Roche Limit - Determining The Roche Limit - Fluid Satellites - Derivation of The Formula

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 and volume 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 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 is given by Kepler's third law:

When M is very much bigger than m, this will be close to

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 we have

We want to determine the shape of the satellite for which the sum of the self-gravitation potential and 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:

where 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

and, surprisingly enough, does not depend on the volume of the satellite.

Although the explicit form of the function f looks complicated, it is clear that 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

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.

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

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,