Hill Sphere Radius
The zone where a body's gravity dominates over its host's.
Formula
Variables
Example
Earth's Hill sphere reaches about 1.5 million km
How it works
The Hill sphere radius estimates how far from a smaller body an orbit can stay bound to that body instead of being pulled away by a much larger one. It comes from balancing the smaller body’s gravity against the tidal pull of the host, with the factor of 3 arising from the three-body dynamics in a circular-orbit approximation. In practice, it marks the region where moons, rings, or satellites can remain stable around a planet, asteroid, or exoplanetary body.
When to use it
Use this when checking whether a moon, spacecraft, or ring particle can stably orbit a planet or asteroid while that object itself orbits a star. It is also useful in exoplanet studies to estimate how large a planet’s long-term satellite zone might be.
Worked example, step by step
Example: Estimate Mars’s Hill sphere around the Sun. Given: a = 1.524 AU = 1.524 × 1.496 × 10^8 km = 2.279 × 10^8 km, m = 6.42 × 10^23 kg, M = 1.99 × 10^30 kg Step 1: Compute the mass ratio inside the cube root: (m / 3M) = (6.42 × 10^23) / (3 × 1.99 × 10^30) = (6.42 × 10^23) / (5.97 × 10^30) ≈ 1.08 × 10^-7 Step 2: Take the cube root: (1.08 × 10^-7)^(1/3) ≈ 4.77 × 10^-3 Step 3: Multiply by orbital radius: r_H ≈ (2.279 × 10^8 km)(4.77 × 10^-3) ≈ 1.09 × 10^6 km Final answer: Mars’s Hill sphere radius is about 1.1 million km.
Common mistakes to avoid
- Using diameter instead of orbital radius a
- Forgetting that m and M must be in the same units
- Assuming the Hill sphere is the same as the full stable-orbit region for all inclinations and eccentricities
Did You Know?
A moon can only orbit stably well inside its planet's Hill sphere — the Moon sits comfortably within Earth's.
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