<p>Here's my answer. I'll try to make it as comprehensive as possible.</p>

<p>It's pretty hard to define the edge of the <a href="http://en.wikipedia.org/wiki/Solar_System" rel="noreferrer">Solar System</a>. Most people would probably define it as where objects are no longer gravitationally bound to the Sun. That just shifts the question a little, though: Where is that dividing line? To try to answer this, I'll go over the regions of the Solar System.</p>

<p>The first region is the domain of the <a href="http://en.wikipedia.org/wiki/Terrestrial_planet" rel="noreferrer">inner planets</a> - basically everything from the <a href="http://en.wikipedia.org/wiki/Asteroid_belt" rel="noreferrer">asteroid belt</a> inwards. It is comprised of Mars, Earth, Venus, Mercury, their moons, and all the smaller objects that surround them. The inner Solar System is very rocky, as one can imagine. The terrestrial planets are primarily made of rock, as are the asteroids and the inner planets' moons.</p>

<p>The second region is the domain of the <a href="http://en.wikipedia.org/wiki/Gas_giant" rel="noreferrer">gas giants</a>. It consists of Jupiter, Saturn, Uranus, Neptune, their moons, ring systems, and assorted smaller bodies, such as Trojan asteroids. The gas giants had a big influence on the Solar System when it was first formed, pulling in chunks of rocks, grabbing moons, and possibly stabilizing or de-stabilizing orbits. Some may have migrated outwards (as per the <a href="http://en.wikipedia.org/wiki/Nice_model" rel="noreferrer">Nice model</a>), but their orbits are currently stable. The gas giants are made largely of gases, but it is thought they have solid or molten cores. The composition of their moons is familiar - more like objects in the inner Solar System.</p>

<p>Next up is the <a href="http://en.wikipedia.org/wiki/Kuiper_belt" rel="noreferrer">Kuiper Belt</a>. It's sometimes introduced as a cousin of the asteroid belt, but that's not accurate. The bodies that make up the Kuiper Belt are chunks of rock and ice. Notable examples of Kuiper Belt bodies and/or trans-Neptunian objects are the dwarf planets Pluto, Sedna, Makemake and Haumea. There are also lots of smaller objects, including some short-period comets (although these are more properly part of the lesser-known "scattered disk"). While there have been theories for years about another planet out there, it is not considered likely. The Belt extends from 30 to 50 AU.</p>

<p>Further out still is the <a href="http://en.wikipedia.org/wiki/Oort_cloud" rel="noreferrer">Oort Cloud</a>, named after Jan Oort. Observations of objects in the Oort Cloud are extremely difficult, if not impossible, so its existence has not yet been verified. It is populated by long-period comets and smaller objects. These are also composed of rock and ice. The Oort Cloud is thought to extend up to an incredible 50,000 AU. While the other regions so far mentioned are roughly in planes, the Oort Cloud is spherical.</p>

<p>Some consider the far edge Oort Cloud to be the edge of the Solar System, because the majority of the mass of the Solar System is within it, but the boundary between the Solar System and interstellar space is actually thought to be within its inner reaches: the heliopause. <em>This</em> is generally accepted as the Solar System's boundary because it is where the solar wind meets the interstellar medium. This is often placed at 121 AU - which is where Voyager 1 passed through in 2013. The heliopause is the far boundary of the <a href="http://en.wikipedia.org/wiki/Heliosphere" rel="noreferrer">heliosphere</a>, beyond which the interstellar medium takes control. Inside "layers" are bounded by the termination shock and the heliosheath.</p>

<p>In summary, while the Solar System is made of many regions, the heliopause is considered to be its outer boundary.</p>

<p>Once again, I welcome any and all input regarding this question and answer.</p>

CC BY-SA 3.0
2014-10-04T16:00:03.917Z
2015-08-26T22:48:49.530Z
2015-08-26T22:48:49.530Z
HDE 226868
<p>Whenever I see this question discussed, it seems that the heliopause, or some variation thereof, is given as an answer -- and then it's mentioned that the Oort Cloud extends beyond it.</p>

<p>A more correct answer, therefore, should be that it ends at that distance at which objects are, for all practical purposes, no longer bound to the solar system barycenter. This is usually defined by the <a href="http://en.m.wikipedia.org/wiki/Hill_sphere" rel="noreferrer">Hill Sphere</a>, which approximates the gravitational sphere of influence.</p>

<blockquote>
  <p>One simple view of the extent of the Solar System is the Hill sphere of the Sun with respect to local stars and the galactic nucleus.(1)</p>
</blockquote>

<p>This extends out to two hundred thirty thousand AU, about 3.6 light years. Again, not a wall. According to (1) <a href="http://adsabs.harvard.edu/full/1965SvA.....8..787C" rel="noreferrer">Cherbatov (1965)</a>, the radii of the gravitational spheres of the sun may be subdivided into:  </p>

<ul>
<li><p>Sphere of attraction out to 4500 AU (sun's attraction > galactic center's attraction),  </p></li>
<li><p>Sphere of action 60,000 AU (more convenient to use sun as central body &amp; galactic center as pertubating body in orbital calculations), and finally  </p></li>
<li><p>Hill sphere 230,000 AU (object must orbit within this limit to be retained by the Sun).</p></li>
</ul>

CC BY-SA 4.0
2020-05-19T05:30:14.487Z
I guess this would depend on whether you're more interested in objects with large mass and small charge (like rocks) or whether you're more interested in objects with low mass/charge ratio, like protons and electrons. At the edge of our solar system there is far more total mass as plasma than as rocks, and far more electromagnetic energy than gravitational energy. I suspect that is the reason why most astronomers and astrophysicists define the solar system boundary along electromagnetic lines, although of course there are good arguments for using the Hill Sphere as the boundary.
craq
CC BY-SA 4.0
2024-06-16T05:42:16.727Z
Worth noting that [Wikipedia](https://en.wikipedia.org/wiki/Hill_sphere) says orbits further than 1/3 to 1/2 of the Hill radius are thought to be unstable. So, although the Sun's gravitational influence dominates within 230 kAU, we should expect the Sun to be unable, as a practical matter, to hold onto anything further away than ~115 kAU for more than a few orbits.
Tom
CC BY-SA 3.0
2014-11-20T17:23:25.337Z
2015-08-26T23:37:47.887Z
2015-08-26T23:37:47.887Z
Jerard Puckett
<p>I believe NASA is stating it is not only when solar wind but gravitational pull shifts... That's not to say the sun has no pull or solar wind but that the influence of the sun is now less than the surrounding environment. To put it simply, when the sun is no longer winning the tug of war.</p>

CC BY-SA 3.0
2015-08-26T23:57:29.150Z
What source(s) do you have for this?
HDE 226868
CC BY-SA 3.0
2015-08-26T06:57:27.417Z
2015-08-26T06:57:27.417Z
R. Tachoir
<p>I consider the edge of <em>any</em> solar system the distance beyond which the central star(s) don't provide for enough light to create a reasonable &quot;day&quot; on the star-facing side. In other words, if you're on a celestial body that has night on all sides (e.g. on an Oort cloud object), you may no longer consider being <em>in</em> that solar system, you're just on a body that is gravitationally attracted/influenced by a certain star. That border would be around an apparent magnitude of minus 12 caused by the central star.</p>
<p>As for this system, the edge would be around 1000 a.u. from the Sun, this is where I set its boundary, which is just beyond the outermost planet Sedna's aphelion. Beyond 1000 a.u.  it is interstellar space.</p>

