<p>The question in the title and the body are different.  There are no tidally locked objects in the Oort cloud. The sun's tidal forces at that distance are not sufficient to lock an object's rotation.  Tidal locking doesn't mean 'co-incidentally' having the same rate of rotation.</p>
<p>Tidal forces decrease with the cube of distance; tidal torques decrease with the sixth power of distance, so are much less significant relative to gravity, which decreases with the square of distance, in the Oort cloud. Moreover, the time for tidal lock to occur increases with the sixth power of the distance. Tidal lock also affects large objects more than small ones (in general) and most Oort cloud object are small.</p>
<p>There are other things that can cause changes in rotation rates: gravitational interactions in close passes with other objects, small impacts, the <a href="https://en.wikipedia.org/wiki/YORP_effect" rel="nofollow noreferrer">YORP effect</a>. All of these are rare or weak in the Oort cloud, but stronger than tidal locking, and so would tend to cause any object which by chance had a synchronous rotation rate to break that co-incidence.</p>

CC BY-SA 4.0
2023-03-24T09:53:45.743Z
Thanks. Ive put the title in quotes in order to avoid this misunderstanding
Rabbi Kaii
CC BY-SA 4.0
2023-03-24T16:31:05.660Z
The other answer calculates that the probability that rotation rate = revolution rate is infinitessimally small. So if this coincidence occurs, it's almost certainly an indication of tidal locking.
Barmar
CC BY-SA 4.0
2023-03-26T01:55:21.287Z
James, this is a good answer. Still, you may want to correct a slip of the pen in it. Tidal torques (which are actually responsible for tidal capture) decrease not with the cube but with the inverse 6-th power of distance, see eqns (23) and (36) in http://iopscience.iop.org/0004-637X/764/1/26  [It is the perturbing potential $W$ that falls of as the inverse cube of distance.]
Michael_1812
CC BY-SA 4.0
2023-03-26T02:00:12.423Z
@Barmar I would disagree with your statement. In my opinion, if a coincidence occurs it will be a coincidence only. As James K rightly noticed, there are too many interactions in the Oort cloud that are much stronger than tidal forces. In fact, the tidal forces caused by the Sun are totally irrelevant there, falling of as the ratio of an object's size to its distance from the Sun, taken to the minus six power.
Michael_1812
CC BY-SA 4.0
2023-03-26T17:01:58.917Z
@Michael_1812 I'm not saying that there's actually any tidal locking. I'm saying that since gravity isn't strong enough, and the probability is so small, it won't happen at all.
Barmar
CC BY-SA 4.0
2023-03-26T17:22:59.413Z
@Barmar But this is how I understood you. You said: "if this coincidence occurs, it's almost certainly an indication of tidal locking."
Michael_1812
CC BY-SA 4.0
2023-03-26T19:01:31.613Z
@Michael_1812 Discovering that the Moon has this coincidence led to the discovery of tidal locking, it was assumed there was a cause, not just coincidenc.
Barmar
CC BY-SA 4.0
2023-03-26T19:26:08.583Z
@Barmar The question was about Oort. There, no coincidence of the kind would originate from solar tides.
Michael_1812
CC BY-SA 4.0
2023-05-02T00:01:14.563Z
I believe this last edit of @Michael_1812 is not correct. Tidal force is the cube - just like James initially writes. If you want to look at the amplitude, you might end up with a different power, but he writes force, so cube is right.
planetmaker
CC BY-SA 4.0
2023-05-02T00:12:46.420Z
That's also quite intuitively clear: Tidal acceleration is the change of the gravitational acceleration caused by the tide-causing body. Considering only axial acceleration $F_{tide} = \frac{\mathrm{d}}{\mathrm{d}r} F_{grav} = \frac{\mathrm{d}}{\mathrm{d}r} \frac{GM}{r^2}$
planetmaker
CC BY-SA 4.0
2023-05-02T03:48:51.727Z
@planetmaker The tidal force is *not* equal to $dF_{grav}/dr$, because it matters how the body deforms in response to one or another component of the disturbing potential   In eqn (5) from https://iopscience.iop.org/article/10.1088/0004-637X/764/1/26 , the leading, quadrupole part corresponds to $l=2$.  It scales as the inverse cube of the distance. Now, how will the deformable body react to it? Eqn (6) says that the corresponding, quadrupole part of the additional tidal potential of the deformed body will scale as the negative 6th power of the distance.
Michael_1812
CC BY-SA 4.0
2023-05-02T03:58:45.400Z
@planetmaker ...which means that I indeed made a mistake  -- though a different one. Just as the polar tidal torque, so the tidal force scales as the inverse 6th power, not 7th as I erroneously wrote. Will now have to correct it. Thank you!
Michael_1812
CC BY-SA 4.0
2023-05-03T11:24:10.393Z
Now that I've long ago edited the title/question, I am still surprised you won't edit your answer to reflect that, nor why this is attracting the majority of votes?
Rabbi Kaii
CC BY-SA 4.0
2023-03-24T06:13:27.470Z
2023-05-02T20:31:45.830Z
2023-05-02T20:31:45.830Z
James K
<p>Suppose you orbit with period <span class="math-container">$P$</span>, and have a angular velocity <span class="math-container">$\omega$</span> randomly chosen (let's start with everything aligned in a plane orthogonal to spin too). Then to keep your face pointed towards the sun for <span class="math-container">$k$</span> periods you need to have <span class="math-container">$\omega=1/P\pm 1/Pk$</span>. So a million-year orbit that achieves accidental locking for a billion years <span class="math-container">$k=1000$</span> need <span class="math-container">$\omega$</span> in <span class="math-container">$10^{-6} \pm 10^{-9}$</span> per year.</p>
<p>How likely is this? <a href="https://www.lpi.usra.edu/books/AsteroidsIII/pdf/3016.pdf" rel="noreferrer">Typical asteroids rotate about 3 times per Earth-day, and for large asteroids the distribution looks Maxwellian</a>. If we assume the Oort cloud is similar, then the probability of being in that tiny range becomes... a number so small my math program just cancels it down to 0 (close to <span class="math-container">$x\approx 0$</span>, the Maxwell distribution <span class="math-container">$\propto x^2$</span>, so these tiny rotation rates get turned into even tinier probabilities, and then the CDF calculation that uses some fancy functions just rounds it to zero). Even if we have trillions of Oort objects this is not enough to get one rotating in the right way by chance.</p>

