3-Body Problem Simulator

chaotic gravity, twin-run divergence, and numerical stability

Worked example — 3-Body Problem Simulator

Kinetic energy	1.2129
Potential energy	−2.4996
Total energy	−1.2867
Quantity	0

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

Loading interactive simulation...

why does it look the same on every refresh? 🖖

The three-body problem is chaotic but not random. Chaos means tiny differences in initial conditions grow exponentially — but refreshing the page changes nothing: same initial positions, same velocities, same dt, same algorithm, same result every time. Deterministic chaos and computational reproducibility are two different things. To actually see the chaos: set perturbation to 0.000001 and watch the dashed twin trajectory. Both trajectories start 10⁻⁶ apart and eventually diverge into completely different paths. That exponential separation is the chaos — the Lyapunov exponent at work. The divergence chart below the orbits plots log₁₀ of that separation over time; a straight rising line means exponential growth.

why three is the hard number 🖖

Two bodies pulling on each other trace simple, endlessly repeating ellipses — that case Newton solved with pen and paper. Add a third body and the neat formula disappears: there is no general equation for where everyone ends up, so the only way forward is to advance the simulation one tiny dt at a time, exactly as this tool does. Click the figure-8 preset to meet one of the rare exceptions — three equal masses forever chasing each other along a single looping path.

the solution that exists but is useless 🖖

People often call the three-body problem 'unsolvable', yet in 1912 Karl Sundman proved a convergent power series in t^1/3 that yields the bodies' positions for all time (barring triple-collision starts). The catch: it converges so slowly that astronomical accuracy would demand on the order of 10^8000000 terms — hopelessly beyond any computer. A formal solution truly exists; stepping the simulation forward, as here, is simply the only usable route.

Problem solved in full

Energy of the figure-8 choreography for 3 equal masses from initial conditions 7 steps
The figure-8 choreography sends 3 equal masses around a single closed curve. Work out its energy at t = 0 from the initial conditions alone. This is the figure-8 orbit preset: G = 1, m₁ = m₂ = m₃ = 1, softening ε = 0. Positions r₁ = (−0.97000436, 0.24308753), r₂ = −r₁, r₃ = (0, 0); velocities v₁ = v₂ = (0.466203685, 0.43236573), and the total momentum is 0.

Open this problem in the tool
1. Those long decimals look arbitrary. Square them and add, and they are not: bodies 1 and 2 each sit exactly 1 unit from the origin, which is where body 3 is.
  
  $∣ r_{1} ∣ = 0.9700043 6^{2} + 0.2430875 3^{2} = 0.940908 + 0.059092 = 1.0000$
2. Body 2 is body 1 reflected through the origin, so at this instant all 3 bodies lie on a single straight line with body 3 at its midpoint. Every separation follows from that, with no second square root.
  
  $r_{2} = - r_{1} ⟹ r_{13} = r_{23} = 1.0000, r_{12} = 2.0000$
3. Newton's law now needs nothing but those separations. Body 3 is 1 unit from each of the others, while bodies 1 and 2 are twice as far apart, so the pull between them is weaker by a factor of 4.
  
  $F_{ij} = \frac{G m _{i} m _{j}}{r _{ij}^{2}} ⟹ F_{12} = \frac{1}{2 ^{2}} = 0.2500, F_{13} = F_{23} = 1.0000$
4. Spend the collinearity. Body 3 is pulled equally in opposite directions, so its net force vanishes exactly. Bodies 1 and 2 are pulled the same way by both partners, because body 3 sits between them, so their 2 forces add as plain numbers with no components to resolve.
  
  $∣ F_{3} ∣ = 1.0000 - 1.0000 = 0.0000, ∣ F_{1} ∣ = 1.0000 + 0.2500 = 1.2500$
5. Zero total momentum, plus bodies 1 and 2 moving identically, pins body 3's velocity at −2v₁. That makes its kinetic energy 4 times theirs, and the sum over all 3 collapses to a single term.
  
  $\sum m_{i} v_{i} = 0, v_{1} = v_{2} ⟹ v_{3} = - 2 v_{1}, K E = \frac{1}{2} (1 + 1 + 4) v_{1}^{2} = 3 v_{1}^{2}$
6. Squaring the 2 components of v₁ and tripling is the whole of the kinetic-energy calculation for 3 bodies.
  
  $v_{1}^{2} = 0.46620368 5^{2} + 0.4323657 3^{2} = 0.404286 ⟹ K E = 1.2129$
7. Potential energy needs no vectors either — just the reciprocals of the 3 separations, and with ε = 0 those reciprocals are exact.
  
  $P E = - G (\frac{m _{1} m _{2}}{2} + \frac{m _{1} m _{3}}{1} + \frac{m _{2} m _{3}}{1}) = - 2.5000$
Answer

E = −1.2871, and every row of the panel's t = 0 table — separations, pair forces, net forces, both energies and the total — came out of those 4 decimals and the vanishing of the total momentum. Watch what happens to the parts: by t = 25.0 the kinetic energy has climbed to 1.305 and the potential has fallen to −2.592, and the total has not moved. The sign is what carries a consequence. A bound trio can still lose a member, but not for free: if 1 body drifts away with nothing to spare, the pair left behind has to hold the entire −1.2871, and 2 unit masses with that energy have a semi-major axis of 1/(2 × 1.2871) = 0.3885 — under 1/5 of the 2.0000 that separates them now. Ejection and hardening are the same event, and the energy budget is why.

$E = 1.2129 - 2.5000 = - 1.2871 ⟹ a_{m a x} = \frac{G m ^{2}}{2∣ E ∣} = \frac{1}{2 ( 1.2871 )} = 0.3885$

References (1)

Sundman's convergent series in t^(1/3): K. F. Sundman, "Mémoire sur le problème des trois corps." Acta Mathematica 36, 105–179, 1913.

YouTube (1)

StarTalk·Neil deGrasse Tyson Explains The Three-Body Problem

By the 42Math team · Reviewed against published sources. Corrections welcome.