<p>Rather than your question (which Sten has already <a href="https://astronomy.stackexchange.com/a/54003">answered</a> well), let me reply to <a href="https://astronomy.stackexchange.com/posts/comments/120988">one of your comments</a> under Sten's answer, since I think that's where the core of your misunderstanding lies:</p>
<blockquote>
<p>In this scenario [where Moon's orbital period = Moon's rotational period = Earth's rotational period = 24 h], the Moon's orbital period is the same with the Earth's rotational period. So the Moon will be only visible to a certain region of Earth. Well, the Moon also rotates every 24 hours. So people of that certain region can see both sides of the Moon (12 hours one side, and another 12 hours the other side)</p>
</blockquote>
<p>No, that's not what would happen in your scenario.  It is, however, what would happen if the Moon did not rotate at all.</p>
<p>As it happens, Wikipedia's article on <a href="https://en.wikipedia.org/wiki/Tidal_locking" rel="noreferrer">tidal locking</a> already has an excellent <a href="https://en.wikipedia.org/wiki/Tidal_locking#/media/File:Tidal_locking_of_the_Moon_with_the_Earth.gif" rel="noreferrer">animation</a> illustrating this, so I'll just borrow it:</p>
<p><a href="https://i.sstatic.net/uyGrW.gif" rel="noreferrer"><img src="https://i.sstatic.net/uyGrW.gif" alt="Two animations of the Moon orbiting the Earth (not to scale, Earth drawn as a black circle). The animation on the left shows the Moon rotating at the same rate as it orbits the Earth, and thus always keeping the same side facing the Earth, while the animation on the right shows the Moon not rotating at all, and thus showing all of its sides to the Earth during each orbit." /></a><br />
<sup><a href="https://commons.wikimedia.org/wiki/File:Tidal_locking_of_the_Moon_with_the_Earth.gif" rel="noreferrer">Animation</a> by Wikipedia user <a href="https://en.wikipedia.org/wiki/User:Stigmatella_aurantiaca" rel="noreferrer">Stigmatella aurantiaca</a>, used under the <a href="https://creativecommons.org/licenses/by-sa/3.0/deed.en" rel="noreferrer">CC By-SA 3.0</a> license.</sup></p>
<p>On the left, you can see the Moon orbiting the Earth (drawn as just a circle, and obviously not to scale) while also rotating at the same rate as it moves around the Earth, so that it completes one orbit around the Earth in the same time as it completes one rotation around its own axis — just as it (on average, neglecting <a href="https://en.wikipedia.org/wiki/Libration" rel="noreferrer">libration</a>) does in reality, and as it would also do in your scenario.  You can clearly see that this results in the same side of the Moon always pointing towards the Earth.</p>
<p>On the right, meanwhile, is what would happen if the Moon did not rotate at all <em>while still orbiting the Earth</em>.  In that case the side of the Moon facing the Earth <em>does</em> change: the Moon doesn't rotate, but <em>it moves around the Earth</em>, so that people watching it from the Earth see it from different directions.</p>
<p>(The Earth itself is drawn as just a featureless circle in this animation, so you'll have to imagine its rotation.  But if you imagine it rotating at the same rate as the Moon in the left animation, so that the same side of the Earth always points towards the Moon, then you'll have exactly the mutual tidal locking scenario you envisioned.)</p>
<hr />
<p>For an alternative visualization, get a friend to help you and have them stand in one spot while you walk a full circle around them, <em>with both of you looking each other in the eyes all the time</em>.  Obviously your friend will have to make a full 360° turn in order to maintain eye contact with you as you walk around them — but so will you!  If you started the turn e.g. with your friend on your west side and went clockwise, then you'd end up looking first west, then north (after walking 90° around your friend), then east (after 180°), then south (after 270°) and finally west again (after walking a full 360° circle).</p>
<p>Or, if staring your friend in the eyes while walking around them feels too creepy and awkward, just hold hands with them instead.  Or just hold your hand against a tree or a light pole while walking around it.</p>
<p>In any case, any time you walk a circle around something while keeping it on the same side of you, you'll inevitably end up making a 360° turn.  And conversely, any time you walk around something <em>and turn a full 360° at the same time</em>, the same side of your body will always face the thing you walk around (at least on average).</p>
<p>Meanwhile, if your try to walk around your friend (or a signpost) without turning at all, and initially face towards them, you'll find that after walking 180° around them you now have you back towards them instead.  (And if you practice your footwork until you can do the full circle like that smoothly without stumbling, you'll now know how to do a basic <a href="https://en.wikipedia.org/wiki/Do-si-do" rel="noreferrer">do-si-do</a> and be one step closer to perfecting your folk dancing skills.  Congratulations.)</p>

