Hill Sphere Radius Calculator
Enter primary mass, secondary mass, orbital separation, and eccentricity to find the Hill sphere radius and stable moon orbit boundaries for any two-body system.
🌐 What Is the Hill Sphere?
The Hill sphere (named after American astronomer George William Hill) is the region around a secondary body within which it can gravitationally dominate over the tidal forces of the primary. A satellite orbiting within the Hill sphere experiences a net gravitational pull toward the secondary body, allowing it to remain bound. A satellite beyond the Hill sphere is pulled more strongly toward the primary and will eventually drift into a solar orbit rather than a planetary orbit.
The Hill sphere radius is given by rH = a(1 − e)(m/3M)1/3, where a is the semi-major axis, e is the orbital eccentricity, m is the secondary body mass, and M is the primary mass. The eccentricity term (1 − e) reflects the fact that tidal forces are strongest at perihelion, where the secondary is closest to the primary, so the gravitational sphere of influence contracts at closest approach. For circular orbits (e = 0) the formula simplifies to rH = a(m/3M)1/3.
The Hill sphere governs which moons are stable around a planet. Prograde moons (orbiting in the same sense as the planet's orbit around its star) are typically stable only within about 0.5 rH. Retrograde moons (orbiting in the opposite direction) can remain stable out to about 0.7 rH because they are less vulnerable to resonant perturbations from the star's gravity. Jupiter's outermost known retrograde moon Neso orbits at about 47 million km, remarkably close to the 0.7 rH stability boundary of 35 million km. Beyond these limits, moons are gradually perturbed onto hyperbolic escape trajectories.
The Hill sphere is closely related to the Lagrange points. The L1 and L2 Lagrange point distances from the secondary body are approximately equal to rH (to first order), both being proportional to (m/3M)1/3. The key difference is that the Hill sphere formula uses m/3M (mass of secondary over three times the mass of the primary) while the Lagrange formula uses the mass ratio mu = m/(M+m). For small m/M, these are nearly identical. The Hill sphere formula also includes the eccentricity correction, which the standard Lagrange point formula omits. This calculator is therefore more appropriate when orbital eccentricity is significant.
📐 Formula
📖 How to Use This Calculator
Steps
💡 Example Calculations
Example 1 — Earth Orbiting the Sun (e = 0.0167)
Sun (333,000 M⊕) and Earth (1 M⊕) at 1 AU with eccentricity e = 0.0167
Example 2 — Jupiter Orbiting the Sun (e = 0.0489)
Sun (333,000 M⊕) and Jupiter (317.8 M⊕) at 5.203 AU with eccentricity e = 0.0489
Example 3 — Pluto Orbiting the Sun (e = 0.2488)
Sun (333,000 M⊕) and Pluto (0.00218 M⊕) at 39.48 AU with eccentricity e = 0.2488
❓ Frequently Asked Questions
🔗 Related Calculators
What is the Hill sphere and how does it differ from the Lagrange point?
The Hill sphere is the roughly spherical region around a secondary body within which it can hold satellites against perturbations from the primary. Its radius r_H = a(1-e)(m/(3M))^(1/3) includes eccentricity because the gravitational influence shrinks at perihelion. The L1 Lagrange point distance uses the slightly different formula r = a(mu/3)^(1/3) with mu = m/(M+m); for small m/M these are nearly identical, but the Hill sphere formula is the standard used for satellite stability analysis.
How far out can a moon orbit stably within the Hill sphere?
In general, prograde moons (orbiting in the same sense as the secondary's orbit around the primary) are stable only within about 0.5 times the Hill sphere radius. Retrograde moons can remain stable out to about 0.7 times r_H because retrograde orbits are less susceptible to resonant perturbations from the primary star. Objects beyond these limits are eventually ejected by tidal forces.
Why does orbital eccentricity shrink the Hill sphere?
At perihelion (closest approach to the primary star), the tidal forces from the primary are strongest. The Hill sphere formula accounts for this by multiplying by (1-e), so a body in a highly elliptical orbit has a smaller effective Hill sphere than one in a circular orbit. Pluto (e=0.2488) has a Hill sphere about 75% the size it would have at the same average distance but on a circular orbit.
What is Earth's Hill sphere radius and how does it compare to the Moon's orbit?
Earth's Hill sphere with eccentricity e=0.0167 has a radius of about 1.47 million km. The Moon orbits at 384,400 km, which is about 26% of Earth's Hill sphere radius, safely inside the stable prograde orbit limit of 0.5 r_H = 735,000 km. This means the Moon is well protected from solar perturbations. Earth could theoretically hold a moon at distances up to about 735,000 km on a prograde orbit.
Why does Jupiter have so many moons?
Jupiter's large mass (317.8 M-Earth) and relatively large heliocentric distance give it a Hill sphere of about 50 million km, one of the largest in the solar system after the Sun. This enormous gravitational sphere of influence allows Jupiter to capture and retain over 95 known moons, including numerous irregular satellites in distant, eccentric, and even retrograde orbits out near the Hill sphere boundary.
How is the Hill sphere different from the Roche limit?
The Roche limit defines how close a satellite can approach a primary before tidal forces tear it apart (roughly 2.44 times the primary's radius for fluid bodies). The Hill sphere defines how far away a satellite can orbit the secondary and still be gravitationally bound to it. Moons must orbit outside the Roche limit (to avoid being disrupted) and inside the Hill sphere (to avoid being captured by the primary). Planetary rings typically lie inside the Roche limit.
Can the Sun capture a body that is outside Earth's Hill sphere?
Yes. Objects beyond Earth's Hill sphere experience stronger gravitational influence from the Sun than from Earth and will eventually be pulled into solar orbit rather than remaining bound to Earth. The Hill sphere defines the boundary of Earth's gravitational dominance. Satellites in lunar-transfer trajectories that temporarily exceed the Hill sphere radius must be actively controlled to return to Earth capture.
What is the Hill sphere of a binary star system?
For a binary star system (two comparable-mass stars), neither star has a Hill sphere in the usual sense because neither mass is much larger than the other. Instead, circumbinary planets (orbiting both stars) can be stable outside about 2 to 4 times the binary separation, while circumstellar planets (orbiting one star) can be stable within about 0.2 to 0.5 times the binary separation. For a binary with a 10:1 mass ratio, the less massive star has a Hill sphere r_H = a(m/(3M))^(1/3) where M is the primary.
How does the Hill sphere relate to the Lagrange point calculator?
The Hill sphere radius and the L1/L2 Lagrange point distance from the secondary are both approximately a(mu/3)^(1/3) for small mass ratios mu = m/(M+m). The Hill sphere formula uses m/(3M) exactly, while the Lagrange formula uses m/(3(M+m)). For small m/M (such as Sun-Earth with m/M = 3e-6), the difference is negligible. The Hill sphere calculator adds the eccentricity correction (1-e), which is important for bodies in elliptical orbits.
What determines whether the Moon is stably bound to Earth?
The Moon orbits at 384,400 km, which is 26% of Earth's Hill sphere radius. The stable prograde orbit limit is 0.5 r_H = 735,000 km, so the Moon orbits well within this boundary. Over billions of years the Moon is gradually receding from Earth at 3.8 cm per year due to tidal energy transfer, but it will not escape Earth's Hill sphere within the Sun's remaining lifetime. The Moon's orbit is also nearly circular (e = 0.055), which helps with long-term stability.