In 1772, Joseph-Louis Lagrange, while studying the restricted three-body problem, discovered five equilibrium points where the gravitational forces of two massive bodies balance the centrifugal force in the rotating reference frame.

What Are Lagrange Points?

In the rotating reference frame co-moving with two massive bodies, there are five points where a body of negligible mass can remain stationary relative to the massive bodies. At these points, the gravitational attraction of the massive bodies and the centrifugal force cancel each other out.

Collinear Points L₁, L₂, L₃

Three points lie on the line connecting the two massive bodies. L₁ is between the bodies, L₂ is beyond the smaller body, L₃ is beyond the larger body. All three are unstable: small perturbations cause the body to drift away. However, L₁ and L₂ are actively used for space missions (James Webb Telescope at L₂, SOHO at L₁).

Triangular Points L₄ and L₅

Two points form equilateral triangles with the two massive bodies. L₄ lies 60° ahead of the smaller body in its orbit, L₅ 60° behind. When the mass ratio μ = m₂/(m₁ + m₂) < 0.0385, these points are stable. In the Solar System, this explains the Trojan asteroids — thousands of asteroids at the L₄ and L₅ points of the Jupiter—Sun system, as well as at other planets.

\mu = \frac{m_2}{m_1 + m_2} = \frac{10}{510} \approx 0.02 < 0.0385

Significance for Space Exploration

L₁ and L₂ are widely used for space observatories and probes due to their stable position relative to Earth and the Sun. The James Webb Space Telescope is at the Sun—Earth L₂ point. L₄ and L₅ are important for understanding Solar System dynamics and planetary system evolution.

Example: L₄ and L₅ in the Simulation

Consider a binary system with masses m₁ = 500 and m₂ = 10. The mass ratio of the smaller body to the total is μ = m₂/(m₁ + m₂) = 10/510 ≈ 0.02. For the triangular Lagrange points to be stable, μ must be less than 0.0385 (Routh's criterion); our value of 0.02 satisfies this condition. The distance between bodies is a = 60. The angular velocity of the system:

\omega = \sqrt{\frac{G(m_1 + m_2)}{a^3}} = \sqrt{\frac{80 \cdot 510}{60^3}} = \sqrt{0.1889} \approx 0.435

Point L₄ is at the vertex of an equilateral triangle with sides a = 60. Its coordinates and velocity in the inertial reference frame:

\begin{aligned} L_4 &= \left(\frac{x_1 + x_2}{2},\; a\frac{\sqrt{3}}{2}\right) = (28.82,\; 51.96) \\ \mathbf{v}_{L_4} &= \boldsymbol{\omega} \times \mathbf{r}_{L_4} = (-22.58,\; 12.53) \end{aligned}