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

<p>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.</p>

<p>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 href="http://www.atlasoftheuniverse.com/12lys.html" rel="noreferrer">a chart of nearby stars</a>) 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.</p>

CC BY-SA 4.0
2018-05-05T08:06:57.143Z
`-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*.
uhoh
CC BY-SA 4.0
2018-05-05T09:51:30.337Z
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.
Steve Linton
CC BY-SA 4.0
2018-05-05T19:26:51.787Z
The answer needs to include Galactic tides , since AFAIK know, this is what sets the limit at around 100,000 au.
ProfRob
CC BY-SA 4.0
2018-10-08T09:46:38.293Z
@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.
userLTK
CC BY-SA 4.0
2018-10-08T11:21:49.483Z
@userltk Yes, it's "about" 100,000 au, but not spherically symmetric.
ProfRob
CC BY-SA 4.0
2018-05-04T10:10:18.070Z
2018-05-04T10:10:18.070Z
Steve Linton
<p>Astronomers define the Solar System as the distance under the influence of gravity from the Sun.</p>

<p>I read that the <a href="https://en.wikipedia.org/wiki/Oort_cloud" rel="nofollow noreferrer">Oort cloud</a> 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 <strong>50,000 AU</strong>, making its total diameter nearly two light years.</p>

<p>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.</p>

CC BY-SA 4.0
2018-05-05T18:34:31.130Z
@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.
userLTK
<orbit><gravity><oort-cloud>
How far out can the sun keep celestial objects revolving?
defalt
# How far out can the sun keep celestial objects revolving?

(Asked by: username_0 on 2018-05-04)

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

I read that the [Oort cloud](https://en.wikipedia.org/wiki/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.

**Tags:** orbit, gravity, oort-cloud
> **Comments:**
> **username_1**: @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.  

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## 1 answers

### Answer by username_2 on 2018-05-04. Score: 7 | Accepted answer

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](http://www.atlasoftheuniverse.com/12lys.html)) 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.

> **Comments:**
> **username_3**: `-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*.  
> **username_2**: 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.  
> **username_4**: The answer needs to include Galactic tides , since AFAIK know, this is what sets the limit at around 100,000 au.  
> **username_1**: @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.  
> **username_4**: @userltk Yes, it's "about" 100,000 au, but not spherically symmetric.  

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