The Two-Body Problem

Two objects pulling on each other by gravity — a star and its planet, a planet and its moon, two stars — make up the two-body problem, and it is the one gravitational problem physics can solve completely and exactly. Both bodies orbit their shared centre of mass, or barycentre, each tracing an ellipse; the heavier body stays on a tighter loop, the lighter one swings wide.^[1]

Mass ratio μ = m₂/(m₁+m₂) = 0.20

Two bodies always orbit their shared centre of mass (the dot at the middle) — the lighter one swings on a wider circle. Ride along in the rotating frame and five points hold still: the Lagrange points, where the two gravities plus the orbital (centrifugal) pull cancel. L1–L3 sit on the line through both bodies and are unstable — a nudge and you drift off, so spacecraft there (SOHO at Sun–Earth L1, JWST at L2) burn fuel to hold station. L4 and L5 lead and trail by 60°, forming equilateral triangles; they are stable when one body is at least ~25× the other (μ < 0.0385), which is why asteroids pile up at the Sun–Jupiter L4/L5 as Trojans. Slide μ and watch the barycentre and every Lagrange point shift.

The trick that makes it solvable is that the two-body problem collapses into a one-body problem. Track only the separation between the masses and the pair behaves like a single object of reduced mass $\mu = \dfrac{m_1 m_2}{m_1 + m_2}$ orbiting a fixed centre — and that is exactly Kepler's problem, whose solutions are the familiar ellipses, parabolas, and hyperbolas.

Deep dive · The five Lagrange points

Put a third, negligibly light object into a two-body system and ask where it could sit still relative to the orbiting pair. In the frame that rotates with them, five such points exist — the Lagrange points — where the two gravitational pulls and the centrifugal effect of the rotation exactly cancel.^[2]

Three of them — L1, L2, L3 — lie on the line joining the two bodies. They are found by balancing gravity against the orbital pull along that line,

x - \frac{(1-\mu)\,(x - x_1)}{|x - x_1|^3} - \frac{\mu\,(x - x_2)}{|x - x_2|^3} = 0,

(written in units where the total mass and separation are 1, with $\mu = m_2/(m_1+m_2)$ ). All three are unstable saddle points: drift a little and you keep drifting, so real spacecraft — SOHO at Sun–Earth L1, the James Webb Space Telescope at L2 — must fire thrusters periodically to stay put.

The other two, L4 and L5, sit 60° ahead of and behind the smaller body, each forming an equilateral triangle with the two masses. Remarkably they are stable whenever one body is at least about 25 times heavier than the other ( $\mu < 0.0385$ ): a nudge sets the object circling the point rather than escaping. That is why swarms of asteroids — the Trojans — collect at Jupiter's L4 and L5.

Add even a third comparable mass, though, and this clean predictability collapses. That is the notorious three-body problem.