Sphere of Influence Calculator

Compute patched-conic SOI from orbital distance and mass ratio, then contrast with Hill-sphere scale.

Worked example — Sphere of Influence Calculator

Mass ratio	3.00×10⁻⁶
Ratio2/5	0.006181
SOI radius	0.006181AU
SOI radius (km)	9.246×10⁵km
Hill radius	0.01000AU
Hill radius (km)	1.497×10⁶km
Hill/SOI ratio	1.619

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

Loading interactive simulation...

why SOI ≠ Hill sphere — and what spacecraft missions actually use 🖖

Two different questions, two different radii. The Hill sphere asks: where can a moon orbit stably? It comes from tidal forces — the balance between the planet's own gravity and the Sun's tidal pull — giving r_H = a(m₂/3m₁)^1/3. The Laplace SOI asks something narrower: where is it a good approximation to pretend only the planet's gravity exists (ignoring the Sun)? That balance gives r_SOI = a(m₂/m₁)^2/5. SOI is always smaller than Hill sphere — for Earth, Hill ≈ 1.5M km but SOI ≈ 925,000 km; for Jupiter, Hill ≈ 53M km but SOI ≈ 48M km. The ratio is not a constant, though: it scales as (m/M)^−1/15, so it runs from 1.10 at Jupiter up to 1.96 at Mercury — the heavier the planet, the closer its SOI sits to its Hill sphere. Mission designers use the SOI boundary as the handoff point for patched-conic trajectories: inside SOI, integrate Earth-centered; outside, switch to heliocentric. It's not physically exact — there is no sharp boundary — but the error is small enough for preliminary mission design. Full precision (for actual navigation) always uses n-body propagation.

gravity's home turf, in plain terms 🖖

The sphere of influence is the invisible bubble around a planet or moon where its gravity is effectively "in charge." Cross the boundary inward and you can safely pretend the Sun isn't there; the planet alone shapes your path. That is why the Moon has its own small sphere of influence, roughly 66,000 km in radius. During Apollo, trajectory calculations switched from an Earth-centered frame to a Moon-centered one right as the spacecraft crossed it.

why distance beats mass 🖖

Look closely at r = a(m₂/m₁)^2/5: the distance a enters linearly, but mass only to the 2/5 power. Being far from the Sun matters far more than being massive. Pluto is about 25 times less massive than Mercury, yet its sphere of influence (~3 million km) dwarfs Mercury's (~112,000 km) — simply because Pluto orbits roughly 100 times farther out.

Problem solved in full

Two standard answers for where Earth's gravity stops winning 5 steps
Where does Earth's gravity stop winning against the Sun's? There are two standard answers, they differ by 62%, and both are correct. Work out both and find what separates them.
1. Everything depends on one number, the mass ratio, and it is small: Earth is about a three-hundred-thousandth of the Sun.
  
  $q = \frac{m}{M} = \frac{1}{332 946} = 3.00 \times 1 0^{- 6}$
2. The first answer is the patched-conic sphere of influence, and its exponent is 2/5. It marks where switching your calculation from heliocentric to geocentric introduces least error — a statement about approximation quality, not about force.
  
  $r_{SOI} = a q^{2/5} = 1 \times (3.00 \times 1 0^{- 6})^{0.4} = 0.006181 AU$
3. The second is the Hill radius, with an exponent of 1/3. It marks where a small third body can stay in orbit indefinitely — a statement about stability, in the rotating frame where centrifugal terms matter.
  
  $r_{H} = a 3 \frac{q}{3} = 3 1.001 \times 1 0^{- 6} = 0.01000 AU$
4. The two exponents are what make the answers differ, and 1.619 is the ratio between them.
  
  $\frac{r _{H}}{r _{SOI}} = \frac{0.01000}{0.006181} = 1.619$
5. The Moon settles the practical question: at 384 400 km it sits at 42% of the sphere of influence, comfortably inside both, which is why it is Earth's moon and not a second planet.
  
  $r_{Moon} = 3.844 \times 1 0^{5} km = 0.416 r_{SOI}$
Answer

The tool prints 0.006181 AU for the sphere of influence, 0.01000 AU for the Hill radius, and a ratio of 1.619. The point is that neither is the boundary — they answer different questions, and using the wrong one is a real navigational error. A mission planner switching reference frames wants the 2/5 figure; someone asking whether a captured asteroid will stay captured wants the 1/3 one. Objects between the two radii are in the awkward band: bound enough to orbit for a while, not bound enough to stay. That is roughly where distant irregular satellites live, and why they are so easily lost.

$r_{SOI} = a q^{2/5} \neq = r_{H} = a 3 q /3 : 1.619 \times apart$

References (1)

The patched-conic method that makes the SOI boundary useful, and its limits: R. R. Bate, D. D. Mueller and J. E. White, Fundamentals of Astrodynamics, §7.4. Dover, 1971. ISBN 978-0-486-60061-1.

By the 42Math team · Updated 2026-07-10

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

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Example problems

Earth around Sun - Earth SOI is roughly 0.006 AU, a standard reference for patched-conic transfers.
Mars around Sun - Mars has a smaller SOI than Earth despite larger orbital distance because of lower planetary mass.
Moon around Earth - Moon SOI around Earth is useful for cislunar trajectory segmentation.
Jupiter around Sun - Jupiter has a very large SOI and strongly dominates local trajectory geometry.