CC BY-SA 4.0
2021-05-23T12:26:51.007Z
Magnitude -12 is very bright (full moon bright).
franz_lyre
CC BY-SA 4.0
2021-05-23T14:35:09.927Z
@f_n_lyre Right, and has a full moon ever made the sky appear blue as in daytime?
Giovanni
CC BY-SA 4.0
2021-05-24T03:09:11.000Z
I misunderstood. I thought you meant that the celestial object at the border would be magnitude -12 from our point of view. Do you mean that at your border of 1000 A.U. the star would be magnitude -12?
franz_lyre
CC BY-SA 4.0
2021-05-24T04:32:27.900Z
@f_n_lyre Yes, I mean the central star's apparent magnitude at that distance.
Giovanni
CC BY-SA 4.0
2021-05-24T20:40:40.517Z
@Giovanni taken by itself "has a full moon ever made the sky appear blue" is an interesting question of its own! It certainly makes the sky bright enough that even if you cover it with your hand the dimmer stars are no longer easy to see, and if you hold your hand out against the sky (when behind a wall blocking the moon) you can see a dark outline of your hand against a brighter sky, but I don't think I've ever seen it bright enough to register as blue. of course the albedo of the Moon is actually pretty low in the blue so it won't be the same color as that for the Sun.
uhoh
CC BY-SA 4.0
2021-05-25T04:50:21.510Z
@uhoh About three years ago I've seen an extremely bright full moon. I don't remember if any stars were visible, the brightest probably were but there was none in the vicinity of the Moon. The sky however was still black. An extremely light black (that's hard to explain) but still the color of the night. I know that bright meteors and nukes can make the night sky appear blue.
Giovanni
CC BY-SA 4.0
2024-06-16T05:52:30.097Z
"Reasonable day" does not seem like an objective measure. The choice of -12 magnitude seems arbitrary, and forces the patently absurd conclusion that _Neptune_ is not part of the solar system. Perhaps there are contexts where this definition is useful, but astronomy is not one of them.
Tom
CC BY-SA 4.0
2021-05-23T10:33:04.487Z
2021-05-24T04:33:25.183Z
2021-05-24T04:33:25.183Z
Giovanni
<p>The Sun is roughly 4 light-years away from the closest star system, the Alpha Centauri system. The planets in our Solar System, however, aren't even close to that far away from the Sun. Where does our Solar System end? Is the edge considered to be the orbit of Neptune, the Kuiper Belt, the Oort Cloud, or something else?</p>
<p>Note: <a href="https://physics.stackexchange.com/questions/25784/how-big-is-the-solar-system">this</a> question on Physics SE is similar, but the answers posted here go in different directions.</p>

CC BY-SA 4.0
2018-11-01T18:06:35.167Z
If we could estimate the frequency of close encounters of distance $D$ by stars of mass $M$ and the probability of such an encounter ejecting an object of  orbital radius $R$ then we could average over billions of years come up with a statement like: "Objects at $R \gt R_L$ have an $80$% chance of being ejected while objects at $R \lt R_L$ have an $80$% chance of not being ejected. Has anything like that been done?
Keith McClary
CC BY-SA 3.0
2014-10-05T10:57:57.767Z
Obligatory [xkcd](http://xkcd.com/1189/).
Sparhawk
CC BY-SA 3.0
2015-07-17T16:55:45.197Z
Not sure about the votes, but it gets a +1 from me. Great question.
fantasia
<the-sun><solar-system><oort-cloud><kuiper-belt>
Where does the Solar System end?
HDE 226868
# Where does the Solar System end?

(Asked by: username_0 on 2014-10-04)

The Sun is roughly 4 light-years away from the closest star system, the Alpha Centauri system. The planets in our Solar System, however, aren't even close to that far away from the Sun. Where does our Solar System end? Is the edge considered to be the orbit of Neptune, the Kuiper Belt, the Oort Cloud, or something else?

Note: [this](https://physics.stackexchange.com/questions/25784/how-big-is-the-solar-system) question on Physics SE is similar, but the answers posted here go in different directions.

**Tags:** the-sun, solar-system, oort-cloud, kuiper-belt
> **Comments:**
> **username_2**: Obligatory [xkcd](http://xkcd.com/1189/).  
> **username_3**: Not sure about the votes, but it gets a +1 from me. Great question.  
> **username_1**: If we could estimate the frequency of close encounters of distance $D$ by stars of mass $M$ and the probability of such an encounter ejecting an object of  orbital radius $R$ then we could average over billions of years come up with a statement like: "Objects at $R \gt R_L$ have an $80$% chance of being ejected while objects at $R \lt R_L$ have an $80$% chance of not being ejected. Has anything like that been done?  

