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.
  
  $|\mathbf{r}_1|=\sqrt{0.97000436^{2}+0.24308753^{2}}=\sqrt{0.940908+0.059092}=1.0000$
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.
  
  $\mathbf{r}_2=-\mathbf{r}_1\quad\Longrightarrow\quad r_{13}=r_{23}=1.0000,\qquad r_{12}=2.0000$
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.
  
  $F_{ij}=\frac{Gm_im_j}{r_{ij}^{2}}\quad\Longrightarrow\quad F_{12}=\frac{1}{2^{2}}=0.2500,\qquad F_{13}=F_{23}=1.0000$
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.
  
  $|\mathbf{F}_3|=1.0000-1.0000=0.0000,\qquad |\mathbf{F}_1|=1.0000+0.2500=1.2500$
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.
  
  ${\textstyle\sum}m_i\mathbf{v}_i=0,\ \mathbf{v}_1=\mathbf{v}_2\ \Longrightarrow\ \mathbf{v}_3=-2\mathbf{v}_1,\ \ KE=\tfrac12(1+1+4)v_1^{2}=3v_1^{2}$
6. Squaring the 2 components of v₁ and tripling is the whole of the kinetic-energy calculation for 3 bodies.
  
  $v_1^{2}=0.466203685^{2}+0.43236573^{2}=0.404286\quad\Longrightarrow\quad KE=1.2129$
7. Potential energy needs no vectors either — just the reciprocals of the 3 separations, and with ε = 0 those reciprocals are exact.
  
  $PE=-G\left(\frac{m_1m_2}{2}+\frac{m_1m_3}{1}+\frac{m_2m_3}{1}\right)=-2.5000$
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.

$E=1.2129-2.5000=-1.2871\quad\Longrightarrow\quad a_{\max}=\frac{Gm^{2}}{2|E|}=\frac{1}{2(1.2871)}=0.3885$

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 7 Raven Room team · Reviewed against published sources. Corrections welcome.

3-Body Problem Simulator

why does it look the same on every refresh? 🖖

why three is the hard number 🖖

the solution that exists but is useless 🖖

why does it look the same on every refresh?

why three is the hard number

the solution that exists but is useless