Preview

My Solar System

Build and simulate your own multi-body solar system

In 1887 King Oscar II of Sweden offered a prize for answering the simplest question in astronomy: given N bodies pulling on each other with Newton's gravity, can you write down their motion forever? Henri Poincaré won the prize in 1890 — by proving the answer is no. Two bodies, Newton had already solved exactly: ellipses, clockwork, a universe you could price with Kepler's laws — and that price list has one signature: a circular orbit's speed carries the attracting mass under a square root, v = √(G·M/r), so more mass runs a faster orbit and a shorter clock, but only by the root, never in proportion. Three bodies, it turns out, is a different universe — no general formula exists, and trajectories diverge so sensitively that prediction itself has a horizon. But here is what Poincaré's no does not mean. It does not mean the books stop balancing: total energy and total angular momentum hold exactly, in the chaos, forever — only the trajectories escape prediction. It does not mean every three-body system flies apart: Jupiter's Trojan asteroids have been parked at the L4 and L5 points for billions of years, and three equal stars can even chase each other around a perfect figure-8. In this lab you get both universes side by side. The two-body clockwork first — a binary whose energy and angular momentum you can price to four decimals before it moves — then a multiplier slider that unbinds a planet the moment its speed crosses √2 times circular, and finally three equal suns tangling into chaos while the ledgers on the panel refuse to blink. The deep pattern under complex motion is not the trajectory. It's the conservation law.

What you'll be able to do

  • Price the two-body closed forms in the sim's own units (AU, years, M☉, G = 4π² — so v_circ at 1 AU around 1 M☉ is exactly 2π AU/yr = 29.7854 km/s, Earth's own speed, UP1 §13.4 Eq. 13.7): the equal-mass binary at d = 4 AU seeds each star at v = √(GM/2d) = π/√2 = 2.2214 AU/yr orbiting the common CM at r = d/2 (§13.4 Example 13.11), clocking T = 4√2 = 5.657 yr with panel books E = −π²/2 = −4.9348 and L = 2π√2 = 8.8858; doubling BOTH stars reseeds v = π (the preset's dead 3.14 literal, resurrected) and the clock falls to 4.000 yr — a ÷√2 from T² = 4π²a³/GM (§13.5 Eq. 13.11), never a halving
  • Read conservation as the deep pattern under complexity: on the chaotic three-body preset the Total Energy and Angular Mom. rows hold their seed values (−13.8766, −0.6000 — node-verified) while the trails tangle, because Poincaré's non-integrability (the HTML's own Historical card) forecloses closed-form trajectories, not conserved ledgers — only each body's PRIVATE orbital energy is fair game for exchange (the HTML quiz Q1's correct answer). The ΔE drift row measures integrator error, not physics ('if you see them drifting, that's integration error — a signal to use smaller time steps', HTML sidebar verbatim) — and at 1-AU-scale separations its unsoftened bookkeeping overstates that error by ~s²/2r² (registered)
  • Replace 'more bodies = more unstable' with structured stability: Jupiter's Trojans ride in two clouds preceding and following the planet — the L4/L5 equilateral configurations, 'controlled by the giant planet's gravity' (Astronomy 2e §13.1, verified) — and equal-mass three-body even admits exact stable choreographies (the 1993 figure-8, Moore PRL, Crossref-confirmed). The escape threshold is the other structural fact: v_esc = √2·v_circ at ANY radius (§13.4: 'exactly √2 times greater, about 40%'), so the Speed Multiplier slider unbinds Earth the moment it crosses 1.41 — the energy row's sign flip from negative (bound ellipse) through zero (parabola) to positive (hyperbola) is §13.5's conic classification read live

Formulas

F12=Gm1m2r122r^12\vec{F}_{12} = -G\frac{m_1 m_2}{r_{12}^2}\hat{r}_{12}
Gravitational force between bodies
a=jiFij/mi\vec{a} = \sum_{j\neq i}\vec{F}_{ij}/m_i
N-body acceleration

Make a prediction

Binary Star preset: two 1.0 M☉ stars, 4 AU apart, each clocking a 4√2 ≈ 5.657-year orbit around the shared center of mass. You drag BOTH mass sliders to 2.0 M☉ — the separation stays 4 AU and the sim reseeds both stars at the new circular speed. What happens to the period?

No grading here — commit to a guess, then scroll down and test it yourself.

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Check

Did the lab agree with you?

Your prediction

You skipped the prediction — jump back to the preview and commit to a guess first; the comparison is the whole point.

