Derivation of the day: hill sphere

Background

The Hill Sphere of an astronomical object defines the maximum radius at which its own gravitational force dominates the pull on a satellite. i.e., this is the distance at which a satellite would be in unstable equilibrium between the object and a second body where a slight perturbation in trajectory could send it into orbit about the second.

Units

a = semimajor axis of the two-body system (planet and star)

R_H = radius of the Hill sphere around the planet
m = mass of satellite

M = mass of planet

M_☉ = mass of sun
G = Newtonian gravitation constant
Ω = angular frequency of the satellite
v_H = Hill velocity

Derivation
We start with Newton's form of Kepler's Third Law:

a³Ω² = GM_☉ where Ω = 2π/Period
R_H³Ω² = GM

Dividing, we find the Hill radius:

R_H³/a³ = M/M_☉

R_H = (M/M_☉)^1/3a

Another useful quantity to find is the Hill velocity, the speed at which the satellite orbits at the Hill radius.  To do this, we equate centripetal force and gravitational force.

GMm/R_H² = mv_H²/R_H
v_H = (GM/R_H)^1/2

Substituting in Kepler's Law,

v_H = (R_H³Ω²/R_H)^1/2  = R_HΩ

We note that this is consistent with Newtonian mechanics' v = r × ω.



References

• http://en.wikipedia.org/wiki/Hill_sphere
• Inke de Pater and Jack J. Lissauer. Planetary Sciences. 2nd. Ed.