Do all stars have an Oort cloud or is it a rare occurence?

  • Do all stars have an Oort cloud like ours that will be filled with comets and other objects? If not, why are they not around every star?


  • HDE 226868

    HDE 226868 Correct answer

    8 years ago

    Awesome question, especially since we know so little of the answer.



    Nobody knows for sure how the Oort Cloud formed - I'll put that out there right now - but the current hypothesis is that it was originally part of the Sun's protoplanetary disk. All of the ice and rock coalesced into small bodies - proto-comets, if you will. While these bodies were much closer in to the Sun than they are today, they were tossed far out by gravitational interactions with the gas giants. Other interstellar comets could also have been captured by the Sun, adding to the population.



    So why is the Oort Cloud spherical? After all, the protoplanetary disk was just a flat disk. Why were the orbits of the objects perturbed? Well, the Oort Cloud objects are only loosely bound to the Sun - relatively, that is. They can be influenced by passing stars or other objects. It appears that galactic-scale tidal forces, combined with the influence of passing stars, molded the Cloud into its current spherical shape.



    So what does this all tell us? Well, we know other stars have protoplanetary disks, right? Some also have exoplanets - gas giants like Jupiter. They are also subject to tidal forces and the passing of nearby stars. So, theoretically, there's no reason why other stars shouldn't have Oort Clouds.



    So can we find them? The answer is, most likely, no. Here's why. According to Wikipedia,




    The outer Oort cloud may have trillions of objects larger than 1 km (0.62 mi), and billions with solar system absolute magnitudes brighter than 11




    An absolute magnitude of a solar system object of 11 is very dim. Now, the object's apparent magnitude is how it would look from a given distance; the absolute magnitude is how it looks from a distance of 1 AU (in the case of Solar System objects, this quantity is denoted $H$). Oort-cloud objects are 2,000 - 50,000 (or more) AU away-- so these objects, to us in the same solar system, have an apparent magnitude much fainter than 11.



    The point of that poorly-explained interlude is that these objects are faint. Very faint. And objects in Oort Clouds around other stars would appear even fainter. Using the distance modulus, we can calculate the apparent magnitude of an object if the distance to that object and its absolute magnitude are known:



    $$m-M=5(\log_{10}d-1)$$



    (from here)



    where $m$ is apparent magnitude, $M$ is a scaling of $H$ normally used for stars, and $d$ is the distance in AU.



    Given an Oort Cloud object $x$ light-years away, you can figure out how bright (or dim) it would appear, given that 1 light-year is 63241 AU. Try this with the distances of nearby stars, and you'll realize how dim objects in these stars' Oort Clouds would be.



    As a final note: We don't know for sure if other Oort Clouds exist. From what I've been able to find, we don't have sufficiently powerful telescopes to observe these hypothetical Clouds, and so we don't (and may never) know if they exist.



    I hope this helps.



    This paper was instrumental in this answer. Start at page 38 for the relevant information. This page, too, has some good information.



    As I found from a link from an answer to this question on Physics, we've found Kuiper-Belt-like disks around other stars. This means it is certainly plausible for these stars to have Oort Clouds, too. And exocomets have been detected, which is another good sign.


    Concerning the assertion that passing stars may disturb Oort cloud objects, this question is in need of elucidation!

    I am pretty sure that absolute magnitude for solar system bodies is calculated at 1 AU (rather than 10 pc; it also says so in the note on Wikipedia; and it's like a factor of 2000000, which in magnitude terms is about 16 magnitudes dimmer)! Because while the 11 absolute magnitude you mentioned is low, there are STARS with that brightness, and there cannot be Oort cloud objects brighter than stars. This is especially important because we cannot even properly 'see' our own Oort cloud. So seeing other stars' Oort cloud (and the need to explain it in detail) is out of question.

    Also, I get a feeling that the dominant balancing forces at the distance are (a) radiation pressure and (b) gravity. Since both are radial, balanced and weak, the Oort cloud has spherical symmetry. And it's also true that the galactic tidal forces do play a role in making it spherical and also in imparting angular momentum to it.

    @Takku Where's the note on the Wikipedia page? (I'll change my answer if it's wrong)

    Note #14. Right next to the term absolute magnitudes.

    @Takku Which page? The absolute magnitude page starts off with the 10-parsecs definition.

    No no. The Oort cloud page. The sentence which you referred to, there's a superscript note #14, which says that here it is the 1 AU definition (just do a Ctrl F for magnitude and you'll land on one of the results). Also on the absolute magnitude page, there is a separate section for solar system bodies for which the absolute magnitude is defined differently, and Oort cloud objects count as solar system bodies.

    I read a graduate student saying he's on a project to detect Solar System Oort cloud objects in Gaia data thanks to microlensing. Could exo-Oort cloud objects be found by microlensing too? It's such a magic technique that I have no sense of what the limits are.

    @LocalFluff I don't know. I haven't read much on the subject that doesn't focus on applications to exoplanets.

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Content dated before 7/24/2021 11:53 AM