A Hill sphere is, roughly, the volume around an astronomical body (such as a planet) where it dominates in attraction of satellites to that body, rather than to a larger body (such as a star) which it orbits. Thus, for a planet to retain a moon, the moon must have an orbit that lies within the Hill sphere of the planet. That moon would, in turn, have a Hill sphere of its own. Any objects within that distance would tend to become satellites of the moon, rather than of the planet itself.
More precisely, 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 upon the work of the French astronomer Édouard Roche. For this reason, it is also known as the Roche sphere.
To illustrate, consider the specific case of the planet Jupiter, which orbits the Sun. For any point in space, one can compute the sum of the following three forces:
- gravity due to the Sun,
- gravity due to Jupiter,
- the centrifugal force experienced by a particle at such a point moving with the same frequency as Jupiter around the Sun.
The Hill sphere for Jupiter is the largest sphere, centered at Jupiter, within which the sum of the three forces is always directed towards Jupiter. In more general terms, it is the sphere around a secondary body in orbit around a primary body within which the net force is a centripetal force directed at the secondary body. Thus, the Hill sphere in our example describes the outer limit that an even smaller object such as a moon or artificial satellite can stably orbit Jupiter rather than simply entering an elliptical orbit of its own around the Sun.
The Hill sphere extends between the Lagrangian points L1 and L2, which lie along the line of centers of the two bodies. 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. Jupiter) 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.
The Roche sphere is not to be confused with the Roche lobe and Roche limit which were also described by Roche. The Roche limit is the distance at which an object held together only by gravity begins to break up due to tidal forces. The Roche lobe describes the limits at which an object which is in orbit around two other objects will be captured by one or the other.
Contents |
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 [1]
![r \approx a (1-e) \sqrt[3]{\frac{m}{3 M}}](http://wpcontent.answers.com/math/5/0/f/50fc57604cad722d8fae9e49a0544d13.png)
When eccentricity is negligible (the most favourable case for orbital stability), this becomes
![r \approx a \sqrt[3]{\frac{m}{3M}}](http://wpcontent.answers.com/math/5/c/f/5cf96e48a1d82de12814b75962e3156a.png)
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. In terms of orbital period: all stable satellites of the Earth 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 these two spheres.
A quick way of estimating the radius of the Hill sphere comes from replacing mass with density in the above equation:
![\frac{r}{R_{secondary}} \approx \frac{a}{R_{primary}} \sqrt[3]{\frac{\rho_{secondary}}{3 \rho_{primary}}} \approx \frac{a}{R_{primary}}](http://wpcontent.answers.com/math/1/d/7/1d7a399c8b32e279840a13630658b494.png)
where ρsecond and ρprimary 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, ![\sqrt[3]{\frac{\rho_{secondary}}{3 \rho_{primary}}}](http://wpcontent.answers.com/math/b/d/4/bd4d6d2d6fa8bfe4407c2a13762205e6.png)
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.
True region of stability
The Hill sphere is but 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 (with retrograde orbits being more stable than prograde orbits).[citation needed]
Further examples
An astronaut could not orbit the Space Shuttle (with mass of 104 tonnes), where the orbit is 300 km above the Earth, since the Hill sphere of the shuttle is only 120 cm in radius, much smaller than the shuttle itself. In fact, in any low Earth orbit, a spherical body must be 800 times denser 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 would need to be more than 5 times denser than lead to support satellites of its own; such a satellite would be 2.5 times denser than iridium, the densest naturally-occurring material on Earth. Only at twice the geostationary distance could a lead sphere possibly support its own satellite; since the moon is more than three times further than the 3-fold geostationary distance necessary, lunar orbits are possible.
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 main belt will have a Hill sphere that can reach 220 000 km (for 1 Ceres), diminishing rapidly with its mass. In the case of (66391) 1999 KW₄, a Mercury-crosser asteroid which has a moon (S/2001 (66391) 1), its Hill sphere measures 22 km in radius.
Derivation
A non-rigorous, but conceptually accurate derivation of the Hill radius can be made by equating the orbital velocity of the orbiter around a body (i.e., a planet) and the orbital velocity of that planet around the host star. This is the radius at which the gravitational influence of the star roughly equals that of the planet. This is accurate to within factors of order unity.
Where RH is the Hill radius, a is the semi-major axis of the planet orbiting the star. With some basic algebra:
Giving a Hill radius of:
See also
External links
References
- ^ D.P. Hamilton & J.A. Burns (1992). "Orbital stability zones about asteroids. II - The destabilizing effects of eccentric orbits and of solar radiation". Icarus 96: 43. doi:. http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1992Icar...96...43H&db_key=AST&data_type=HTML&format=&high=444b66a47d16486.
This entry is from Wikipedia, the leading user-contributed encyclopedia. It may not have been reviewed by professional editors (see full disclaimer)