CC BY-SA 4.0
2023-03-24T15:03:32.433Z
A bit back of the envelope but I'd say pretty conclusive, thanks for taking the time
Rabbi Kaii
CC BY-SA 4.0
2023-03-24T15:02:19.560Z
2023-03-24T15:02:19.560Z
Anders Sandberg
<p>Is the Oort cloud spherical enough and are there enough objects in it that it is likely there is at least one in there that it has shown the same face to the sun for billions of years and will continue to do so for billions of years? Just based on chance spin axis and rate?</p>
<p>Don't know why I am wondering what the most distant &quot;tidally locked&quot; body is... I like using things like this to give some context to the scale of things...</p>

<tidal-locking><oort-cloud>
"Tidally locked" Oort Cloud object
Rabbi Kaii
# "Tidally locked" Oort Cloud object

(Asked by: username_0 on 2023-03-24)

Is the Oort cloud spherical enough and are there enough objects in it that it is likely there is at least one in there that it has shown the same face to the sun for billions of years and will continue to do so for billions of years? Just based on chance spin axis and rate?

Don't know why I am wondering what the most distant "tidally locked" body is... I like using things like this to give some context to the scale of things...

**Tags:** tidal-locking, oort-cloud


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## 2 answers

### Answer by username_5 on 2023-03-24. Score: 9 | Accepted answer

Suppose you orbit with period $P$, and have a angular velocity $\omega$ randomly chosen (let's start with everything aligned in a plane orthogonal to spin too). Then to keep your face pointed towards the sun for $k$ periods you need to have $\omega=1/P\pm 1/Pk$. So a million-year orbit that achieves accidental locking for a billion years $k=1000$ need $\omega$ in $10^{-6} \pm 10^{-9}$ per year.

How likely is this? [Typical asteroids rotate about 3 times per Earth-day, and for large asteroids the distribution looks Maxwellian](https://www.lpi.usra.edu/books/AsteroidsIII/pdf/3016.pdf). If we assume the Oort cloud is similar, then the probability of being in that tiny range becomes... a number so small my math program just cancels it down to 0 (close to $x\approx 0$, the Maxwell distribution $\propto x^2$, so these tiny rotation rates get turned into even tinier probabilities, and then the CDF calculation that uses some fancy functions just rounds it to zero). Even if we have trillions of Oort objects this is not enough to get one rotating in the right way by chance.

