<p>You've answered you own question. If there are cohesive forces beyond that of simple self-gravitation, then objects can survive intact inside the self-gravitating Roche limit - as does every solid item on the surface of the Earth for example.</p>
<p>Saturn's rings are made of ice, not rocks.</p>
<p>The tensile strength of ice is about <span class="math-container">$10^6$</span> N m<span class="math-container">$^{-2}$</span>.</p>
<p>For a self-gravitating body, the Roche limiting radius is about
<span class="math-container">$$ R_L \simeq 1.4R_p\left(\frac{\rho_p}{\rho_s}\right)^{1/3},$$</span>
where <span class="math-container">$R_p$</span> is the radius of the planet and <span class="math-container">$\rho_p$</span> and <span class="math-container">$\rho_s$</span> are the densities of the planet and satellite respectively. This is for a rigid object. The leading constant would be bit higher at about 2.4, for a fluid.</p>
<p>Thus the self-gravitating Roche limit <strong>does not</strong> depend on the size of the satellite, it just depends weakly on its density.</p>
<p>However this limit becomes much larger if the cohesive forces holding a body together exceed the gravitational forces. This becomes true for smaller objects because the gravity at the surface of an object <em>does</em> depend on its size, whereas the cohesive forces do not.</p>
<p>A critical size will depend on the geometry, density and tensile strength. Roughly speaking, gravity and the cohesive forces will be equivalent for an satellite of radius
<span class="math-container">$$ r \sim \frac{T}{ \rho_s  g},$$</span>
where <span class="math-container">$T$</span> is the tensile strength and <span class="math-container">$g$</span> the surface gravity. But
<span class="math-container">$$ g \simeq GM_s/r^2 = \frac{4\pi G}{3}\rho_s r$$</span>
Thus
<span class="math-container">$$r = \left(\frac{3T}{4\pi G}\right)^{1/2}\rho_s^{-1}$$</span></p>
<p>For ice of density <span class="math-container">$1000$</span> kg m<span class="math-container">$^{-3}$</span> and <span class="math-container">$T\sim 10^6$</span> N m<span class="math-container">$^{-2}$</span>, then <span class="math-container">$r = 60$</span> km.</p>
<p>This rough and ready analysis suggests that ice objects smaller than 60 km could survive inside the self-gravitating Roche limit.</p>
<p>FYI, the innermost named moon of Saturn, <a href="https://en.m.wikipedia.org/wiki/Pan_(moon)" rel="nofollow noreferrer">Pan,</a> has a radius of about 15 km, an icy composition and is inside the Roche limit. There are other satellites in the ring system (Epithemius, Pandora) with radii of 30-60 km. Whereas the largest inner moon, Mimas with a radius of 200 km, is in an orbit at just over 3 Saturnian radii and outside the Roche limit.</p>

CC BY-SA 4.0
2021-01-08T00:09:08.480Z
I am struggling with the example of solid items on the surface of the Earth.  If the Earth had no atmosphere, a lightly glued card castle could presumably stand up on the surface. But wouldn't it be pulled apart in an orbit within the Roche boundary?
Connor Garcia
CC BY-SA 4.0
2021-01-08T00:17:09.193Z
Whether an object is on the surface or in low Earth orbit orbit does not affect the tidal forces.  All objects at the surface of the Earth are inside the Roche limit. The cohesive forces in them are easily sufficient to exceed the tidal force. @ConnorGarcia And the answer is no, it wouldn't be pulled apart, as indeed pieces of card on the international space station are not.
ProfRob
CC BY-SA 4.0
2021-01-07T12:01:42.843Z
2021-01-08T00:38:47.397Z
2021-01-08T00:38:47.397Z
ProfRob
<h2>tl;dr: Yes, but they must be small.</h2>
<p>The reason why is because large bodies, like major moons, break apart at a certain limit from their hosts, using the formula for the Roche Limit: <span class="math-container">$$d=r\Big{(}2\dfrac{M}{m}\Big{)}^\frac{1}{3}$$</span></p>
<p><span class="math-container">$d$</span> is the Roche Limit, <span class="math-container">$r$</span> is the radius of the satellite, and <span class="math-container">$M$</span> and <span class="math-container">$m$</span> are the masses of the host and satellite objects, respectively.</p>
<p>If <span class="math-container">$M$</span> remains constant, then as <span class="math-container">$m$</span> gets smaller, the Roche Limit gets smaller, while if <span class="math-container">$m$</span> gets bigger, so does the Roche Limit. The reason why Saturn's rings cannot coalesce into a moon is because they are inside Saturn's Roche Limit for a massive body. But, the ringlets cannot fall apart because the same driving equation prevents the tiny rocks from disintegrating from tidal forces.</p>
<p>I hope this helps.</p>

