<p>The <a href="https://en.wikipedia.org/wiki/Roche_lobe" rel="noreferrer">Roche lobe</a> is a gravitational potential well of a two-body configuration. <a href="https://comp-astrophys-cosmol.springeropen.com/articles/10.1186/s40668-015-0008-8" rel="noreferrer">Technically</a>, it is the potential energy per unit test mass which is orbiting the center-of-mass of a binary-star system at the same rate as the two stars. It is often depicted with equipotential surfaces and the 5 <a href="https://en.wikipedia.org/wiki/Lagrange_point" rel="noreferrer">Lagrange points</a>. The Roche lobe includes gravitational and centripetal energy, so it's derivative is the acceleration on the test mass. Thus, it is useful, among other things, for determining when a particle is gravitationally bound to a star in a binary system. You can think of the Roche limit of a body in a binary as the limit beyond which particles are no longer &quot;part&quot; of that body.</p>
<p>The <a href="https://en.wikipedia.org/wiki/Hill_sphere" rel="noreferrer">Hill sphere</a> is an approximation of the perturbations of the larger body in a binary system onto the smaller (i.e. less massive) body in the binary system. So, you can see how it is distinct, necessarily, from the Roche lobe. You can consider the Hill sphere as the limit where gravitational perturbations from a more massive body become weak compared to the self gravity of a less massive body.</p>
<p>Here is a figure that nicely depicts the difference of these boundaries from the wiki article about the Hill sphere:</p>
<p><a href="https://i.sstatic.net/YeyvJ.png" rel="noreferrer"><img src="https://i.sstatic.net/YeyvJ.png" alt="enter image description here" /></a></p>
<p>The <a href="https://en.wikipedia.org/wiki/Sphere_of_influence_(astrodynamics)" rel="noreferrer">sphere of influence</a> is an approximation where you only consider gravity of nearby bodies, and not of more massive ones that are further away. As shown by the figure below, the SoI is a decent approximation to the Hill sphere for Solar System bodies that are not too close to the Sun.</p>
<p><a href="https://i.sstatic.net/Vfj8L.png" rel="noreferrer"><img src="https://i.sstatic.net/Vfj8L.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2021-12-24T11:47:29.030Z
Hi, thanks for the response! I'm afraid I don't quite get your explanation. You say that the Hill sphere is "the limit where gravitational perturbations from a more massive body become weak compared to the self gravity of a less massive body", but why is this not the same as the boundary of the region where a unit mass would be gravitationally bound to the less massive body (the Roche lobe)?
YiFan
CC BY-SA 4.0
2021-12-24T11:49:04.083Z
By the way, the picture you attached shows the Roche limit and not the Roche lobe, and unless I'm mistaken these are distinct, right?
YiFan
CC BY-SA 4.0
2021-12-24T12:11:34.343Z
@YiFan IF I understand your question correctly, it is because a test mass does not have self-gravity. The Roche limit, or Roche radius, is computed from the Roche lobe which is a gravitational potential of two self-gravitating bodies.
Daddy Kropotkin
CC BY-SA 4.0
2021-12-24T13:01:36.103Z
The Roche lobe defines the region around a star in a binary star system where a test mass would be gravitationally bound to the star, right? Are you saying this is the same as the Roche limit, the distance at which an asteroid (or whatever body only held together by its own gravity) would disintegrate, the Rouche limit? I don't see why these two are equivalent.
YiFan
CC BY-SA 4.0
2021-12-24T13:06:34.403Z
Consider a binary star system, with masses say M and m, and we *don't* assume M>>m. Now I put a test mass m' (e.g. an asteroid) much smaller than m and M near the system. Am I right to say that (i) the Roche lobe defines the region around m (or M) where m' would be gravitationally bound to m; and (ii) the Hill sphere defines the region where m' would orbit m? If yes, then why are (i) and (ii) different? If not, why?
YiFan
CC BY-SA 4.0
2021-12-24T20:36:48.317Z
@YiFan I know that you want a simple answer, but one must be careful to not oversimplify. The phrase "gravitationally bound" is not synonymous with "orbits," necessarily. For example, a particle on a hyperbolic orbit is not gravitationally bound to the object that it orbits. In the first figure in my answer, notice how the Hill sphere radius is larger than the Roche limit for the three bodies - the sun, the earth and the moon - because a test particle can be gravitationally perturbed without being gravitationally bound.
Daddy Kropotkin
CC BY-SA 4.0
2021-12-25T02:28:10.483Z
Interesting. Just to be sure, are you saying that both (i) and (ii) from my previous comment are correct, the issue is that being gravitationally bound to m is not the same as orbiting m, because of e.g. hyperbolic orbits as you mentioned? So, the notion of being gravitationally bound to an object is strictly stronger than that of simply orbiting an object (which is to be mainly "gravitationally perturbed" by it)?
YiFan
CC BY-SA 4.0
2021-12-25T22:54:11.730Z
Yes, I think that is an accurate statement. The Roche lobe defines the limit where a body can lose mass to mass transfer, like in a stellar binary, which is much closer to the surface of the star than is the Hill radius
Daddy Kropotkin
CC BY-SA 4.0
2021-12-26T02:51:07.413Z
Okay, thank you for the help. :)
YiFan
CC BY-SA 4.0
2021-12-26T09:33:15.280Z
My pleasure!!!!
Daddy Kropotkin
CC BY-SA 4.0
2021-12-24T11:21:37.893Z
2021-12-24T11:21:37.893Z
Daddy Kropotkin
<p>I am just beginning to look into this topic, so apologies if there are any striking misconceptions in the following.</p>
<p>From Wikipedia, the Roche lobe is &quot;the region around a star in a binary system within which orbiting material is gravitationally bound to that star&quot;. So, to my understanding, this is the region around the star where matter is able to &quot;remain near&quot; the star and not get &quot;sucked away&quot;, for example by the other star in the system. Roche lobe overflow seems to be the term used to describe the case where matter falls out of the Roche lobe of one star and hence is &quot;lost&quot; by the star.</p>
<p>On the other hand, the Roche sphere (or Hill sphere) seems to be a confusingly similar concept! Wikipedia again states, &quot;to be retained by a planet, a moon must have an orbit that lies within the planet's Hill sphere&quot;. But isn't this the same as the interpretation of the Roche lobe? The Wikipedia article also mentions stability in the face of perturbations from the gravity of a larger body, but I don't really understand what this means.</p>
<p>Could someone explain in simple terms the difference between the two? Thanks in advance.</p>
<p>(Note: I have seen the question <a href="https://astronomy.stackexchange.com/q/6348/30945">What is the difference between Sphere of Influence and Hill sphere?</a>, but I think that is a distinct question since the SoI is not the same as the Roche lobe; please correct me if I am mistaken!)</p>