CC BY-SA 4.0
2023-06-17T13:13:17.410Z
That feels funny to me instead of creepy and awkward :) Thank yoi for your detailed answer.
Snack Exchange
CC BY-SA 4.0
2023-06-17T12:55:42.483Z
2023-06-17T12:55:42.483Z
Ilmari Karonen
<p>I agree with the answer provided by Sten, but canot help from adding several details.</p>
<ol>
<li>Historically, it was Immanuel Kant who proposed, on qualitative grounds, that eventually the tidal synchronism of the Earth and the Moon will become mutual -- i.e., that the Earth also will be showing the same face to the Moon.</li>
</ol>
<p>Even among the scholars of Kant's heritage few people are aware of that work of his, because Kant published it not in a scientific journal or book, but in two consequtive issues of a local newspaper.</p>
<ol start="2">
<li>Whether the mutual synchronism will actually be achieved depends on how far the Moon will have to recede for that. If it has to tidally recede too far, the solar gravity will take over, and the Moon will fly away. In other words, the synchronism must be achieved while the Moon is still within the Hill sphere of the Earth. And this is not the end of story -- because in real life a satellite should not approach the Hill radius too closely, lest it motion becomes unstable. So the Moon should actually be inside the <em>reduced Hill sphere</em>, whose radius (for prograde satellites) is about 0.49 of the Hill radius.</li>
</ol>
<p>It is possible to demonstrate that, yes, the mutual synchronism will be attained with the Moon still slightly below the reduced Hill radius.</p>
<ol start="3">
<li>In principle, mutual synchronism can be achieved not only by a tidally receding moon, but also by a tidally decending moon, provided it is sufficiently heavy, and the synchronism is reached before the moon crosses the planet's Roche radius.</li>
</ol>

CC BY-SA 4.0
2023-06-17T00:20:52.540Z
2023-06-17T00:20:52.540Z
Michael_1812
<p>Currently, the Moon is tidally locked to the Earth but the Earth is <em>not</em> tidally locked to the Moon.</p>
<p>In your scenario (which could be accomplished by raising the mass of the Earth), where the Moon has an orbital and rotational period of 24 hours, the two bodies are mutually tidally locked.</p>
<p>Indeed, the Earth-Moon system is gradually moving toward such a state due to tidal energy exchange -- albeit through slowing of the Earth's rotation rather than shortening of the Moon's orbit.</p>

CC BY-SA 4.0
2023-06-16T11:32:17.083Z
In my scenario isn't the Moon's far side also facing the Earth?
Snack Exchange
CC BY-SA 4.0
2023-06-16T11:37:03.733Z
@SnackExchange If the far side were facing the Earth, it wouldn't be the far side! Sorry, it's not clear to me what you mean there. If the Moon's orbital and rotational periods are the same, then only one side can face the Earth, unless I'm misinterpreting something.
Sten
CC BY-SA 4.0
2023-06-16T11:45:02.707Z
In this scenario, the Moon's orbital period is the same with the Earth's rotational period. So the Moon will be only visible to a certain region of Earth. Well, the Moon also rotates every 24 hours. So people of that certain region can see both sides of the Moon (12 hours one side, and another 12 hours the other side)
Snack Exchange
CC BY-SA 4.0
2023-06-16T11:55:42.373Z
@SnackExchange The Moon would indeed be stationary on the sky (Earth's libration aside). But people on Earth would only see one side. If the Moon's orbital and rotation periods are equal, it can only present one face to the Earth, regardless of how the Earth's rotational period compares.
Sten
CC BY-SA 4.0
2023-06-16T11:56:44.040Z
@SnackExchange Or are you thinking of a retrograde-rotating Moon? That would indeed not be tidally locked, even if its rotational period were equal to its orbital period.
Sten
CC BY-SA 4.0
2023-06-16T12:41:41.517Z
Why? Also, does a retrograde-orbiting moon remain tidally locked if its orbital and rotational period are the same? What is an example of a retrograde-rotating moon?
Snack Exchange
CC BY-SA 4.0
2023-06-16T13:00:10.573Z
@SnackExchange This is going into definitions, so I had to look up what the precise definitions are. Tidal locking really just means that the rotation rate has become stable with respect to tidal energy exchange. That's the case for the Moon now. For a circular orbit, the only stable configuration is the Moon's, where rotation is in the same direction as the orbit with the same period. For eccentric orbits, resonant tidal locking is possible, where the rotation period is a rational multiple of the orbital period -- but I don't know if resonant effects could stabilize retrograde rotation.
Sten
CC BY-SA 4.0
2023-06-16T14:54:20.140Z
Terminology in astronomy is often vague and confusing.  Technically, the Earth-Moon system is partially locked and what you described is fully locked.  But sometimes partially locked means the two periods are close but not exactly the same.
eshaya
CC BY-SA 4.0
2023-06-16T11:16:27.437Z
2023-06-16T11:38:43.477Z
2023-06-16T11:38:43.477Z
Sten
<p>We always are hearing this: &quot;If a moon has equal rotational and orbital periods it's tidally locked to its host planet and always one side of it will face to the planet.&quot;</p>
<p>But what if, for example, the Moon had an orbital and rotational period of 24 hours? (with the same distance and the same rotational period of the Earth)</p>
<p>Would it be called tidally locked again?</p>

<the-moon><earth><tidal-locking>
Is this definition of tidal locking really satisfying?
Snack Exchange
# Is this definition of tidal locking really satisfying?

(Asked by: username_0 on 2023-06-16)

We always are hearing this: "If a moon has equal rotational and orbital periods it's tidally locked to its host planet and always one side of it will face to the planet."

But what if, for example, the Moon had an orbital and rotational period of 24 hours? (with the same distance and the same rotational period of the Earth)

Would it be called tidally locked again?

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

---

## 3 answers

### Answer by username_3 on 2023-06-16. Score: 7 | Accepted answer

Currently, the Moon is tidally locked to the Earth but the Earth is *not* tidally locked to the Moon.

In your scenario (which could be accomplished by raising the mass of the Earth), where the Moon has an orbital and rotational period of 24 hours, the two bodies are mutually tidally locked.

Indeed, the Earth-Moon system is gradually moving toward such a state due to tidal energy exchange -- albeit through slowing of the Earth's rotation rather than shortening of the Moon's orbit.

> **Comments:**
> **username_0**: In my scenario isn't the Moon's far side also facing the Earth?  
> **username_3**: @SnackExchange If the far side were facing the Earth, it wouldn't be the far side! Sorry, it's not clear to me what you mean there. If the Moon's orbital and rotational periods are the same, then only one side can face the Earth, unless I'm misinterpreting something.  
> **username_0**: In this scenario, the Moon's orbital period is the same with the Earth's rotational period. So the Moon will be only visible to a certain region of Earth. Well, the Moon also rotates every 24 hours. So people of that certain region can see both sides of the Moon (12 hours one side, and another 12 hours the other side)  
> **username_3**: @SnackExchange The Moon would indeed be stationary on the sky (Earth's libration aside). But people on Earth would only see one side. If the Moon's orbital and rotation periods are equal, it can only present one face to the Earth, regardless of how the Earth's rotational period compares.  
> **username_3**: @SnackExchange Or are you thinking of a retrograde-rotating Moon? That would indeed not be tidally locked, even if its rotational period were equal to its orbital period.  
> **username_0**: Why? Also, does a retrograde-orbiting moon remain tidally locked if its orbital and rotational period are the same? What is an example of a retrograde-rotating moon?  
> **username_3**: @SnackExchange This is going into definitions, so I had to look up what the precise definitions are. Tidal locking really just means that the rotation rate has become stable with respect to tidal energy exchange. That's the case for the Moon now. For a circular orbit, the only stable configuration is the Moon's, where rotation is in the same direction as the orbit with the same period. For eccentric orbits, resonant tidal locking is possible, where the rotation period is a rational multiple of the orbital period -- but I don't know if resonant effects could stabilize retrograde rotation.  
> **username_4**: Terminology in astronomy is often vague and confusing.  Technically, the Earth-Moon system is partially locked and what you described is fully locked.  But sometimes partially locked means the two periods are close but not exactly the same.  

---

### Answer by username_1 on 2023-06-17. Score: 7

Rather than your question (which Sten has already [answered](https://astronomy.stackexchange.com/a/54003) well), let me reply to [one of your comments](https://astronomy.stackexchange.com/posts/comments/120988) under Sten's answer, since I think that's where the core of your misunderstanding lies:

> In this scenario [where Moon's orbital period = Moon's rotational period = Earth's rotational period = 24 h], the Moon's orbital period is the same with the Earth's rotational period. So the Moon will be only visible to a certain region of Earth. Well, the Moon also rotates every 24 hours. So people of that certain region can see both sides of the Moon (12 hours one side, and another 12 hours the other side)

No, that's not what would happen in your scenario. It is, however, what would happen if the Moon did not rotate at all.