CC BY-SA 4.0
2021-01-07T14:14:25.367Z
Thanks! I had not known that that Roche Limit also depends on the mass of the satellite. I had just assumed (foolishly) that it is constant.
Astro Noob
CC BY-SA 4.0
2021-01-07T14:36:27.867Z
@AstroNoob If you'd like check this Wikipedia page about the Roche Limit: https://en.wikipedia.org/wiki/Roche_limit

If you think my answer was helpful, you can click on the check mark below the up and down arrows on the left.
WarpPrime
CC BY-SA 4.0
2021-01-07T23:27:40.983Z
@AstroNoob I don't think this is correct. As you can see from the formula, the Roche limit in fact only depends on the  density of the orbiting object (to the power of 1/3). So it doesn't matter if you have a large or small lump of something unless their densities are very different (and to first order, they aren't).
ProfRob
CC BY-SA 4.0
2021-01-07T23:29:26.970Z
@ProfRob Wasn't there two formulae: the one mentioned here and the other one relating densities on Wikipedia? It seems like both are correct.
WarpPrime
CC BY-SA 4.0
2021-01-07T23:30:13.843Z
How do you propose to change the mass of your satellite without changing its radius?
ProfRob
CC BY-SA 4.0
2021-01-07T23:35:05.490Z
@Profrob Increase its density, I suppose.
WarpPrime
CC BY-SA 4.0
2021-01-07T23:44:18.810Z
And thus the Roche limit does not depend on size. It depends very weakly on density. So it matters not what size the object is.
ProfRob
CC BY-SA 4.0
2021-01-07T13:53:22.863Z
2021-01-07T13:53:22.863Z
WarpPrime
<p>Saturn's rings are inside its own Roche Limit - the limit beyond which bodies start disintegrating due to the tidal forces of the parent body. But as every rock in the ring can be regarded as a satellite of Saturn, why are they not disintegrated to smaller constituents? Also, is it possible for a body to hold itself together even within the Roche Limit (maybe due to some cohesive forces holding it together)? I thought it may be the reason for the rocks in Saturn's rings holding themselves together- but then again, I have no reason to believe it may be the case any way. Can someone provide an estimate for what the cohesive forces may be for a metallic rock as a fraction of the internal gravitational forces in it?</p>

<planet><tidal-forces><planetary-ring><roche-limit><moonlet>
Can moons exist inside a planet's roche limit?
Astro Noob
# Can moons exist inside a planet's roche limit?

(Asked by: username_0 on 2021-01-07)

Saturn's rings are inside its own Roche Limit - the limit beyond which bodies start disintegrating due to the tidal forces of the parent body. But as every rock in the ring can be regarded as a satellite of Saturn, why are they not disintegrated to smaller constituents? Also, is it possible for a body to hold itself together even within the Roche Limit (maybe due to some cohesive forces holding it together)? I thought it may be the reason for the rocks in Saturn's rings holding themselves together- but then again, I have no reason to believe it may be the case any way. Can someone provide an estimate for what the cohesive forces may be for a metallic rock as a fraction of the internal gravitational forces in it?

**Tags:** planet, tidal-forces, planetary-ring, roche-limit, moonlet

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## 2 answers

### Answer by username_1 on 2021-01-07. Score: 3 | Accepted answer

You've answered you own question. If there are cohesive forces beyond that of simple self-gravitation, then objects can survive intact inside the self-gravitating Roche limit - as does every solid item on the surface of the Earth for example.

Saturn's rings are made of ice, not rocks.

The tensile strength of ice is about $10^6$ N m$^{-2}$.

