4
$\begingroup$

Astronomers define the Solar System as the distance under the influence of gravity from the Sun.

I read that the Oort cloud is probably the farthest object that surrounds the sun. The sphere of this icy cloud is thought to surround the Sun at up to a distance of 50,000 AU, making its total diameter nearly two light years.

How far can the sun's gravity extend to hold celestial bodies around it? The Oort cloud may be the known farthest body but its edge doesn't mark the end of the sun's effective gravitational pull.

Improve this question
$\endgroup$
4
  • $\begingroup$ Looks like someone had a quick go at this, might be a start point?: forum.cosmoquest.org/… . It's a very rough estimate though; we have many stars of varying masses nearby. $\endgroup$
    – user10106
    Commented May 4, 2018 at 10:07
  • $\begingroup$ If you rephrase this question as "where in the galaxy/universe is the Sun's gravity stronger than the gravity of any other star", you could use a Voronoi diagram to come up w/ a mathematically precise answer. As others note, it's 2.25ish ly in the direction of Alpha Centauri, but could be bigger in other directions. Someone should do the math. $\endgroup$
    – user21
    Commented May 5, 2018 at 16:47
  • 1
    $\begingroup$ @barrycarter The Alpha Centauri trinary is a bit over 2.1 solar masses. That puts the "trojan point" if I can call it that, closer to 1.75 lya not 2.25. Small nitpick. $\endgroup$
    – userLTK
    Commented May 5, 2018 at 18:34
  • $\begingroup$ @userLTK Good point. I vaguely remember reading that Alpha Centauri is about the same size as the sun. While that's true, I forgot that it's a triple star. $\endgroup$
    – user21
    Commented May 6, 2018 at 16:27

1 Answer 1

Reset to default
7
$\begingroup$

The suns gravitational pull extends unlimitedly far and, if the universe were otherwise empty, a body could orbit it (very slowly) at unlimited distance.

However, the universe is NOT otherwise empty. The nearest other stars are about 4 light years from the Sun at the moment, so a body more than two lightyears away will typically feel equally strong attraction from other stars as from the Sun, and so will not really orbit the sun.

That's a bit of an over-simplification, because (a) the sun is not surrounded by other stars 4 l.y. away in all directions (here is a chart of nearby stars) so a body might orbit in such a way that it didn't pass between the sun and any of its closest neighbours; (b) not all stars have the same mass, many are smaller than the Sun but some (eg Sirius) much bigger, so their gravitational influences vary and (c) the stars are all moving, so that from time to time one comes quite a bit closer, disrupting orbits.

Improve this answer
$\endgroup$
5
  • $\begingroup$ -1 This doesn't really answer the question "How far out can the sun keep celestial objects revolving?" nor does it give any information where the Oort cloud is thought to end. All this does is say it's an interesting problem and here are some things to think about if you were going to try to answer it. $\endgroup$
    – uhoh
    Commented May 5, 2018 at 8:06
  • $\begingroup$ I'd claim that my answer is 2ly with considerable caveats, but basically you're right. The question would need some refining -- crucially, how long do you want the orbit to be stable for? At 1ly, orbital period is 10s of millions of years, so there stellar neighbourhood will change quite a lot over a single orbit. I guess you could try and compute the probability of completeing a single orbit without major perturbations from another star, as a function of the size of the orbit, or something like that. $\endgroup$
    – Steve Linton
    Commented May 5, 2018 at 9:51
  • 2
    $\begingroup$ The answer needs to include Galactic tides , since AFAIK know, this is what sets the limit at around 100,000 au. $\endgroup$
    – ProfRob
    Commented May 5, 2018 at 19:26
  • $\begingroup$ @RobJeffries minor point, but 100,000 au is more than 1 light year. I think you meant to say it was closer than 1 light year. $\endgroup$
    – userLTK
    Commented Oct 8, 2018 at 9:46
  • $\begingroup$ @userltk Yes, it's "about" 100,000 au, but not spherically symmetric. $\endgroup$
    – ProfRob
    Commented Oct 8, 2018 at 11:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .