Roche Limit Calculator
Find the Roche limit for fluid, rubble-pile or rigid satellites from the planet's radius and both densities, plus the density needed to survive at any orbit.
Roche Limit Calculator
Background.
Bring the Moon in towards Earth and, somewhere around 18,000 km from Earth's centre, it stops being a moon. The pull on its near side is now so much stronger than the pull on its far side that its own gravity can no longer hold it together, and the pieces spread into a ring. That distance is the Roche limit, named after Édouard Roche, who worked it out for a fluid satellite in 1848.
Only three numbers decide it: the radius of the planet and the bulk densities of planet and satellite. A dense satellite can come closer; a dense planet pushes the limit outward. The satellite's size drops out altogether, so a 10 km moonlet and a 1,000 km moon of the same density break up at the same distance.
What changes the answer most is the kind of body you assume. A satellite that can flow into a stretched, egg-like shape hands the tide a longer lever and fails far out; a perfectly rigid sphere holds on much closer in. For Earth and a Moon-density satellite the four models offered here span 18,472 km down to 9,483 km, nearly a factor of two, so a Roche limit quoted without its model is half an answer. All four are shown together, with the one you selected as the headline.
The second result turns the question round: given an orbit, how dense must a body be to survive there? Ring scientists use this critical density because a ring's radius can be measured directly while its particles' density cannot.
Behind every figure sits one assumption: the satellite is held together by gravity alone. Rock strength is ignored, which is why spacecraft, boulders and small solid moons can sit well inside the limit without coming apart.
What is roche limit calculator?
The Roche limit is the orbital distance, measured from the primary's centre, inside which a satellite bound only by its own gravity is pulled apart by tides. Gravity weakens with distance, so across an extended satellite the near side is pulled harder than the centre and the centre harder than the far side. That difference stretches the satellite along the line to the planet. At the Roche limit the stretching just equals the self-gravity holding a surface particle down. Inside it, a loose body breaks up and loose material cannot gather into a moon, which is why planetary rings are found close to their planets.
How to use this calculator.
- Choose the satellite model: fluid for a molten, icy or very weak body; rubble pile for an aggregate of loose fragments, as many small moons and asteroids are; rigid synchronous for a solid sphere that keeps one face to its planet; rigid tidal-term-only to match the 1.26 figure in most textbooks.
- Enter the primary's mean radius in kilometres. Use the volumetric mean radius rather than the equatorial one, because that is the radius consistent with a published bulk density; JPL's planetary physical parameters list both.
- Enter the two bulk densities. Only their ratio enters, so any unit works as long as both match; g/cm³ is what agencies tabulate. Water ice is about 0.9, porous comet nuclei 0.3 to 0.6, solid iron 7.9.
- Enter the satellite's orbital distance from the centre of the primary, not its height above the surface, to get the inside-or-outside verdict and the critical density.
- Compare the four limits. If your orbit falls between them, the verdict depends on what the body is made of, and the range is the result worth quoting.
The formula.
Consider a small particle on the surface of a satellite of radius r, a distance d from a planet of mass M_p. The tide trying to lift it off is about 2GM_p r / d³; the satellite's gravity holding it down is GM_s / r². Write each mass as density times (4/3)πR³ and set the two equal: the radius r cancels and d³ = 2R_p³ ρ_p / ρ_s, so d = 1.260 R_p (ρ_p / ρ_s)^(1/3). That is the textbook rigid-sphere result.
A satellite this close is tidally locked, and in its rotating frame a centrifugal term of GM_p r / d³ joins the tide, raising the 2 to a 3 and the coefficient to 3^(1/3) = 1.442. Tiscareno, Hedman, Burns and Castillo-Rogez (2013) write every case as d = R_p (4π ρ_p / (γ ρ_s))^(1/3), so C = (4π/γ)^(1/3). Their shape factors are γ = 4π/3 for a rigid sphere (C = 1.442), γ ≈ 1.6 for an elongated rubble pile after Porco and colleagues (C = 1.988), and γ ≈ 0.85 for an incompressible fluid after Chandrasekhar's work on equilibrium ellipsoids (C = 2.454). The tidal-term-only case corresponds to γ = 2π.
Rearranged for density, ρ_Roche = ρ_p (C R_p / d)³: the density a body must exceed to hold together at distance d. Because of the cube, halving the distance multiplies the required density by eight.
The primary is treated as a point mass, the satellite as uniform throughout and the orbit as circular. All four inputs must be greater than zero.
A worked example.
