Hill sphere

The Hill sphere or Roche sphere of an astronomical body is the region in which it dominates the attraction of satellites. The outer shell of that region constitutes a zero-velocity surface. To be retained by a planet, a moon must have an orbit that lies within the planet's Hill sphere. That moon would, in turn, have a Hill sphere of its own. Any object within that distance would tend to become a satellite of the moon, rather than of the planet itself. One simple view of the extent of the Solar System is the Hill sphere of the Sun with respect to local stars and the galactic nucleus.[1]

A contour plot of the effective potential of a two-body system due to gravity and inertia at one point in time. The Hill spheres are the circular regions surrounding the two large masses.

In more precise terms, the Hill sphere approximates the gravitational sphere of influence of a smaller body in the face of perturbations from a more massive body. It was defined by the American astronomer George William Hill, based on the work of the French astronomer Édouard Roche. For this reason, it is also known as the "Roche sphere" (not to be confused with the Roche limit or Roche Lobe).

In the example to the right, Earth's Hill sphere extends between the Lagrangian points L1 and L2, which lie along the line of centers of the two bodies (the Earth and the Sun). The region of influence of the second body is shortest in that direction, and so it acts as the limiting factor for the size of the Hill sphere. Beyond that distance, a third object in orbit around the second (e.g. the Moon) would spend at least part of its orbit outside the Hill sphere, and would be progressively perturbed by the tidal forces of the central body (e.g. the Sun), eventually ending up orbiting the latter.

Formula and examples

Comparison of the Hill spheres and Roche limits of the Sun-Earth-Moon system (not to scale) with shaded regions denoting stable orbits of satellites of each body

If the mass of the smaller body (e.g. the Earth) is ${\displaystyle m}$ , and it orbits a heavier body (e.g. the Sun) of mass ${\displaystyle M}$  with a semi-major axis ${\displaystyle a}$  and an eccentricity of ${\displaystyle e}$ , then the radius ${\displaystyle r_{\mathrm {H} }}$  of the Hill sphere of the smaller body, calculated at the pericenter, is approximately[2]

${\displaystyle r_{\mathrm {H} }\approx a(1-e){\sqrt[{3}]{\frac {m}{3M}}}.}$

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

${\displaystyle r_{\mathrm {H} }\approx a{\sqrt[{3}]{\frac {m}{3M}}}.}$

In the Earth-Sun example, the Earth (5.97×1024 kg) orbits the Sun (1.99×1030 kg) at a distance of 149.6 million km, or one astronomical unit (AU). 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 seven months.

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

${\displaystyle 3{\frac {r_{\mathrm {H} }^{3}}{a^{3}}}\approx {\frac {m}{M}}.}$

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.

Derivation

The expression for the Hill radius can be found by equating gravitational and centrifugal forces acting on a test particle (of mass much smaller than ${\displaystyle m}$ ) orbiting the secondary body. Assume that the distance between masses ${\displaystyle M}$  and ${\displaystyle m}$  is ${\displaystyle r}$ , and that the test particle is orbiting at a distance ${\displaystyle r_{\mathrm {H} }}$  from the secondary. When the test particle is on the line connecting the primary and the secondary body, the force balance requires that

${\displaystyle {\frac {Gm}{r_{\mathrm {H} }^{2}}}-{\frac {GM}{(r-r_{\mathrm {H} })^{2}}}+\Omega ^{2}(r-r_{\mathrm {H} })=0,}$

where ${\displaystyle G}$  is the gravitational constant and ${\displaystyle \Omega ={\sqrt {\frac {GM}{r^{3}}}}}$  is the (Keplerian) angular velocity of the secondary about the primary (assuming that ${\displaystyle m\ll M}$ ). The above equation can also be written as

${\displaystyle {\frac {m}{r_{\mathrm {H} }^{2}}}-{\frac {M}{r^{2}}}\left(1-{\frac {r_{\mathrm {H} }}{r}}\right)^{-2}+{\frac {M}{r^{2}}}\left(1-{\frac {r_{\mathrm {H} }}{r}}\right)=0,}$

which, through a binomial expansion to leading order in ${\displaystyle r_{\mathrm {H} }/r}$ , can be written as