As it happens, Wikipedia's article on [tidal locking](https://en.wikipedia.org/wiki/Tidal_locking) already has an excellent [animation](https://en.wikipedia.org/wiki/Tidal_locking#/media/File:Tidal_locking_of_the_Moon_with_the_Earth.gif) illustrating this, so I'll just borrow it:

[![Two animations of the Moon orbiting the Earth (not to scale, Earth drawn as a black circle). The animation on the left shows the Moon rotating at the same rate as it orbits the Earth, and thus always keeping the same side facing the Earth, while the animation on the right shows the Moon not rotating at all, and thus showing all of its sides to the Earth during each orbit.](https://i.sstatic.net/uyGrW.gif)](https://i.sstatic.net/uyGrW.gif)  
[Animation](https://commons.wikimedia.org/wiki/File:Tidal_locking_of_the_Moon_with_the_Earth.gif) by Wikipedia user [Stigmatella aurantiaca](https://en.wikipedia.org/wiki/User:Stigmatella_aurantiaca), used under the [CC By-SA 3.0](https://creativecommons.org/licenses/by-sa/3.0/deed.en) license.

On the left, you can see the Moon orbiting the Earth (drawn as just a circle, and obviously not to scale) while also rotating at the same rate as it moves around the Earth, so that it completes one orbit around the Earth in the same time as it completes one rotation around its own axis — just as it (on average, neglecting [libration](https://en.wikipedia.org/wiki/Libration)) does in reality, and as it would also do in your scenario. You can clearly see that this results in the same side of the Moon always pointing towards the Earth.

On the right, meanwhile, is what would happen if the Moon did not rotate at all *while still orbiting the Earth*. In that case the side of the Moon facing the Earth *does* change: the Moon doesn't rotate, but *it moves around the Earth*, so that people watching it from the Earth see it from different directions.

(The Earth itself is drawn as just a featureless circle in this animation, so you'll have to imagine its rotation. But if you imagine it rotating at the same rate as the Moon in the left animation, so that the same side of the Earth always points towards the Moon, then you'll have exactly the mutual tidal locking scenario you envisioned.)

---

For an alternative visualization, get a friend to help you and have them stand in one spot while you walk a full circle around them, *with both of you looking each other in the eyes all the time*. Obviously your friend will have to make a full 360° turn in order to maintain eye contact with you as you walk around them — but so will you! If you started the turn e.g. with your friend on your west side and went clockwise, then you'd end up looking first west, then north (after walking 90° around your friend), then east (after 180°), then south (after 270°) and finally west again (after walking a full 360° circle).

Or, if staring your friend in the eyes while walking around them feels too creepy and awkward, just hold hands with them instead. Or just hold your hand against a tree or a light pole while walking around it.

In any case, any time you walk a circle around something while keeping it on the same side of you, you'll inevitably end up making a 360° turn. And conversely, any time you walk around something *and turn a full 360° at the same time*, the same side of your body will always face the thing you walk around (at least on average).

Meanwhile, if your try to walk around your friend (or a signpost) without turning at all, and initially face towards them, you'll find that after walking 180° around them you now have you back towards them instead. (And if you practice your footwork until you can do the full circle like that smoothly without stumbling, you'll now know how to do a basic [do-si-do](https://en.wikipedia.org/wiki/Do-si-do) and be one step closer to perfecting your folk dancing skills. Congratulations.)

> **Comments:**
> **username_0**: That feels funny to me instead of creepy and awkward :) Thank yoi for your detailed answer.  

---

### Answer by username_2 on 2023-06-17. Score: 3

I agree with the answer provided by Sten, but canot help from adding several details.

1. Historically, it was Immanuel Kant who proposed, on qualitative grounds, that eventually the tidal synchronism of the Earth and the Moon will become mutual -- i.e., that the Earth also will be showing the same face to the Moon.

Even among the scholars of Kant's heritage few people are aware of that work of his, because Kant published it not in a scientific journal or book, but in two consequtive issues of a local newspaper.

2. Whether the mutual synchronism will actually be achieved depends on how far the Moon will have to recede for that. If it has to tidally recede too far, the solar gravity will take over, and the Moon will fly away. In other words, the synchronism must be achieved while the Moon is still within the Hill sphere of the Earth. And this is not the end of story -- because in real life a satellite should not approach the Hill radius too closely, lest it motion becomes unstable. So the Moon should actually be inside the *reduced Hill sphere*, whose radius (for prograde satellites) is about 0.49 of the Hill radius.

It is possible to demonstrate that, yes, the mutual synchronism will be attained with the Moon still slightly below the reduced Hill radius.

3. In principle, mutual synchronism can be achieved not only by a tidally receding moon, but also by a tidally decending moon, provided it is sufficiently heavy, and the synchronism is reached before the moon crosses the planet's Roche radius.


---

