N-Body Gravity (2D)

Newton's law of gravitation

Every body attracts every other: ai = G Σj≠i mj (rj - ri) / (|rj - ri|² + ε²)3/2. The page sums all N(N-1)/2 pairs directly each step, so cost grows as N². The softening length ε (0 by default for small systems) replaces the singular 1/r² force at very small separations; it is how cluster codes survive close encounters, and the potential -G mimj/sqrt(r²+ε²) is changed consistently so energy is still conserved.

Conserved quantities

Total energy E = K + U, total momentum P = Σ m v (so the centre of mass moves in a straight line) and angular momentum L = Σ m r × v are conserved by the exact equations. How well a numerical method keeps them is the quality measure shown in the graphs. The virial ratio 2K/|U| equals 1 for a bound system in equilibrium.

Integrators

Leapfrog (kick-drift-kick) is second order, time-reversible and symplectic: its energy error oscillates instead of growing, which is why it is the workhorse for orbits. Yoshida 4th order composes three leapfrog steps with coefficients w1 = 1/(2-21/3), w0 = -21/3 w1 for fourth-order accuracy and the same symplectic structure. Runge-Kutta 4 is very accurate per step but not symplectic: over very long runs energy drifts slowly. Explicit Euler is first order and adds energy every step: planets spiral outwards, which you can watch by selecting it. Use "Reverse time" to see that leapfrog retraces its path almost exactly.

Famous solutions

The figure-eight (Chenciner and Montgomery, 2000) is a periodic orbit of three equal masses chasing each other along one curve with period T = 6.32591398 in units G = m = 1. The Lagrange triangle rotates rigidly with ω² = 3Gm/s³. The Pythagorean 3-4-5 problem starts three bodies at rest at the corners of a 3-4-5 triangle and ends in a chaotic ejection: tiny changes in the time step change the outcome, which is a sign of the sensitivity of the three-body problem (it has no general closed-form solution). The solar-system preset uses AU, years and solar masses, where G = 4π².

Model checks (run in the page)

1. A circular two-body orbit must have the Kepler period T = 2π sqrt(a³/(G(M+m))). 2. An eccentric orbit (e = 0.6) must keep energy and angular momentum and have perihelion a(1-e). 3. The figure-eight must return to its starting positions after one period, with zero total momentum.

Honest simplifications

Bodies are point masses (no collisions, tides, relativity or non-gravitational forces); the fixed time step can be too coarse for very close encounters, where energy errors jump, and you can see this in the error graph. The solar-system preset ignores the other planets and the Moon, and 3D inclinations are small. Display sizes are exaggerated and not to scale. The Plummer cluster is a random statistical model, not a specific real cluster.