Hill Sphere Calculator

Compute Hill radius from orbit size, masses, and eccentricity, then estimate conservative stable-orbit bands.

Worked example — Hill Sphere Calculator

Mass ratio	1.00×10⁻⁶
Cube-root factor	0.01000
At perihelion	0.9833AU
Hill radius	0.009837AU
Hill radius (km)	1.472×10⁶km
Stable prograde	7.36×10⁵km
Stable retrograde	1.03×10⁶km

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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why hot Jupiters have no moons — and where JWST lives 🖖

A hot Jupiter at 0.04 AU has a Hill sphere of only ~450,000 km. The stable prograde zone ends at ~225,000 km — smaller than the Moon's 384,400 km orbit around Earth. Any moon that existed was swept away during the planet's inward migration, and no new one could survive there. This is why no confirmed exomoon has ever been found around a hot Jupiter. The same geometry explains where space telescopes park: the Lagrange points L1 and L2 sit at almost exactly one Hill sphere radius from Earth (~1.5 million km). JWST orbits L2, which places it just inside Earth's gravitational boundary — far enough to escape atmospheric and thermal interference, close enough that Earth's gravity keeps it bound against solar perturbations. The Hill sphere is not just a theoretical limit; it is the address of our most powerful observatory.

Where a planet's grip beats the Sun's 🖖

A moon is caught in a tug-of-war: its planet pulls it close while the far more massive Sun tries to pry it loose. The Hill sphere marks where the planet wins — inside it orbits can be stable, while outside a satellite drifts off onto its own path around the Sun. The radius grows only with the cube root of the mass ratio, so it changes slowly. As a rule of thumb, a moon on a normal prograde orbit stays safe only out to about half the Hill radius, which is why real moons huddle well inside the boundary.

Backward-orbiting moons survive farther out 🖖

Counterintuitively, a moon circling its planet the wrong way — retrograde, opposite the planet's motion — stays stable out to roughly 0.7 of the Hill radius, while a prograde moon breaks free past about 0.5. The reversed motion softens the Sun's repeated tug over each orbit. This is no curiosity: nearly all of Jupiter's and Saturn's distant outer moons orbit retrograde, precisely because those are the only wide orbits that last.

Problem solved in full

Limit where something can still orbit Earth rather than the Sun 5 steps
How far from Earth can something still orbit Earth rather than being taken by the Sun — and is the Moon safely inside that limit? Earth: m = 5.972 × 10²⁴ kg, a = 1 AU, e = 0.0167. Sun: M = 1.989 × 10³⁰ kg.
1. The Hill radius is where a planet’s pull on a small body matches the Sun’s tidal tug at that distance. The cube root is the signature of that balance: it is a ratio of a mass to a mass, taken to the one-third power because gravity falls as an inverse square while the tidal term does not.
  
  $r_{H} \approx a (1 - e) 3 \frac{m}{3 M}$
2. Work out the mass ratio first — this is the quantity the calculator above prints as “mass ratio”, and note it is m/3M, not m/M.
  
  $\frac{m}{3 M} = \frac{5.972 \times 1 0 ^{24}}{3 ( 1.989 \times 1 0 ^{30} )} = 1.00 \times 1 0^{- 6}$
3. Take the cube root. Notice how brutally the cube root compresses: a millionth becomes a hundredth, so Earth’s sphere of influence is about one hundredth of its distance from the Sun even though Earth is a millionth of the Sun’s mass.
  
  $3 1.00 \times 1 0^{- 6} = 0.01000$
4. Multiply by the perihelion distance — the tightest point of the orbit, because that is where the Sun competes hardest and so where the limit is smallest.
  
  $r_{H} = 1 AU \times (1 - 0.0167) \times 0.01000 = 0.009837 AU = 1.472 \times 1 0^{6} km$
5. Now test the Moon against it. Prograde orbits are only stable out to roughly half the Hill radius, so that is the line the Moon has to beat.
  
  $r_{Moon} = 3.84 \times 1 0^{5} km = 0.26 r_{H} < 0.5 r_{H}$
Answer

1.472 × 10⁶ km, and the Moon at 3.84 × 10⁵ km sits at 0.26 of it — comfortably inside even the stricter half-Hill limit for prograde orbits. The cube root is why this works at all: it is such a compressive function that a planet a millionth of its star’s mass still commands a region one hundredth as wide as its orbit. It is also why capture is rare and why distant irregular moons are all retrograde — retrograde orbits stay stable out to about 0.7 r_H, further than prograde ones.

$r_{H} = a (1 - e) 3 \frac{m}{3 M} = 1.472 \times 1 0^{6} km$

References (2)

By the 42Math team · Updated 2026-07-08

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

All Astronomy & Space lessons →

Example problems

Earth around Sun - Earth around Sun gives a Hill radius near 0.01 AU, setting a rough bound for stable satellites.
Jupiter around Sun - Jupiter has a very large Hill sphere, supporting many distant moons.
Moon around Earth - The Moon around Earth has a much smaller Hill sphere, limiting long-term sub-satellite stability.
Hot Jupiter close-in - Close-in hot Jupiters have compact Hill spheres, making wide moon orbits hard to retain.