## Derivation of the Equation for Roche Limit

Initial Equations:

Gravitational Acceleration: a= GM/r^{2}

Tidal Acceleration:

a= GM/(d-r)^{2}-GM/r^{2}

a= (GM/d^{2})*{(1-r/d)^{2}-1}

a~ (GM/d^{2})*(2r/d)

a~ 2GMr/d^{3}

Centrifugal Acceleration: a= (4π
^{2}r)/t^{2}

Orbital Period: t= (2π
r)/(GM/r)^{1/2} = (2π
r^{3/2})/(GM)^{1/2}

The Roche Limit occurs when the forces pulling material away from the satelite equal the forces pulling material onto the satelite. That is, when Tide and Centrifugal Force from the satelite's rotation equal the satelite's surface gravity.

Given that d= the distance between the centers of mass of the planet and its satelite, R= the radius of the planet, M= the mass of the planet, r= the radius of the satelite, m= the mass of the satelite, and A= the centrifugal force caused by the rotation of the satelite, then d= the Roche Limit when

2GMr/d^{3}+A = Gm/r^{2}

Given that the satelite is tidelocked, its orbital and rotational periods are equal. Therefore:

A= (4π^{2}r)/t^{2}
t^{2}= (4π
^{2}d^{3})/(Gm)

A= (4π
^{2}r)/{(4π
^{2}d^{3})/(Gm)}

A= Gmr/d^{3}

And

2GMr/d^{3}+A = 3GMr/d^{3}

Solving for d:

d^{3}/3GMr = r^{2}/Gm

d^{3}= 3GMr^{3}/Gm

Substituting Density for Mass:

d^{3} = 3{P_{planet}*(4/3)π
R^{3}}/{P_{sat.}*(4/3)π
r^{3}}*r^{3}

d^{3} = 3(P_{planet}*R^{3})/P_{sat.}

d^{3} = R^{3}*3(P_{planet}/P_{sat.})

d = R*{3(P_{planet}/P_{sat.})}^{1/3}

d ~ R*1.44*(P_{planet}/P_{sat.})^{1/3}

This assumes a rigid body and is, therefore, valid only for rocky satelites of less than ~500km radius which will not deform into ellipsoids, because the constant cube-root of 3 changes if the satelite is elastic. For a liquid body, it is near 2.44, giving the equation

d = R*2.44*(P_{planet}/P_{sat.})^{1/3}