Lagrange Point Stability Explorer

small perturbations reveal stable and unstable equilibria in rotating-frame dynamics

Worked example — Lagrange Point Stability Explorer

Mass ratio μ	0.01215
Stability criterion	0.3241<1
Discriminant D	0.6759
Initial conditions	(0.02,0)
Peak displacement	1.784
Perturbation doubling	1.08orbits
Behavior	diverging

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

Loading interactive simulation...

L4 is a hilltop that traps you anyway 🖖

L4 and L5 are not basins. In the rotating frame they sit at maxima of the effective potential — set a marble there and it rolls away. What holds a Trojan is the Coriolis force, which bends the escape into a path that closes on itself, and the tool's own output carries the signature: a stable triangular point has two libration periods, not one. Choose Sun–Earth and the short one is 1.00 orbit while the long one is 222 orbits — a 222-year drift around a point the particle is supposedly sitting still at. For Earth–Moon the pair is 1.05 and 3.35 orbits. Both systems qualify because stability requires 27μ(1−μ) < 1, a mass ratio above 24.96, and Earth outweighs the Moon 81 to 1.

Reading a nudge in the rotating frame 🖖

Imagine watching from a merry-go-round that turns along with the two bodies, so the Lagrange point sits perfectly still. This tool gives a test particle a tiny push (ε) plus a small sideways speed, then traces where it goes. At a stable point the path curls into slow loops that never wander far; at an unstable one the distance keeps doubling until the particle escapes. Try a nudge at L4 versus one at L1 to feel the difference.

When instability becomes a free highway 🖖

The collinear points here are unstable — yet that very flaw is exactly why spacecraft love them. Because a saddle point has an escape direction, a probe can drift away along it using almost no fuel, riding invisible tubes called the Interplanetary Transport Network. NASA's Genesis mission surfed these low-energy pathways through the Sun-Earth L1 and L2 points, trading speed for enormous fuel savings.

Common wrong intuition

Lagrange points are not all equally stable. L1/L2/L3 generally need station-keeping; L4/L5 can be naturally stable in suitable mass ratios.

Problem solved in full

Linear stability of the triangular points L4 and L5 5 steps
Earth is 5.972 × 10²⁴ kg and the Moon 7.346 × 10²² kg. Routh's criterion decides whether the triangular points L4 and L5 of a two-body system are linearly stable, and it asks for exactly one number about the pair. Work it out for the Earth-Moon system, then find the mass ratio at which the criterion stops holding.
1. The two masses enter only through the fraction of the total that the smaller one carries. The Moon is a heavy secondary by planetary standards — over 1% of the pair — so this is not a case where μ can be treated as negligible.
  
  $μ = \frac{M _{2}}{M _{1} + M _{2}} = \frac{7.346 \times 1 0 ^{22}}{6.045 \times 1 0 ^{24}} = 0.01215$
2. Linearise the rotating-frame equations of motion about L4 and the four eigenvalues λ satisfy a quartic with no odd powers. That is a quadratic in disguise: put s = λ² and there are two roots to think about instead of four.
  
  $λ^{4} + λ^{2} + \frac{27}{4} μ (1 - μ) = 0 s = λ^{2} s^{2} + s + \frac{27}{4} μ (1 - μ) = 0$
3. λ = ±√s, so a negative s gives a purely imaginary pair and e^λt neither grows nor decays. Read what the coefficients already guarantee: the roots sum to −1 and multiply to a positive number, so if they are real, both are negative with no further work. Only the square root can spoil it.
  
  $s_{\pm} = \frac{- 1 \pm 1 - 27 μ ( 1 - μ )}{2}, s_{+} + s_{-} = - 1, s_{+} s_{-} > 0$
4. So the whole criterion is one inequality — 27μ(1−μ) must stay below 1. The Earth-Moon pair puts it at 0.3241, 32% of the limit.
  
  $27 μ (1 - μ) = 27 \times 0.01215 \times 0.9879 = 0.3241 < 1$
5. What is left under the square root is the discriminant, 0.6759, and it is positive: the two roots for s come out real, at −0.0889 and −0.9111. Both negative, as the coefficients promised.
  
  $D = 1 - 0.3241 = 0.6759 ⟹ s = - 0.0889, - 0.9111$
Answer

μ = 0.01215, 27μ(1−μ) = 0.3241, D = 0.6759. The question worth asking is where that 32% runs out. Set 27μ(1−μ) = 1 and you have a quadratic in μ: 27μ² − 27μ + 1 = 0, whose relevant root is (27 − √621)/54 = 0.03852. As a mass ratio, the primary must outweigh the secondary by 24.96 to 1. Earth outweighs the Moon by 81.30 to 1, so the Moon could be 3.26 times heavier before the criterion failed — the margin is large, but it is a margin, not an inequality that holds for every pair. Pluto and Charon are the pair that breaks it: 8.2 to 1 gives μ = 0.1085, 27μ(1−μ) = 2.612 and a discriminant of −1.612. A negative discriminant makes both roots for s complex, and the four square roots of a complex conjugate pair always include one with a positive real part.

$27 μ (1 - μ) = 1 ⟹ μ_{crit} = \frac{27 - 621}{54} = 0.03852 ⟹ \frac{M _{1}}{M _{2}} > 24.96$

References (3)

The Routh criterion and the two libration frequencies at L4/L5: C. D. Murray and S. F. Dermott, Solar System Dynamics, §3.9. Cambridge University Press, 1999. ISBN 978-0-521-57597-3.
The restricted three-body problem in full, including why L4/L5 are potential maxima: V. Szebehely, Theory of Orbits: The Restricted Problem of Three Bodies. Academic Press, 1967.
The low-energy transfer network named in the third block: W. S. Koon, M. W. Lo, J. E. Marsden and S. D. Ross, "Heteroclinic connections between periodic orbits and resonance transitions in celestial mechanics." Chaos 10(2), 427–469, 2000.

By the 42Math team · Updated 2026-07-15

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

All Astronomy & Space lessons →

Example problems

Earth-Moon L1 - L1 perturbations typically diverge from equilibrium.
Earth-Moon L4 - L4 perturbations can remain bounded and circulate nearby.
Earth-Sun L2 - L2 is dynamically useful but intrinsically unstable without station-keeping.
Earth-Sun L5 - L5 often shows bounded motion under small perturbations.

YouTube

StarTalk·What are the Points of Lagrange?