---

## 4 answers

### Answer by username_0 on 2014-10-04. Score: 21

Here's my answer. I'll try to make it as comprehensive as possible.

It's pretty hard to define the edge of the [Solar System](http://en.wikipedia.org/wiki/Solar_System). Most people would probably define it as where objects are no longer gravitationally bound to the Sun. That just shifts the question a little, though: Where is that dividing line? To try to answer this, I'll go over the regions of the Solar System.

The first region is the domain of the [inner planets](http://en.wikipedia.org/wiki/Terrestrial_planet) - basically everything from the [asteroid belt](http://en.wikipedia.org/wiki/Asteroid_belt) inwards. It is comprised of Mars, Earth, Venus, Mercury, their moons, and all the smaller objects that surround them. The inner Solar System is very rocky, as one can imagine. The terrestrial planets are primarily made of rock, as are the asteroids and the inner planets' moons.

The second region is the domain of the [gas giants](http://en.wikipedia.org/wiki/Gas_giant). It consists of Jupiter, Saturn, Uranus, Neptune, their moons, ring systems, and assorted smaller bodies, such as Trojan asteroids. The gas giants had a big influence on the Solar System when it was first formed, pulling in chunks of rocks, grabbing moons, and possibly stabilizing or de-stabilizing orbits. Some may have migrated outwards (as per the [Nice model](http://en.wikipedia.org/wiki/Nice_model)), but their orbits are currently stable. The gas giants are made largely of gases, but it is thought they have solid or molten cores. The composition of their moons is familiar - more like objects in the inner Solar System.

Next up is the [Kuiper Belt](http://en.wikipedia.org/wiki/Kuiper_belt). It's sometimes introduced as a cousin of the asteroid belt, but that's not accurate. The bodies that make up the Kuiper Belt are chunks of rock and ice. Notable examples of Kuiper Belt bodies and/or trans-Neptunian objects are the dwarf planets Pluto, Sedna, Makemake and Haumea. There are also lots of smaller objects, including some short-period comets (although these are more properly part of the lesser-known "scattered disk"). While there have been theories for years about another planet out there, it is not considered likely. The Belt extends from 30 to 50 AU.

Further out still is the [Oort Cloud](http://en.wikipedia.org/wiki/Oort_cloud), named after Jan Oort. Observations of objects in the Oort Cloud are extremely difficult, if not impossible, so its existence has not yet been verified. It is populated by long-period comets and smaller objects. These are also composed of rock and ice. The Oort Cloud is thought to extend up to an incredible 50,000 AU. While the other regions so far mentioned are roughly in planes, the Oort Cloud is spherical.

Some consider the far edge Oort Cloud to be the edge of the Solar System, because the majority of the mass of the Solar System is within it, but the boundary between the Solar System and interstellar space is actually thought to be within its inner reaches: the heliopause. *This* is generally accepted as the Solar System's boundary because it is where the solar wind meets the interstellar medium. This is often placed at 121 AU - which is where Voyager 1 passed through in 2013. The heliopause is the far boundary of the [heliosphere](http://en.wikipedia.org/wiki/Heliosphere), beyond which the interstellar medium takes control. Inside "layers" are bounded by the termination shock and the heliosheath.

In summary, while the Solar System is made of many regions, the heliopause is considered to be its outer boundary.

Once again, I welcome any and all input regarding this question and answer.


---

### Answer by username_4 on 2014-11-20. Score: 13

Whenever I see this question discussed, it seems that the heliopause, or some variation thereof, is given as an answer -- and then it's mentioned that the Oort Cloud extends beyond it.

A more correct answer, therefore, should be that it ends at that distance at which objects are, for all practical purposes, no longer bound to the solar system barycenter. This is usually defined by the [Hill Sphere](http://en.m.wikipedia.org/wiki/Hill_sphere), which approximates the gravitational sphere of influence.

