<p>There isn't a simple way to do this. It depends on what you mean by &quot;percentage of fluidity&quot;. A viscous body could start to break up at 2.44, but the process would be very slow.</p>
<p>The &quot;rigid&quot; calculation assumes that the body will stay perfectly spherical, but otherwise has no tensile strength, so will break up as soon as tides from the planet exceed gravity.</p>
<p>The &quot;fluid&quot; approximation assumes that the body will immediately flow into an &quot;oval&quot; or &quot;teardrop&quot; shape under the effect of the tide, this shape is longer than a sphere, and so more greatly affected by the tides, so it breaks up at a larger ratio.</p>
<p>A body with tensile strength can orbit within the roche limit. The space station does! It is held together by the strength of the materials, not by gravity.</p>
<p>The rigid calculation can be done exactly, but a body that is rigid but with no tensile strength is impossible. The fluid calculation is based on approximation or numerical simulation. A body that consisted of a perfectly rigid shell, with a fluid interior would not deform, and so could be treated like a rigid body.  A body with a fluid &quot;ocean&quot; surrounding a rigid core, would deform, so would behave more like a fluid moon.</p>
<p>Actual bodies like the moon may respond in complex ways to tidal forces, but at the scale of the moon, even solid rock will &quot;flow&quot; or at least &quot;deform&quot;, so a value closer to 2.44 is more likely to be correct. But what would actually happen as the moon approached and passed this point would be complex and dynamic. It wouldn't break up instantaneously as it reached 2.44, but the process of break-up could begin then. The process would be faster if the moon came closer.</p>

CC BY-SA 4.0
2024-03-01T18:33:06.623Z
muchas gracias for the info.
I don't suppose you could direct me towards code for those numerical calculations? I've tried to find some, but I think I'm using all the wrong search terms.
Joshua Knicely
CC BY-SA 4.0
2024-03-01T06:36:23.907Z
2024-03-01T06:36:23.907Z
James K
<p>I want to calculate the Roche limit for the Earth/Moon system. The general equation given is:</p>
<p><span class="math-container">$$d = C \cdot r_p \cdot (\rho_p / \rho_s)^{1/3}$$</span></p>
<p>where <span class="math-container">$d$</span> is the roche limit, <span class="math-container">$r_p$</span> is the radius of the primary (Earth), <span class="math-container">$\rho_p$</span> and <span class="math-container">$\rho_s$</span> are the densities of the primary and secondary respectively.</p>
<p><span class="math-container">$C$</span> is a constant that is <span class="math-container">$1.26$</span> for rigid bodies and <span class="math-container">$2.44$</span> for completely fluid bodies.</p>
<p>Do I just linearly vary <span class="math-container">$C$</span> between <span class="math-container">$1.26$</span> and <span class="math-container">$2.44$</span> for different percentages of fluidity? For example, for a half fluid system, <span class="math-container">$C = (2.44+1.26)/2$</span> ? That seems way too simple. Am I missing something?</p>

<roche-limit>
how do I calculate the Roche limit for a body that is only partially fluid?
Joshua Knicely
# how do I calculate the Roche limit for a body that is only partially fluid?

(Asked by: username_0 on 2024-02-28)

I want to calculate the Roche limit for the Earth/Moon system. The general equation given is:

$$d = C \cdot r\_p \cdot (\rho\_p / \rho\_s)^{1/3}$$

where $d$ is the roche limit, $r\_p$ is the radius of the primary (Earth), $\rho\_p$ and $\rho\_s$ are the densities of the primary and secondary respectively.

$C$ is a constant that is $1.26$ for rigid bodies and $2.44$ for completely fluid bodies.

Do I just linearly vary $C$ between $1.26$ and $2.44$ for different percentages of fluidity? For example, for a half fluid system, $C = (2.44+1.26)/2$ ? That seems way too simple. Am I missing something?

**Tags:** roche-limit


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## 1 answers

### Answer by username_1 on 2024-03-01. Score: 1

There isn't a simple way to do this. It depends on what you mean by "percentage of fluidity". A viscous body could start to break up at 2.44, but the process would be very slow.

The "rigid" calculation assumes that the body will stay perfectly spherical, but otherwise has no tensile strength, so will break up as soon as tides from the planet exceed gravity.

The "fluid" approximation assumes that the body will immediately flow into an "oval" or "teardrop" shape under the effect of the tide, this shape is longer than a sphere, and so more greatly affected by the tides, so it breaks up at a larger ratio.

A body with tensile strength can orbit within the roche limit. The space station does! It is held together by the strength of the materials, not by gravity.

The rigid calculation can be done exactly, but a body that is rigid but with no tensile strength is impossible. The fluid calculation is based on approximation or numerical simulation. A body that consisted of a perfectly rigid shell, with a fluid interior would not deform, and so could be treated like a rigid body. A body with a fluid "ocean" surrounding a rigid core, would deform, so would behave more like a fluid moon.

Actual bodies like the moon may respond in complex ways to tidal forces, but at the scale of the moon, even solid rock will "flow" or at least "deform", so a value closer to 2.44 is more likely to be correct. But what would actually happen as the moon approached and passed this point would be complex and dynamic. It wouldn't break up instantaneously as it reached 2.44, but the process of break-up could begin then. The process would be faster if the moon came closer.

> **Comments:**
> **username_0**: muchas gracias for the info. I don't suppose you could direct me towards code for those numerical calculations? I've tried to find some, but I think I'm using all the wrong search terms.  

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