The
Roche Limit ( ), sometimes referred to as the
Roche Radius, is the distance within which a
celestial body, held together only by its own
gravity, will disintegrate due to a second celestial
body's
tidal forces exceeding the first
body's gravitational self-attraction. Inside the Roche limit,
orbiting material will tend to
disperse and form rings, while outside the limit, material will
tend to
coalesce.
The term
is named after Édouard Roche, the
French astronomer who first
calculated this theoretical limit in 1848.
Typically, the Roche limit applies to a
satellite disintegrating due to tidal forces
induced by its
primary, the body about which it
orbits. Some real satellites, both
natural and
artificial, can orbit within their
Roche limits because they are held together by forces other than
gravitation.
Jupiter's moon
Metis and
Saturn's moon
Pan are examples of such satellites,
which hold together because of their
tensile strength. In extreme cases, objects
resting on the surface of such a satellite could actually be lifted
away by tidal forces. A weaker satellite, such as a
comet, could be broken up when it passes within its
Roche limit.
Since tidal forces overwhelm gravity within the Roche limit, no
large satellite can coalesce out of smaller particles within that
limit. Indeed, almost all known
planetary
rings are located within their Roche limit, Saturn's
E-Ring and
Phoebe ring being notable
exceptions. They could either be remnants from the planet's
proto-planetary
accretion disc that
failed to coalesce into moonlets, or conversely have formed when a
moon passed within its Roche limit and broke apart.
It is also worth considering that the Roche limit is not the only
factor that causes comets to break apart. Splitting by
thermal stress, internal
gas pressure and rotational splitting are a
more likely way for a comet to split under stress.
Determining the Roche limit
The Roche limit depends on the rigidity of the satellite. At one
extreme, a completely rigid satellite will maintain its shape until
tidal forces break it apart. At the other extreme, a highly fluid
satellite gradually deforms leading to increased tidal forces,
causing the satellite to elongate, further compounding the tidal
forces and causing it to break apart more readily. Most real
satellites are somewhere between these two extremes, with internal
friction,
viscosity, and tensile strength rendering the
satellite neither perfectly rigid nor perfectly fluid.
Rigid satellites
To calculate the
rigid body Roche limit for a spherical
satellite, the cause of the rigidity is neglected but the body is
assumed to maintain its
spherical shape while
being held together only by its own self-gravity. Other effects are
also neglected, such as tidal deformation of the primary, rotation
of the satellite, and its irregular shape. These somewhat
unrealistic assumptions greatly simplify the Roche limit
calculation.
The Roche limit, d, for a rigid spherical satellite orbiting a
spherical primary is
- d = R\left( 2\;\frac {\rho_M} {\rho_m} \right)^{\frac{1}{3}}
,
where R is the
radius of the primary, \rho_M
is the
density of the primary, and \rho_m is
the density of the satellite.
If the satellite is more than twice as dense as the primary, as can
easily be the case for a rocky moon orbiting a gas giant, then the
Roche limit will be inside the primary and hence not
relevant.
Derivation of the formula
Derivation of the Roche limit.
In order to determine the Roche limit, we consider a small mass u
on the surface of the satellite closest to the primary. There are
two forces on this mass u: the gravitational pull towards the
satellite and the gravitational pull towards the primary. Since the
satellite is already in orbital
free fall
around the primary, the
tidal force is
the only relevant term of the gravitational attraction of the
primary.
The gravitational pull F_G on the mass u towards the satellite with
mass m and radius r can be expressed according to
Newton's law of
gravitation.
- F_G = \frac{Gmu}{r^2}
The
tidal force F_T on the mass u
towards the primary with radius R and mass M, at a distance d
between the centers of the two bodies, can be expressed
approximately as
- F_T = \frac{2GMur}{d^3}.
To obtain this approximation, find the difference in the primary's
gravitational pull on the center of the satellite and on the edge
of the satellite closest to the primary:
- F_T = \frac{GMu}{(d-r)^2}-\frac{GMu}{d^2}
- F_T = GMu\frac{d^2-(d-r)^2}{d^2(d-r)^2}
- F_T = GMu\frac{2dr-r^2}{d^4-2d^3r+r^2d^2}
In the approximation where r<<R and="" Rr^2 in the numerator
and every term with r in the denominator goes to zero, which gives
us:
- F_T = GMu\frac{2dr}{d^4}
- F_T = \frac{2GMur}{d^3}
The Roche limit is reached when the gravitational force and the
tidal force balance each other out.
- F_G = F_T \;
or
- \frac{Gmu}{r^2} = \frac{2GMur}{d^3},
which gives the Roche limit, d, as
- d = r \left( 2\;\frac{M}{m} \right)^{\frac{1}{3}} .
However, we don't really want the radius of the satellite to appear
in the expression for the limit, so we re-write this in terms of
densities.
For a sphere the mass M can be written as
- M = \frac{4\pi\rho_M R^3}{3} where R is the radius of the
primary.
