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Hill Sphere Calculator

Find how far a body's gravity holds satellites.

How it is calculated

The Hill radius is rH = a × (1 − e) × (m / 3M)1/3. Here a is the semi-major axis of the orbit in kilometres, e the orbital eccentricity, m the mass of the smaller body and M the mass of the body it orbits, both in kilograms. The factor (1 − e) shrinks the radius for an elliptical orbit by using the closest approach. If eccentricity is left empty the tool uses 0 (a circular orbit); a negative value or one of 1 or more is not a bound orbit and gives no result. Axis and both masses must be positive. The result is shown in kilometres and in millions of kilometres. It is an approximation for a small body orbiting a much heavier one; stable satellite orbits sit well inside this boundary, not at its edge.

Worked example

Use Earth-like values: a = 149,600,000 km, e = 0.0167, m = 5.972e24 kg and M = 1.989e30 kg. The factor 1 − e = 0.9833 gives a closest distance of about 147,101,680 km. Next m / 3M = 5.972e24 / 5.967e30 ≈ 1.0008×10⁻⁶, whose cube root is about 0.0100028. Multiplying gives 1,471,428 km, or 1.4714 million km.

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Common mistakes

  • Entering the distance in astronomical units or metres; the tool expects kilometres.
  • Using masses in tonnes or solar masses; both masses must be in kilograms.
  • Swapping the two masses so that the heavier body is entered as m.
  • Typing eccentricity as a percentage (1.67), which is 1 or more and is silently replaced by 0.
  • Treating the Hill radius as the limit for a stable orbit; real stable orbits stay at a fraction of it.

About this calculator

The Hill sphere (or Roche sphere) is the region around a celestial body — a planet orbiting a star, or a moon orbiting a planet — within which its own gravity dominates over the pull of the larger body it orbits, so that it can hold onto satellites. It's approximated by r ≈ a·(m / 3M)^(1/3), where a is the orbital distance between the smaller body (mass m) and the larger one (mass M).

Astronomers use the Hill sphere to determine whether a planet's candidate moon can remain gravitationally bound, to set the outer stability limit for exoplanet moons ("exomoons") in habitability studies, and to explain why moons far from a planet (like an outer irregular satellite) are more easily stripped away by the Sun's gravity. Earth's Hill sphere, for comparison, extends to about 1.5 million km — well beyond the Moon's orbit, which is why the Moon remains stably bound to Earth rather than the Sun.

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Comments (1)

  • Editorial team

    The Hill sphere is the outer limit of a body's gravitational dominance, and stable satellite orbits sit well inside it, not out to its edge. Distance goes in kilometres and both masses in kilograms. For a sanity check, Earth around the Sun (149.6 million km, 5.972e24 kg and 1.989e30 kg) should come out near 1.5 million km. Leave eccentricity empty for a circular orbit; a value of 1 or more is not an orbit and gets no result. Which planet and moon system are you checking?

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