Answer: It falls to 4.000 yr — a ÷√2, not a halving: each star now pulls 2× harder, the circular speed rises √2× (2.2214 → π = 3.1416 AU/yr), and the same 4 AU circle closes in 5.657/√2 years

Each star orbits the shared center of mass at r = d/2 = 2 AU ('For equal masses, the center of mass is exactly half way between them' — OpenStax UP1 §13.4, Example 13.11), and the circular condition prices the speed: v = √(G·M/2d). Doubling M multiplies v by √2 — from 2.2214 to π = 3.1416 AU/yr, which is literally the binary preset's own dead 3.14 literal: it was the correct seed for m1 = 2 all along. The period T = 2πr/v therefore divides by √2: 4√2 = 5.657 → 4.000 yr exactly — the same T ∝ 1/√M that Kepler's third law states, T² = 4π²a³/GM (UP1 §13.5, Eq. 13.11). Option B is the linear instinct — and remarkably, the sim's own quiz Q3 institutionalizes it: it marks 'Be halved' correct with the explanation 'multiplies T by 1/√2, which halves it,' a sentence that contradicts itself (1/√2 ≈ 0.707 ≠ 0.5; a registered bug this gate now teaches correctly). The pull really does double per star — force goes as M², acceleration as M — but speed answers with a square root, so the clock moves by √2, not 2. Option C overcorrects from a true fact: the ORBITING body's mass cancels out of v = √(GM/r) ('not upon the mass of the object being acted upon' — §13.4). But the mass you doubled is the source of gravity itself, and the source is very much in the law. Watch the panel as you drag: the energy row deepens −4.9348 → −19.7392 (×4, E ∝ M²) and the angular momentum row climbs 8.8858 → 25.1327 (×2√2, L ∝ M^1.5) — the books reprice instantly, by powers of √M.

Quiz (0/3)

Create a stable figure-8 orbit with three equal masses. What initial conditions are needed?

Add Jupiter (large planet) to a solar system with Earth. How does it perturb Earth's orbit?

Why are chaotic 3-body systems practically unpredictable over long timescales?

You can now

  • Price the two-body closed forms in the sim's own units (AU, years, M☉, G = 4π² — so v_circ at 1 AU around 1 M☉ is exactly 2π AU/yr = 29.7854 km/s, Earth's own speed, UP1 §13.4 Eq. 13.7): the equal-mass binary at d = 4 AU seeds each star at v = √(GM/2d) = π/√2 = 2.2214 AU/yr orbiting the common CM at r = d/2 (§13.4 Example 13.11), clocking T = 4√2 = 5.657 yr with panel books E = −π²/2 = −4.9348 and L = 2π√2 = 8.8858; doubling BOTH stars reseeds v = π (the preset's dead 3.14 literal, resurrected) and the clock falls to 4.000 yr — a ÷√2 from T² = 4π²a³/GM (§13.5 Eq. 13.11), never a halving
  • Read conservation as the deep pattern under complexity: on the chaotic three-body preset the Total Energy and Angular Mom. rows hold their seed values (−13.8766, −0.6000 — node-verified) while the trails tangle, because Poincaré's non-integrability (the HTML's own Historical card) forecloses closed-form trajectories, not conserved ledgers — only each body's PRIVATE orbital energy is fair game for exchange (the HTML quiz Q1's correct answer). The ΔE drift row measures integrator error, not physics ('if you see them drifting, that's integration error — a signal to use smaller time steps', HTML sidebar verbatim) — and at 1-AU-scale separations its unsoftened bookkeeping overstates that error by ~s²/2r² (registered)
  • Replace 'more bodies = more unstable' with structured stability: Jupiter's Trojans ride in two clouds preceding and following the planet — the L4/L5 equilateral configurations, 'controlled by the giant planet's gravity' (Astronomy 2e §13.1, verified) — and equal-mass three-body even admits exact stable choreographies (the 1993 figure-8, Moore PRL, Crossref-confirmed). The escape threshold is the other structural fact: v_esc = √2·v_circ at ANY radius (§13.4: 'exactly √2 times greater, about 40%'), so the Speed Multiplier slider unbinds Earth the moment it crosses 1.41 — the energy row's sign flip from negative (bound ellipse) through zero (parabola) to positive (hyperbola) is §13.5's conic classification read live

Step-by-step

  1. Add bodies by clicking.
  2. Set mass and initial velocity using the control panel.
  3. Run the simulation and watch orbits form.
  4. Add a third body to observe orbital perturbations.
  5. Try creating a stable binary star with two equal masses.

Key formulas

  • F12=Gm1m2r122r^12\vec{F}_{12} = -G\frac{m_1 m_2}{r_{12}^2}\hat{r}_{12}Gravitational force between bodies
  • a=jiFij/mi\vec{a} = \sum_{j\neq i}\vec{F}_{ij}/m_iN-body acceleration

Frequently asked questions

Create a stable figure-8 orbit with three equal masses. What initial conditions are needed?
Specific symmetric initial conditions; try equal masses at 120° with tangential velocities.
Add Jupiter (large planet) to a solar system with Earth. How does it perturb Earth's orbit?
Jupiter's gravity slightly shifts Earth's orbital elements — this is real! Jupiter acts as a 'vacuum cleaner'.
Why are chaotic 3-body systems practically unpredictable over long timescales?
Lyapunov exponents grow exponentially; initial precision required grows exponentially.