And likewise
- m = \frac{4\pi\rho_m r^3}{3} where r is the radius of the
satellite.
Substituting for the masses in the equation for the Roche limit,
and cancelling out 4\pi/3 gives
- d = r \left( \frac{ 2 \rho_M R^3 }{ \rho_m r^3 } \right)^{1/3}
,
which can be simplified to the Roche limit:
- d = R\left( 2\;\frac {\rho_M} {\rho_m} \right)^{\frac{1}{3}}
.
Fluid satellites
A more accurate approach for calculating the Roche Limit takes the
deformation of the satellite into account. An extreme example would
be a
tidally locked liquid satellite
orbiting a planet, where any force acting upon the satellite would
deform it into a prolate
spheroid.
The calculation is complex and its result cannot be represented in
an exact algebraic formula. Roche himself derived the following
approximate solution for the Roche Limit:
- d \approx 2.44R\left( \frac {\rho_M} {\rho_m}
\right)^{1/3}
However, a better approximation that takes into account the
primary's oblateness and the satellite's mass is:
- d \approx 2.423 R\left( \frac {\rho_M} {\rho_m} \right)^{1/3}
\left( \frac{(1+\frac{m}{3M})+\frac{c}{3R}(1+\frac{m}{M})}{1-c/R}
\right)^{1/3}
where c/R is the
oblateness of the
primary. The numerical factor is calculated with the aid of a
computer.
The fluid solution is appropriate for bodies that are only loosely
held together, such as a comet. For instance, comet
Shoemaker-Levy 9's decaying orbit
around Jupiter passed within its Roche limit in July 1992, causing
it to fragment into a number of smaller pieces. On its next
approach in 1994 the fragments crashed into the planet.
Shoemaker-Levy 9 was first observed in 1993, but its orbit
indicated that it had been captured by Jupiter a few decades prior.
[9699]
Derivation of the formula
As the fluid satellite case is more delicate than the rigid one,
the satellite is described with some simplifying assumptions.
First, assume the object consists of incompressible fluid that has
constant density \rho_m and volume V that do not depend on external
or internal forces.
Second, assume the satellite moves in a circular orbit and it
remains in
synchronous
rotation. This means that the angular speed \omega at which it
rotates around its center of mass is the same as the angular speed
at which it moves around the overall system
barycenter.
The angular speed \omega is given by
Kepler's third law:
- \omega^2 = G \, \frac{M + m}{d^3}.
When M is very much bigger than m, this will be close to
- \omega^2 = G \, \frac{M}{d^3}.
The synchronous rotation implies that the liquid does not move and
the problem can be regarded as a static one. Therefore, the
viscosity and
friction of the liquid in this model do not play a
role, since these quantities would play a role only for a moving
fluid.
Given these assumptions, the following forces should be taken into
account:
- The force of gravitation due to the main body;
- the centrifugal
force in the rotary reference system; and
- the self-gravitation field of the satellite.
Since all of these forces are conservative, they can be expressed
by means of a potential. Moreover, the surface of the satellite is
an equipotential one. Otherwise, the differences of potential would
give rise to forces and movement of some parts of the liquid at the
surface, which contradicts the static model assumption. Given the
distance from the main body, our problem is to determine the form
of the surface that satisfies the equipotential condition.
As the orbit has been assumed circular, the total gravitational
force and centrifugal force acting on the main body cancel.
Therefore, the force that affects the particles of the liquid is
the tidal force, which depends on the position with respect to the
center of mass, already considered in the rigid model. For small
bodies, the distance of the liquid particles from the center of the
body is small in relation to the distance
d to the main
body. Thus the tidal force can be linearized, resulting in the same
formula for
F_{T} as given above.While this force
in the rigid model depends only on the radius
r of the
satellite, in the fluid case we need to consider all the points on
the surface and the tidal force depends on the distance
Δd
from the center of mass to a given particle projected on the line
joining the satellite and the main body. We call
Δd the
radial distance. Since the tidal force is linear in
Δd, the related potential is proportional to the square of
the variable and for m\ll M we have
- V_T = - \frac{3 G M }{2 d^3}\Delta d^2 \,
We want to determine the shape of the satellite for which the sum
of the self-gravitation potential and V_T is constant on the
surface of the body. In general, such a problem is very difficult
to solve, but in this particular case, it can be solved by a
skillful guess due to the square dependence of the tidal potential
on the radial distance
Δd
Since the potential
V_{T} changes only in one
direction,
i.e. the direction toward the main body, the
satellite can be expected to take an axially symmetric form. More
precisely, we may assume that it takes a form of a
solid of revolution. The self-potential
on the surface of such a solid of revolution can only depend on the
radial distance to the center of mass. Indeed, the intersection of
the satellite and a plane perpendicular to the line joining the
bodies is a disc whose boundary by our assumptions is a circle of
constant potential. Should the difference between the
self-gravitation potential and
V_{T} be constant,
both potentials must depend in the same way on
Δd. In
other words, the self-potential has to be proportional to the
square of
Δd. Then it can be shown that the equipotential
solution is an ellipsoid of revolution. Given a constant density
and volume the self-potential of such body depends only on the
eccentricity ε
of the ellipsoid:
- V_s = V_{s_{0}} + G \pi \rho_m \cdot f (\epsilon) \cdot \Delta
d^2,
where V_{s_0} is the constant self-potential on the intersection of
the circular edge of the body and the central symmetry plane given
by the equation
Δd=0.