How close could the Moon come before tides tore it apart? JPL gives Earth a mean radius of 6,371.0084 km and a bulk density of 5.5134 g/cm³, and the Moon a bulk density of 3.344 g/cm³. The density ratio is 5.5134 ÷ 3.344 = 1.6487, and its cube root is 1.1814. For a fluid body the coefficient is (4π ÷ 0.85)^(1/3) = 2.4543. Their product, 2.4543 × 1.1814 = 2.8994, is the limit in Earth radii, and 2.89944 × 6,371.0084 km puts the Roche limit at 18,472.382 km from Earth's centre, about 12,100 km above the surface. The rubble-pile model gives 14,960.837 km, the rigid synchronous sphere 10,855.079 km and the tidal-term-only sphere 9,482.785 km. The real Moon orbits at 384,400 km, 60.3358 Earth radii and about 20.8 times the fluid limit. At that distance the critical density is 5.5134 × (2.4543 × 6,371.0084 ÷ 384,400)³ = 3.711 × 10⁻⁴ g/cm³, so a satellite there would need to be roughly 9,000 times less dense than the Moon before tides could pull it apart. The Moon is in no danger, and it is in fact slowly receding.
Frequently asked questions.
Why is the satellite's own size not one of the inputs?
Which satellite model should I choose?
Why do some books give 2.456 for the fluid coefficient instead of 2.454?
How do astronauts, spacecraft and small rocks survive inside the Roche limit?
What happens to a comet that passes inside a planet's Roche limit?
Does a satellite with a dense core break up at the same distance?
References& sources.
- [1]Tiscareno, M.S., Hedman, M.M., Burns, J.A. and Castillo-Rogez, J. (2013). 'Compositions and origins of outer planet systems: insights from the Roche critical density'. The Astrophysical Journal Letters 765, L28. Equation 1, a_Roche = R_p(4πρ_p/γρ)^(1/3); Equation 3, ρ_Roche = 3M_p/γa³; γ = 4π/3 ≈ 4.2 for a sphere, γ ≈ 1.6 for an elongated body after Porco et al. (2007), γ ≈ 0.85 for an incompressible fluid after Chandrasekhar (1969). Primary source for all four coefficients on this page. Peer-reviewed, open access. Retrieved 2026-07-29.
- [2]Leinhardt, Z.M., Ogilvie, G.I., Latter, H.N. and Kokubo, E. (2012). 'Tidal disruption of satellites and formation of narrow rings'. Monthly Notices of the Royal Astronomical Society; doi:10.1111/j.1365-2966.2012.21328.x. N-body result quoted on this page: 'the Roche limit for a rubble pile is closer to the planet than for a fluid body of the same mean density. The Roche limit for a differentiated body is also closer to the planet than for a homogeneous satellite of the same mean density.' Peer-reviewed; abstract open access, journal version paywalled. Retrieved 2026-07-29.
- [3]NASA JPL Solar System Dynamics, 'Planetary Physical Parameters': Earth volumetric mean radius 6371.0084 ± 0.0001 km and bulk density 5.5134 ± 0.0003 g/cm³; Saturn mean radius 58 232 km and bulk density 0.6871 g/cm³. Independent, open access. Retrieved 2026-07-29.
- [4]NASA JPL Solar System Dynamics, 'Planetary Satellite Physical Parameters': Earth's Moon, mean radius 1737.4 ± 0.1 km and mean density 3.344 ± 0.001 g/cm³, from ephemeris DE440. Independent, open access. Retrieved 2026-07-29.
- [5]NASA Science, 'Moon Facts' (page updated 12 February 2026): 'The Moon is an average of 238,855 miles (384,400 kilometers) away.' Source of the default orbital distance. Independent, open access. Retrieved 2026-07-29.
- [6]Chandrasekhar, S. (1969). Ellipsoidal Figures of Equilibrium. New Haven: Yale University Press — the equilibrium-ellipsoid analysis behind the fluid coefficient. Print/bibliographic reference, cited here through Tiscareno et al. (2013) rather than consulted directly.
How this page was produced
- Published by
- Quanta Calculator
- Overseen by
- Daniel, Quanta Calculator Researcher
- Primary sources
- 6 cited below
- Method
- d = R_p · (4π ρ_p ⁄ (γ ρ_s))^(1⁄3) = C · R_p · (ρ_p ⁄ ρ_s)^(1⁄3), C = (4π ⁄ γ)^(1⁄3), ρ_Roche = ρ_p (C·R_p ⁄ d)³
- Published
- Last verified
Built with AI assistance and verified by automated tests against the cited sources — every worked example on this page is computed by the same code that runs the calculator. How we build and check calculators.
In this category
Embed
Quanta Pro
Paid features are coming later.
- All 1560 calculators remain free
- No billing is enabled