> **Comments:**
> **username_0**: A bit back of the envelope but I'd say pretty conclusive, thanks for taking the time  

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### Answer by username_1 on 2023-03-24. Score: 20

The question in the title and the body are different. There are no tidally locked objects in the Oort cloud. The sun's tidal forces at that distance are not sufficient to lock an object's rotation. Tidal locking doesn't mean 'co-incidentally' having the same rate of rotation.

Tidal forces decrease with the cube of distance; tidal torques decrease with the sixth power of distance, so are much less significant relative to gravity, which decreases with the square of distance, in the Oort cloud. Moreover, the time for tidal lock to occur increases with the sixth power of the distance. Tidal lock also affects large objects more than small ones (in general) and most Oort cloud object are small.

There are other things that can cause changes in rotation rates: gravitational interactions in close passes with other objects, small impacts, the [YORP effect](https://en.wikipedia.org/wiki/YORP_effect). All of these are rare or weak in the Oort cloud, but stronger than tidal locking, and so would tend to cause any object which by chance had a synchronous rotation rate to break that co-incidence.

> **Comments:**
> **username_0**: Thanks. Ive put the title in quotes in order to avoid this misunderstanding  
> **username_2**: The other answer calculates that the probability that rotation rate = revolution rate is infinitessimally small. So if this coincidence occurs, it's almost certainly an indication of tidal locking.  
> **username_3**: James, this is a good answer. Still, you may want to correct a slip of the pen in it. Tidal torques (which are actually responsible for tidal capture) decrease not with the cube but with the inverse 6-th power of distance, see eqns (23) and (36) in http://iopscience.iop.org/0004-637X/764/1/26  [It is the perturbing potential $W$ that falls of as the inverse cube of distance.]  
> **username_3**: @username_2 I would disagree with your statement. In my opinion, if a coincidence occurs it will be a coincidence only. As username_1 rightly noticed, there are too many interactions in the Oort cloud that are much stronger than tidal forces. In fact, the tidal forces caused by the Sun are totally irrelevant there, falling of as the ratio of an object's size to its distance from the Sun, taken to the minus six power.  
> **username_2**: @username_3 I'm not saying that there's actually any tidal locking. I'm saying that since gravity isn't strong enough, and the probability is so small, it won't happen at all.  
> **username_3**: @username_2 But this is how I understood you. You said: "if this coincidence occurs, it's almost certainly an indication of tidal locking."  
> **username_2**: @username_3 Discovering that the Moon has this coincidence led to the discovery of tidal locking, it was assumed there was a cause, not just coincidenc.  
> **username_3**: @username_2 The question was about Oort. There, no coincidence of the kind would originate from solar tides.  
> **username_4**: I believe this last edit of @username_3 is not correct. Tidal force is the cube - just like James initially writes. If you want to look at the amplitude, you might end up with a different power, but he writes force, so cube is right.  
> **username_4**: That's also quite intuitively clear: Tidal acceleration is the change of the gravitational acceleration caused by the tide-causing body. Considering only axial acceleration $F_{tide} = \frac{\mathrm{d}}{\mathrm{d}r} F_{grav} = \frac{\mathrm{d}}{\mathrm{d}r} \frac{GM}{r^2}$  
> **username_3**: @username_4 The tidal force is *not* equal to $dF_{grav}/dr$, because it matters how the body deforms in response to one or another component of the disturbing potential   In eqn (5) from https://iopscience.iop.org/article/10.1088/0004-637X/764/1/26 , the leading, quadrupole part corresponds to $l=2$.  It scales as the inverse cube of the distance. Now, how will the deformable body react to it? Eqn (6) says that the corresponding, quadrupole part of the additional tidal potential of the deformed body will scale as the negative 6th power of the distance.  
> **username_3**: @username_4 ...which means that I indeed made a mistake  -- though a different one. Just as the polar tidal torque, so the tidal force scales as the inverse 6th power, not 7th as I erroneously wrote. Will now have to correct it. Thank you!  
> **username_0**: Now that I've long ago edited the title/question, I am still surprised you won't edit your answer to reflect that, nor why this is attracting the majority of votes?  

---

