Hill Sphere - Formula and Examples

Formula and Examples

If the mass of the smaller body (e.g. Earth) is m, and it orbits a heavier body (e.g. Sun) of mass M with a semi-major axis a and an eccentricity of e, then the radius r of the Hill sphere for the smaller body (e.g. Earth) is, approximately

When eccentricity is negligible (the most favourable case for orbital stability), this becomes

In the Earth example, the Earth (5.97×1024 kg) orbits the Sun (1.99×1030 kg) at a distance of 149.6 million km. The Hill sphere for Earth thus extends out to about 1.5 million km (0.01 AU). The Moon's orbit, at a distance of 0.384 million km from Earth, is comfortably within the gravitational sphere of influence of Earth and it is therefore not at risk of being pulled into an independent orbit around the Sun. All stable satellites of the Earth (those within the Earth's Hill sphere) must have an orbital period shorter than 7 months.

The previous (eccentricity-ignoring) formula can be re-stated as follows:

This expresses the relation in terms of the volume of the Hill sphere compared with the volume of the second body's orbit around the first; specifically, the ratio of the masses is three times the ratio of the volume of these two spheres.

A quick way of estimating the radius of the Hill sphere comes from replacing mass with density in the above equation:

where and are the densities of the primary and secondary bodies, and and are their radii. The second approximation is justified by the fact that, for most cases in the Solar System, happens to be close to one. (The Earth–Moon system is the largest exception, and this approximation is within 20% for most of Saturn's satellites.) This is also convenient, since many planetary astronomers work in and remember distances in units of planetary radii.