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I have reached a confusion when searching for derivations of the Roche Limit, which differ in value and explanation.

All equations are given of the form:

$ P = r_{M} \sqrt[3]{f\frac{\rho_{M}}{\rho_{S}}} $

where

$P$ : Roche Limit

$r_M$ : radius of large body

$\rho_{M}$ : density of the large body

$\rho_{S}$ : density of the small satellite

$f$ : what I am calling the `Roche Factor'


My issue is the value of $f$, which seems to vary depending on where one looks. I am interested in its value for a satellite that is a pile of rubble (or sand) with no elasticity or internal tension.

I would appreciate any corrections to my understanding of the justification for the various values, which I will outline below.


$f = 2$

From the Wikipedia page, the justification is that, in the accelerating frame of the satellite, there are two opposing forces. There is the tidal force, $F_T$ resulting for the differing field strength across satellite, and there is the gravitational attraction $F_G$ between all the pieces of the satellite.

The derivation states that, for a small particle to remain on the surface of the satellite, $F_G > F_T$.

Here, $f=2$ essentially because there is a 2 in the tidal force equation.

Note: I haven't seen it mentioned, but although $g$ diminishes linearly inside the satellite, the tidal force will also diminish by the same factor. Hence, passing the Roche limit will cause the whole rubble pile to disintegrate simultaneously, not stating from the outer edge as one might presume.


$f = 3$

As above, but an extra term is included since moons are not only in free-fall but also will be rotating upon their own axes.

In the accelerating frame of a particle on the surface of the satellite, there will also be a centrifugal force $F_C$, outwards from the centre of the satellite. If the satellite is tidally locked, then $F_C$ will always point towards the large body. Hence $F_G > F_T + F_C$. Seemingly coincidentally, $F_C = \frac{1}{2}F_T$, giving $f=3$

A rotating body will be more fragile due to the centrifugal forces, and hence its Roche limit occurs further from the large body.


$f = 16$

For example

Instead of considering a small mass on the surface of the satellite, these derivations touch together two spheres of equal radii and mass, and then find the distance where $F_G > F_T$. Since the distance separating the centres of the two spheres is $2r$, this produces an extra factor of 2 in the tidal force, and an extra factor of $\frac{1}{2^2}$ in the gravitational force. Hence, $f=2\times2\times2^2=16$.

This seems an arbitrary and probably incorrect derivation that is attempting to ape the value that Roche derived (below).


$f = 2.2423^3$ or = $2.2455^3$

Again from the Wikipedia page, it seems that these values are the limit when the pieces of rubble in the satellite become infinitesimal small. When the rubble is large, is stays approximately circular until its Roche limit. When the rubble is very small, the tidal force causes the satellite to become an ellipsoid well before the Roche limit. The values of f in this case cannot be obtained algebraically.

The former value was obtained by Roche, but does not consider centrifugal force due to axial rotation, which the latter values corrects.


So how does one choose the correct value?

It seems that $f=2$ is more applicable to incoming projectiles that are rubble piles (such as Shoemaker-Levy 9 striking Jupiter).

Whereas $f=3$ is better when providing limits for rubble pile moons (but is this contradicted by the fact that Phobos and Saturn's inner moons are said to be inside their Roche limits?).

$f = 2.2455^3$ is useful for explaining orbits that arose during the birth of the solar system, when everything was liquid, however these objects have since acquired tensions from the solid rock and now $f$ is far lower. So they could now inhabit regions that were prohibited during their formation.

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  • $\begingroup$ I would think that Phobos is not just a rubble pile and has some structural strength. $\endgroup$ – fibonatic Jul 27 '16 at 1:20

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