but it's not a requirement.    If you had a moon that was perfectly symmetrical and 100% inflexible, (there's no such material), then it wouldn't be subject to tidal locking or rotational slowing.
userLTK
CC BY-SA 3.0
2017-07-06T21:06:20.557Z
@userLTK  Interesting.  So what about the other moons of Saturn and certain moons of Jupiter whose orbital periods are not matched with their rotational period?  Would this then indicate a more inflexible and symmetric content of the moon?  Or maybe younger moon since it takes a certain time to reach this equilibrium?
0tyranny0poverty
CC BY-SA 3.0
2017-07-06T22:48:25.410Z
@0tyranny0poverty I suspect it's because it's a complex system with lots of moons, as well as the rings. This results in a chaotic mixture of gravitational effects, and it takes some time and luck for a satellite to synchronize.
Barmar
CC BY-SA 3.0
2017-07-07T12:55:35.810Z
@0tyranny0poverty It depends how close or far the moons are from Jupiter (further away the tidal force drops off by the 3rd power - in other words very fast).  It also depends on the size of the moon.   Smaller objects are less symmetrical on average but the tidal force on a tiny object is also lower.    Look at large moons and small moons and compare which ones are tidally locked.   Large moons enter the 1:1 resonance of tidal locking more easily.  The tidal bulge on small moons is less, and for similar reasons, the tides on lakes is tiny.  (if that's not clear, maybe ask a new question)
userLTK
CC BY-SA 3.0
2017-07-07T16:08:46.467Z
@userLTK  why would the tidal force drop off by 3rd power and not the square?  Which formula are you using?
0tyranny0poverty
CC BY-SA 3.0
2017-07-07T17:48:53.950Z
@0tyranny0poverty Because tidal forces are a result of *differential* gravitational forces. Basically you need to differentiate Newton's law of gravity, resulting in the tidal force being proportional to $r^{-3}$. See [the wiki page](https://en.wikipedia.org/wiki/Tidal_force) for more details.
zephyr
CC BY-SA 3.0
2017-07-06T14:47:14.577Z
2017-07-06T14:47:14.577Z
Rumplestillskin
<p>Our moon and Saturn's moon, Titan, have this feature.  Because of this we only observe one side (hemispere) of our moon. Why is this?  What is the Newtonian or astrophysics that would explain this?  It seems that it would be some kind of equilibrium that is arrived at over time, but I don't understand it.</p>

CC BY-SA 3.0
2017-07-06T14:41:22.113Z
Youtube links aren't recommended, but it's a nice simple explanation:   https://www.youtube.com/watch?v=6jUpX7J7ySo    Look up Tidal Locking for more info.
userLTK
CC BY-SA 3.0
2017-07-06T19:39:23.640Z
I was just wondering this yesterday, weird
MCMastery
CC BY-SA 3.0
2017-07-07T13:44:09.787Z
It's not just our moon and Titan. To the best of my knowledge, every large moon in the solar system exhibits this behavior. Tidal locking is a powerful phenomenon.
Mark Foskey
<the-moon><astrophysics><rotation><orbital-mechanics><titan>
Why do certain moons have their rotational period equal to their orbital period?
0tyranny0poverty
# Why do certain moons have their rotational period equal to their orbital period?

(Asked by: username_0 on 2017-07-06)

Our moon and Saturn's moon, Titan, have this feature. Because of this we only observe one side (hemispere) of our moon. Why is this? What is the Newtonian or astrophysics that would explain this? It seems that it would be some kind of equilibrium that is arrived at over time, but I don't understand it.

**Tags:** the-moon, astrophysics, rotation, orbital-mechanics, titan

> **Comments:**
> **username_1**: Youtube links aren't recommended, but it's a nice simple explanation:   https://www.youtube.com/watch?v=6jUpX7J7ySo    Look up Tidal Locking for more info.  
> **username_2**: I was just wondering this yesterday, weird  
> **username_3**: It's not just our moon and Titan. To the best of my knowledge, every large moon in the solar system exhibits this behavior. Tidal locking is a powerful phenomenon.  

---

## 2 answers

### Answer by username_4 on 2017-07-06. Score: 15

The answer to this is certainly tidal forces, but that doesn't explain the exact mechanism for how tidal forces result in *tidal locking*, i.e., an orbiting body showing the same face to the central body as it orbits due to the rotation rate and revolution rate being equal. I'll describe this mechanism using the Earth-Moon system so I can be specific, but it equally applies for any system.