CC BY-SA 4.0
2021-12-25T01:35:42.807Z
...and a "[Roche motel](https://en.wikipedia.org/wiki/Roach_Motel)"
uhoh
<orbit><gravity><astrophysics><orbital-mechanics>
What's the difference between the Roche lobe and Roche sphere?
YiFan
# What's the difference between the Roche lobe and Roche sphere?

(Asked by: username_0 on 2021-12-24)

I am just beginning to look into this topic, so apologies if there are any striking misconceptions in the following.

From Wikipedia, the Roche lobe is "the region around a star in a binary system within which orbiting material is gravitationally bound to that star". So, to my understanding, this is the region around the star where matter is able to "remain near" the star and not get "sucked away", for example by the other star in the system. Roche lobe overflow seems to be the term used to describe the case where matter falls out of the Roche lobe of one star and hence is "lost" by the star.

On the other hand, the Roche sphere (or Hill sphere) seems to be a confusingly similar concept! Wikipedia again states, "to be retained by a planet, a moon must have an orbit that lies within the planet's Hill sphere". But isn't this the same as the interpretation of the Roche lobe? The Wikipedia article also mentions stability in the face of perturbations from the gravity of a larger body, but I don't really understand what this means.

Could someone explain in simple terms the difference between the two? Thanks in advance.

(Note: I have seen the question [What is the difference between Sphere of Influence and Hill sphere?](https://astronomy.stackexchange.com/q/6348/30945), but I think that is a distinct question since the SoI is not the same as the Roche lobe; please correct me if I am mistaken!)