> One simple view of the extent of the Solar System is the Hill sphere of the Sun with respect to local stars and the galactic nucleus.(1)

This extends out to two hundred thirty thousand AU, about 3.6 light years. Again, not a wall. According to (1) [Cherbatov (1965)](http://adsabs.harvard.edu/full/1965SvA.....8..787C), the radii of the gravitational spheres of the sun may be subdivided into:

* Sphere of attraction out to 4500 AU (sun's attraction > galactic center's attraction),
* Sphere of action 60,000 AU (more convenient to use sun as central body & galactic center as pertubating body in orbital calculations), and finally
* Hill sphere 230,000 AU (object must orbit within this limit to be retained by the Sun).

> **Comments:**
> **username_5**: I guess this would depend on whether you're more interested in objects with large mass and small charge (like rocks) or whether you're more interested in objects with low mass/charge ratio, like protons and electrons. At the edge of our solar system there is far more total mass as plasma than as rocks, and far more electromagnetic energy than gravitational energy. I suspect that is the reason why most astronomers and astrophysicists define the solar system boundary along electromagnetic lines, although of course there are good arguments for using the Hill Sphere as the boundary.  
> **username_6**: Worth noting that [Wikipedia](https://en.wikipedia.org/wiki/Hill_sphere) says orbits further than 1/3 to 1/2 of the Hill radius are thought to be unstable. So, although the Sun's gravitational influence dominates within 230 kAU, we should expect the Sun to be unable, as a practical matter, to hold onto anything further away than ~115 kAU for more than a few orbits.  

---

### Answer by username_7 on 2015-08-26. Score: 1

I believe NASA is stating it is not only when solar wind but gravitational pull shifts... That's not to say the sun has no pull or solar wind but that the influence of the sun is now less than the surrounding environment. To put it simply, when the sun is no longer winning the tug of war.

> **Comments:**
> **username_0**: What source(s) do you have for this?  

---

### Answer by username_8 on 2021-05-23. Score: 0

I consider the edge of *any* solar system the distance beyond which the central star(s) don't provide for enough light to create a reasonable "day" on the star-facing side. In other words, if you're on a celestial body that has night on all sides (e.g. on an Oort cloud object), you may no longer consider being *in* that solar system, you're just on a body that is gravitationally attracted/influenced by a certain star. That border would be around an apparent magnitude of minus 12 caused by the central star.

As for this system, the edge would be around 1000 a.u. from the Sun, this is where I set its boundary, which is just beyond the outermost planet Sedna's aphelion. Beyond 1000 a.u. it is interstellar space.

> **Comments:**
> **username_9**: Magnitude -12 is very bright (full moon bright).  
> **username_8**: @f_n_lyre Right, and has a full moon ever made the sky appear blue as in daytime?  
> **username_9**: I misunderstood. I thought you meant that the celestial object at the border would be magnitude -12 from our point of view. Do you mean that at your border of 1000 A.U. the star would be magnitude -12?  
> **username_8**: @f_n_lyre Yes, I mean the central star's apparent magnitude at that distance.  
> **username_10**: @Giovanni taken by itself "has a full moon ever made the sky appear blue" is an interesting question of its own! It certainly makes the sky bright enough that even if you cover it with your hand the dimmer stars are no longer easy to see, and if you hold your hand out against the sky (when behind a wall blocking the moon) you can see a dark outline of your hand against a brighter sky, but I don't think I've ever seen it bright enough to register as blue. of course the albedo of the Moon is actually pretty low in the blue so it won't be the same color as that for the Sun.  
> **username_8**: @uhoh About three years ago I've seen an extremely bright full moon. I don't remember if any stars were visible, the brightest probably were but there was none in the vicinity of the Moon. The sky however was still black. An extremely light black (that's hard to explain) but still the color of the night. I know that bright meteors and nukes can make the night sky appear blue.  
> **username_6**: "Reasonable day" does not seem like an objective measure. The choice of -12 magnitude seems arbitrary, and forces the patently absurd conclusion that _Neptune_ is not part of the solar system. Perhaps there are contexts where this definition is useful, but astronomy is not one of them.  

---