To start, tidal forces are a result of differential gravitational forces across the distributed mass of a body. The Moon is not a point mass, it has an extended size. The force of gravity on the Moon by the Earth is dependent upon distance (as is the force of gravity for anything). What this means is that on the Earth-facing side of the Moon, the force of gravity is stronger and as you progress through the Moon to the Earth-opposing side, the force of gravity gets weaker. This means the Earth-facing side of the Moon gets pulled stronger and closer to the Earth while the Earth-opposing side, while still getting pulled towards the Earth, doesn't get pulled as strongly or as close. Ultimately, this results in the Moon being *deformed* such that it becomes slightly oblate and stretched out in the direction of the Earth. This flexing of the surface is known as tides.

Now let's assume the Moon is not currently tidally locked with the Earth, and in fact spins slightly faster than it orbits. The Earth is causing tides on the Moon and the Moon is spinning on its axis. The tides, caused by the tidal forces, want to remain aligned with the Earth-Moon line since that's the direction the tidal forces are applied. However, it takes time and a lot of energy to deform the Moon. Once the Moon is deformed, it's going to spin along and try to pull that tidal deformity along with it, effectively moving the tidal bulge ahead of the Earth-Moon line. The Earth is still applying the tidal force along that Earth-Moon line to try and pull the tidal bulge back in line. This constant force trying to pull the tidal bulge back (or ahead if the Moon is spinning too slow) allows for a transfer of momentum to slow down the Moon (or again, speed it up if it is too slow). The main point here is that tidal locking is a equilibrium state because if the Moon is spinning too slow or too fast, the Earth trying to pull the tidal bulge into the Earth-Moon line will change the rotation rate of the Moon until it becomes tidally locked. Once it is tidally locked, that tidal bulge will always be along the Earth-Moon line and this force will disappear.

That covers about half of the answer though. When trying to tidal lock the Moon to the Earth, you have to consider two time periods. The first, discussed in the preceding paragraph, is the rotation time of the Moon around its axis. The other is the revolution time of the Moon around the Earth. Both must match. The previous paragraph described how the rotation time of the Moon around its axis could be affected, but there's also a means of affecting the revolution time of the Moon around the Earth. Fortunately, this is by a nearly identical mechanism as above. In effect, the Moon also causes tidal bulges on the Earth and since the Earth is spinning these tidal bulges won't be directly aligned with the Earth-Moon line. This unaligned tidal bulge on Earth acts to transfer energy to the Moon's orbital speed, causing it to speed up or slow down. Incidentally, through conservation of angular momentum, this necessarily causes the Moon to drift away from us at a small but persistent rate.

In summation, tidal forces cause tidal locking, but it happens through intricate and slow forces over a long time period affecting both the Moon's orbital velocity and rotation rate until a equilibrium is found. That equilibrium is tidal locking.

> **Comments:**
> **username_5**: If this goes for any system, does that mean Earth will eventually be tidal locked to the Sun?  
> **username_6**: @username_5 Yes, although it will take a lot longer. In fact, the Earth will become tidally locked to the Moon first.  
> **username_7**: I wonder which countries will get to see the moon... what a tourist attraction! :)  
> **username_4**: @username_5 Technically it could happen if there was only the Earth/Sun. The problem is we have lots of other planets like Jupiter which have noticeable gravitational perturbations on the Earth that prevent us from ever reaching that equilibrium of tidal locking. But, we definitely do see planet's tidally locking to their stars, e.g. Mercury (albeit at a 2:3 resonance rather than a 1:1 like the Moon) or the planets in the recently announced TRAPPIST-1 system.  
> **username_4**: @username_7 Probably none of them. It's predicted that the Earth will take upwards of 50 billion years to tidally lock to the Moon (the Moon's affects on the Earth are just too small to do it sooner) and we know the Solar System will be long gone by then. The Sun will go Red Giant in about 5 or so billion years.  
> **username_8**: As already mentioned in username_4's excellent answer above this is about gravity having a weaker pull on the far side of an orbiting object than the near side. If you think about it, this is why tidal locking is more about the distance from the "mother object" than the mass of the objects. This enlightened me as to why tidally locked habitable zone planets around M dwarfs are more likely than around sun-like stars. Remember, the force of gravity is proportional to the mass of the objects, but inversely proportional to the square of the distance.  
> **username_8**: A note about the comment by username_6. This is obvious if you think about ocean tides. The sun effects tides, too, but not nearly as much as the moon. The length of a day increased 1.7 milliseconds over the last century so I think it will be a little while before we become tidally locked to the moon. (I haven't done the math, but I'm guessing it will be well after the Earth becomes uninhabitable.)  
> **username_8**: According to this guy https://www.universetoday.com/128350/will-earth-lock-moon/ .. or whomever did the math, we're looking at 50 billion years (a little longer than I thought).  
> **username_9**: I don't think it's right to say if a moon has equal rotational and orbital periods it is tidally locked to its host planet. For example, if the Moon had an orbital and rotational period of 24 hours would it be called tidally locked again?  

---

### Answer by username_10 on 2017-07-06. Score: 3

The simple answer is: tidal forces, which are a secondary effect of gravity. In the same way that the Moon causes low and high tides of the oceans here on Earth, the Earth also has a similar effect on the Moon.

The force is of the same origin, however far stronger due to the mass of the Earth. These tidal forces cause a torque on the rotation on the Moon and this is why it only shows the same face to the Earth.

Fun fact, Pluto and it's only moon Charon show only one face during their orbit due to the tidal force. Also the tidal forces are slowing down the rotation of the Earth too.

> **Comments:**
> **username_0**: Do tidal forces depend on asymmetry?   IOW, if the moon were more perfectly symmetrical all around, would tidal forces still exist or have the same effect?  
> **username_10**: They depend upon mass and the  non constant nature of the gravitational force. That is, the gravitational attraction between two bodies is more strong at the sides facing each other. Hope that helps.  
> **username_11**: @username_0 Sort of. If you could make the Moon's mass distribution perfectly symmetrical (the lunar mass distribution is [notoriously uneven](https://en.wikipedia.org/wiki/Mass_concentration_%28astronomy%29)) it would reduce the rate of change of the Moon's rotational period, but tidal forces themselves create an asymmetry in the surface: the tidal bulge. To eliminate the effect of tidal locking you'd need to have a perfectly rigid body, and that's physically impossible.  
> **username_11**: FWIW, the Moon also creates a tidal bulge in the solid matter of Earth, but it's not as noticeable as the tidal bulge it creates in the oceans.  
> **username_1**: @username_0  Symmetry doesn't matter as long as the object is flexible.   Tidal forces will bend a sphere into an oval shape and the oval is asymmetric.    The Moon is very asymmetrical and that may have sped up it's tidal locking but it's not a requirement.    If you had a moon that was perfectly symmetrical and 100% inflexible, (there's no such material), then it wouldn't be subject to tidal locking or rotational slowing.  
> **username_0**: @username_1  Interesting.  So what about the other moons of Saturn and certain moons of Jupiter whose orbital periods are not matched with their rotational period?  Would this then indicate a more inflexible and symmetric content of the moon?  Or maybe younger moon since it takes a certain time to reach this equilibrium?  
> **username_12**: @username_0 I suspect it's because it's a complex system with lots of moons, as well as the rings. This results in a chaotic mixture of gravitational effects, and it takes some time and luck for a satellite to synchronize.  
> **username_1**: @username_0 It depends how close or far the moons are from Jupiter (further away the tidal force drops off by the 3rd power - in other words very fast).  It also depends on the size of the moon.   Smaller objects are less symmetrical on average but the tidal force on a tiny object is also lower.    Look at large moons and small moons and compare which ones are tidally locked.   Large moons enter the 1:1 resonance of tidal locking more easily.  The tidal bulge on small moons is less, and for similar reasons, the tides on lakes is tiny.  (if that's not clear, maybe ask a new question)  
> **username_0**: @username_1  why would the tidal force drop off by 3rd power and not the square?  Which formula are you using?  
> **username_4**: @username_0 Because tidal forces are a result of *differential* gravitational forces. Basically you need to differentiate Newton's law of gravity, resulting in the tidal force being proportional to $r^{-3}$. See [the wiki page](https://en.wikipedia.org/wiki/Tidal_force) for more details.  

---