For a self-gravitating body, the Roche limiting radius is about
$$ R\_L \simeq 1.4R\_p\left(\frac{\rho\_p}{\rho\_s}\right)^{1/3},$$
where $R\_p$ is the radius of the planet and $\rho\_p$ and $\rho\_s$ are the densities of the planet and satellite respectively. This is for a rigid object. The leading constant would be bit higher at about 2.4, for a fluid.

Thus the self-gravitating Roche limit **does not** depend on the size of the satellite, it just depends weakly on its density.

However this limit becomes much larger if the cohesive forces holding a body together exceed the gravitational forces. This becomes true for smaller objects because the gravity at the surface of an object *does* depend on its size, whereas the cohesive forces do not.

A critical size will depend on the geometry, density and tensile strength. Roughly speaking, gravity and the cohesive forces will be equivalent for an satellite of radius
$$ r \sim \frac{T}{ \rho\_s g},$$
where $T$ is the tensile strength and $g$ the surface gravity. But
$$ g \simeq GM\_s/r^2 = \frac{4\pi G}{3}\rho\_s r$$
Thus
$$r = \left(\frac{3T}{4\pi G}\right)^{1/2}\rho\_s^{-1}$$

For ice of density $1000$ kg m$^{-3}$ and $T\sim 10^6$ N m$^{-2}$, then $r = 60$ km.

This rough and ready analysis suggests that ice objects smaller than 60 km could survive inside the self-gravitating Roche limit.

FYI, the innermost named moon of Saturn, [Pan,](https://en.m.wikipedia.org/wiki/Pan_(moon)) has a radius of about 15 km, an icy composition and is inside the Roche limit. There are other satellites in the ring system (Epithemius, Pandora) with radii of 30-60 km. Whereas the largest inner moon, Mimas with a radius of 200 km, is in an orbit at just over 3 Saturnian radii and outside the Roche limit.

> **Comments:**
> **username_2**: I am struggling with the example of solid items on the surface of the Earth.  If the Earth had no atmosphere, a lightly glued card castle could presumably stand up on the surface. But wouldn't it be pulled apart in an orbit within the Roche boundary?  
> **username_1**: Whether an object is on the surface or in low Earth orbit orbit does not affect the tidal forces.  All objects at the surface of the Earth are inside the Roche limit. The cohesive forces in them are easily sufficient to exceed the tidal force. @ConnorGarcia And the answer is no, it wouldn't be pulled apart, as indeed pieces of card on the international space station are not.  

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### Answer by username_3 on 2021-01-07. Score: 1

## tl;dr: Yes, but they must be small.

The reason why is because large bodies, like major moons, break apart at a certain limit from their hosts, using the formula for the Roche Limit: $$d=r\Big{(}2\dfrac{M}{m}\Big{)}^\frac{1}{3}$$

$d$ is the Roche Limit, $r$ is the radius of the satellite, and $M$ and $m$ are the masses of the host and satellite objects, respectively.

If $M$ remains constant, then as $m$ gets smaller, the Roche Limit gets smaller, while if $m$ gets bigger, so does the Roche Limit. The reason why Saturn's rings cannot coalesce into a moon is because they are inside Saturn's Roche Limit for a massive body. But, the ringlets cannot fall apart because the same driving equation prevents the tiny rocks from disintegrating from tidal forces.

I hope this helps.

> **Comments:**
> **username_0**: Thanks! I had not known that that Roche Limit also depends on the mass of the satellite. I had just assumed (foolishly) that it is constant.  
> **username_3**: @AstroNoob If you'd like check this Wikipedia page about the Roche Limit: https://en.wikipedia.org/wiki/Roche_limit  If you think my answer was helpful, you can click on the check mark below the up and down arrows on the left.  
> **username_1**: @AstroNoob I don't think this is correct. As you can see from the formula, the Roche limit in fact only depends on the  density of the orbiting object (to the power of 1/3). So it doesn't matter if you have a large or small lump of something unless their densities are very different (and to first order, they aren't).  
> **username_3**: @ProfRob Wasn't there two formulae: the one mentioned here and the other one relating densities on Wikipedia? It seems like both are correct.  
> **username_1**: How do you propose to change the mass of your satellite without changing its radius?  
> **username_3**: @Profrob Increase its density, I suppose.  
> **username_1**: And thus the Roche limit does not depend on size. It depends very weakly on density. So it matters not what size the object is.  

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