The dimensionless function
f is to be determined from the
accurate solution for the potential of the ellipsoid
- f(\epsilon) = \frac{1 - \epsilon^2}{\epsilon^3} \cdot \left[
\left(3-\epsilon^2 \right) \cdot \mathrm{arsinh}
\left(\frac{\epsilon}{\sqrt{1-\epsilon^2}} \right) -3 \epsilon
\right]
and, surprisingly enough, does not depend on the volume of the
satellite.
The graph of the dimensionless
function
f which indicates how the strength of the tidal
potential depends on the eccentricity
ε of the
ellipsoid
Although the explicit form of the function
f looks
complicated, it is clearthat we may and do choose the value of
ε so that the potential
V_{T} is equal to
V_{S} plus a constant independent of the variable
Δd. By inspection, this occurs when
- \frac{2 G \pi \rho_M R^3}{d^3} = G \pi \rho_m f(\epsilon)
This equation can be solved numerically. The graph indicates that
there are two solutions and thus the smaller one represents the
stable equilibrium form (the ellipsoid with the smaller
eccentricity). This solution determines the eccentricity of the
tidal ellipsoid as a function of the distance to the main body. The
derivative of the function
f has a zero where the maximal
eccentricity is attained. This corresponds to the Roche limit.
The derivative of
f
determines the maximal eccentricity.
This gives the Roche limit.
More precisely, the Roche limit is determined by the fact that the
function
f, which can be regarded as a nonlinear measure
of the force squeezing the ellipsoid towards a spherical shape, is
bounded so that there is an eccentricity at which this contracting
force becomes maximal. Since the tidal force increases when the
satellite approaches the main body, it is clear that there is a
critical distance at which the ellipsoid is torn up.
The maximal eccentricity can be calculated numerically as the zero
of the derivative of
f'. One obtains
- \epsilon_\textrm{max}\approx 0{.}86
which corresponds to the ratio of the ellipsoid axes 1:1.95.
Inserting this into the formula for the function
f one can
determine the minimal distance at which the ellipsoid exists. This
is the Roche limit,
- d \approx 2{.}423 \cdot R \cdot \sqrt[3]{ \frac {\rho_M}
{\rho_m} } \,.
Roche limits for selected examples
The table below shows the mean density and the equatorial radius
for selected objects in our
solar
system.
Primary |
Density (kg/m³) |
Radius (m) |
Sun |
1,408 |
696,000,000 |
Jupiter |
1,326 |
71,492,000 |
Earth |
5,513 |
6,378,137 |
Moon |
3,346 |
1,738,100 |
Saturn |
687.3 |
60,268,000 |
Uranus |
1,318 |
25,559,000 |
Neptune |
1,638 |
24,764,000 |
Using these data, the Roche Limits for rigid and fluid bodies can
easily be calculated. The average density of
comets is taken to be around 500 kg/m³.
The table below gives the Roche limits expressed in metres and in
primary radii. The true Roche Limit for a satellite depends on its
density and rigidity.
Body |
Satellite |
Roche limit (rigid) |
Roche limit (fluid) |
Distance (km) |
R |
Distance (km) |
R |
Earth |
Moon |
9,496 |
1.49 |
18,261 |
2.86 |
Earth |
average Comet |
17,880 |
2.80 |
34,390 |
5.39 |
Sun |
Earth |
554,400 |
0.80 |
1,066,300 |
1.53 |
Sun |
Jupiter |
890,700 |
1.28 |
1,713,000 |
2.46 |
Sun |
Moon |
655,300 |
0.94 |
1,260,300 |
1.81 |
Sun |
average Comet |
1,234,000 |
1.78 |
2,374,000 |
3.42 |
If the primary is less than half as dense as the satellite, the
rigid-body Roche Limit is less than the primary's radius, and the
two bodies may collide before the Roche limit is reached.
How close are the solar system's moons to their Roche limits? The
table below gives each inner satellite's orbital radius divided by
its own Roche radius. Both rigid and fluid body calculations are
given. Note
Pan,
Metis, and
Naiad in
particular, which may be quite close to their actual break-up
points.
In practice, the densities of most of the inner satellites of giant
planets are not known. In these cases, shown in
italics,
likely values have been assumed, but their
actual Roche
limit can vary from the value shown.
See also
References
Other uses
- Roche
Limit is the name of a Canadian Electronic pop band.
Sources
- Édouard Roche: La figure d'une masse fluide soumise à
l'attraction d'un point éloigné, Acad. des sciences de
Montpellier, Vol. 1 (1847–50) p. 243
External links