**Tags:** orbit, gravity, astrophysics, orbital-mechanics

> **Comments:**
> **username_1**: ...and a "[Roche motel](https://en.wikipedia.org/wiki/Roach_Motel)"  

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## 1 answers

### Answer by username_2 on 2021-12-24. Score: 5 | Accepted answer

The [Roche lobe](https://en.wikipedia.org/wiki/Roche_lobe) is a gravitational potential well of a two-body configuration. [Technically](https://comp-astrophys-cosmol.springeropen.com/articles/10.1186/s40668-015-0008-8), it is the potential energy per unit test mass which is orbiting the center-of-mass of a binary-star system at the same rate as the two stars. It is often depicted with equipotential surfaces and the 5 [Lagrange points](https://en.wikipedia.org/wiki/Lagrange_point). The Roche lobe includes gravitational and centripetal energy, so it's derivative is the acceleration on the test mass. Thus, it is useful, among other things, for determining when a particle is gravitationally bound to a star in a binary system. You can think of the Roche limit of a body in a binary as the limit beyond which particles are no longer "part" of that body.

The [Hill sphere](https://en.wikipedia.org/wiki/Hill_sphere) is an approximation of the perturbations of the larger body in a binary system onto the smaller (i.e. less massive) body in the binary system. So, you can see how it is distinct, necessarily, from the Roche lobe. You can consider the Hill sphere as the limit where gravitational perturbations from a more massive body become weak compared to the self gravity of a less massive body.

Here is a figure that nicely depicts the difference of these boundaries from the wiki article about the Hill sphere:

[![enter image description here](https://i.sstatic.net/YeyvJ.png)](https://i.sstatic.net/YeyvJ.png)

The [sphere of influence](https://en.wikipedia.org/wiki/Sphere_of_influence_(astrodynamics)) is an approximation where you only consider gravity of nearby bodies, and not of more massive ones that are further away. As shown by the figure below, the SoI is a decent approximation to the Hill sphere for Solar System bodies that are not too close to the Sun.

[![enter image description here](https://i.sstatic.net/Vfj8L.png)](https://i.sstatic.net/Vfj8L.png)

> **Comments:**
> **username_0**: Hi, thanks for the response! I'm afraid I don't quite get your explanation. You say that the Hill sphere is "the limit where gravitational perturbations from a more massive body become weak compared to the self gravity of a less massive body", but why is this not the same as the boundary of the region where a unit mass would be gravitationally bound to the less massive body (the Roche lobe)?  
> **username_0**: By the way, the picture you attached shows the Roche limit and not the Roche lobe, and unless I'm mistaken these are distinct, right?  
> **username_2**: @username_0 IF I understand your question correctly, it is because a test mass does not have self-gravity. The Roche limit, or Roche radius, is computed from the Roche lobe which is a gravitational potential of two self-gravitating bodies.  
> **username_0**: The Roche lobe defines the region around a star in a binary star system where a test mass would be gravitationally bound to the star, right? Are you saying this is the same as the Roche limit, the distance at which an asteroid (or whatever body only held together by its own gravity) would disintegrate, the Rouche limit? I don't see why these two are equivalent.  
> **username_0**: Consider a binary star system, with masses say M and m, and we *don't* assume M>>m. Now I put a test mass m' (e.g. an asteroid) much smaller than m and M near the system. Am I right to say that (i) the Roche lobe defines the region around m (or M) where m' would be gravitationally bound to m; and (ii) the Hill sphere defines the region where m' would orbit m? If yes, then why are (i) and (ii) different? If not, why?  
> **username_2**: @username_0 I know that you want a simple answer, but one must be careful to not oversimplify. The phrase "gravitationally bound" is not synonymous with "orbits," necessarily. For example, a particle on a hyperbolic orbit is not gravitationally bound to the object that it orbits. In the first figure in my answer, notice how the Hill sphere radius is larger than the Roche limit for the three bodies - the sun, the earth and the moon - because a test particle can be gravitationally perturbed without being gravitationally bound.  
> **username_0**: Interesting. Just to be sure, are you saying that both (i) and (ii) from my previous comment are correct, the issue is that being gravitationally bound to m is not the same as orbiting m, because of e.g. hyperbolic orbits as you mentioned? So, the notion of being gravitationally bound to an object is strictly stronger than that of simply orbiting an object (which is to be mainly "gravitationally perturbed" by it)?  
> **username_2**: Yes, I think that is an accurate statement. The Roche lobe defines the limit where a body can lose mass to mass transfer, like in a stellar binary, which is much closer to the surface of the star than is the Hill radius  
> **username_0**: Okay, thank you for the help. :)  
> **username_2**: My pleasure!!!!  

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