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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...

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2 Answers 2

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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. 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.

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    $\begingroup$ Thanks. Ive put the title in quotes in order to avoid this misunderstanding $\endgroup$
    – Rabbi Kaii
    Mar 24, 2023 at 9:53
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    $\begingroup$ 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. $\endgroup$
    – Barmar
    Mar 24, 2023 at 16:31
  • $\begingroup$ 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 iopscience.iop.org/0004-637X/764/1/26 [It is the perturbing potential $W$ that falls of as the inverse cube of distance.] $\endgroup$ Mar 26, 2023 at 1:55
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    $\begingroup$ @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. $\endgroup$ Mar 26, 2023 at 2:00
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    $\begingroup$ @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. $\endgroup$
    – Barmar
    Mar 26, 2023 at 17:01
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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. 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.

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    $\begingroup$ A bit back of the envelope but I'd say pretty conclusive, thanks for taking the time $\endgroup$
    – Rabbi Kaii
    Mar 24, 2023 at 15:03

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