${\displaystyle {\frac {m}{r_{\mathrm {H} }^{2}}}-{\frac {M}{r^{2}}}\left(1+2{\frac {r_{\mathrm {H} }}{r}}\right)+{\frac {M}{r^{2}}}\left(1-{\frac {r_{\mathrm {H} }}{r}}\right)={\frac {m}{r_{\mathrm {H} }^{2}}}-{\frac {M}{r^{2}}}\left(3{\frac {r_{\mathrm {H} }}{r}}\right)\approx 0.}$

Hence, the relation stated above

${\displaystyle {\frac {r_{\mathrm {H} }}{r}}\approx {\sqrt[{3}]{\frac {m}{3M}}}.}$

If the orbit of the secondary about the primary is elliptical, the Hill radius is maximum at the apocenter, where ${\displaystyle r}$  is largest, and minimum at the pericenter of the orbit. Therefore, for purposes of stability of test particles (for example, of small satellites), the Hill radius at the pericenter distance needs to be considered.[2] To leading order in ${\displaystyle r_{\mathrm {H} }/r}$ , the Hill radius above also represents the distance of the Lagrangian point L1 from the secondary.

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

${\displaystyle {\frac {r_{\mathrm {H} }}{R_{\mathrm {secondary} }}}\approx {\frac {a}{R_{\mathrm {primary} }}}{\sqrt[{3}]{\frac {\rho _{\mathrm {secondary} }}{3\rho _{\mathrm {primary} }}}}\approx {\frac {a}{R_{\mathrm {primary} }}},}$

where ${\displaystyle \rho _{\mathrm {second} }}$  and ${\displaystyle \rho _{\mathrm {primary} }}$  are the average densities of the primary and secondary bodies, and ${\displaystyle R_{\mathrm {secondary} }}$  and ${\displaystyle R_{\mathrm {primary} }}$  are their radii. The second approximation is justified by the fact that, for most cases in the Solar System, ${\displaystyle {\sqrt[{3}]{\frac {\rho _{\mathrm {secondary} }}{3\rho _{\mathrm {primary} }}}}}$  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, because many planetary astronomers work in and remember distances in units of planetary radii.

True region of stability

The Hill sphere is only an approximation, and other forces (such as radiation pressure or the Yarkovsky effect) can eventually perturb an object out of the sphere. This third object should also be of small enough mass that it introduces no additional complications through its own gravity. Detailed numerical calculations show that orbits at or just within the Hill sphere are not stable in the long term; it appears that stable satellite orbits exist only inside 1/2 to 1/3 of the Hill radius. The region of stability for retrograde orbits at a large distance from the primary is larger than the region for prograde orbits at a large distance from the primary. This was thought to explain the preponderance of retrograde moons around Jupiter; however, Saturn has a more even mix of retrograde/prograde moons so the reasons are more complicated.[3]

Further examples

An astronaut could not have orbited the Space Shuttle (with mass of 104 tonnes), where the orbit was 300 km above the Earth, because its Hill sphere at that altitude was only 120 cm in radius, much smaller than the shuttle itself. A sphere of this size and mass would be denser than lead. In fact, in any low Earth orbit, a spherical body must be more dense than lead in order to fit inside its own Hill sphere, or else it will be incapable of supporting an orbit. A spherical geostationary satellite, however, would only need to be more than 6% of the density of water to support satellites of its own.[citation needed]

Within the Solar System, the planet with the largest Hill radius is Neptune, with 116 million km, or 0.775 au; its great distance from the Sun amply compensates for its small mass relative to Jupiter (whose own Hill radius measures 53 million km). An asteroid from the asteroid belt will have a Hill sphere that can reach 220,000 km (for 1 Ceres), diminishing rapidly with decreasing mass. The Hill sphere of 66391 Moshup, a Mercury-crossing asteroid that has a moon (named Squannit), measures 22 km in radius.[4]

A typical extrasolar "hot Jupiter", HD 209458 b,[5] has a Hill sphere radius of 593,000 km, about eight times its physical radius of approx 71,000 km. Even the smallest close-in extrasolar planet, CoRoT-7b,[6] still has a Hill sphere radius (61,000 km), six times its physical radius (approx 10,000 km). Therefore, these planets could have small moons close in, although not within their respective Roche limits.[citation needed]

Solar System

The following logarithmic plot shows the Hill radius (in km) of some bodies of the Solar System: