Why planets speed up near the star

One rule — gravity, exactly, at every instant — decides the whole path. Watch the planet rocket past the star and then crawl through the far side of its loop: the imaginary line joining star and planet sweeps out equal patches of ground no matter where it is. Push the shape dial far enough, though, and the planet stops coming back at all.

What you're seeing

A planet feels one force: gravity from the star, pulling straight toward it, weaker the farther away it is. Nothing else — no friction, no steering. Out of that one rule falls the whole shape of the path, and a law that sounds fussy but is really just bookkeeping: draw a line from the star to the planet, and as the planet moves, that line sweeps out area. Watch the eight wedges on the stage — each one is timed to take exactly the same amount of time to sweep as the others. The wedges near the star are short and fat; the ones on the far side are long and thin. Same area, wildly different shapes. That's only possible if the planet moves fast where the wedge is short and slow where it's long — which is exactly what you'll see it do.

Now drag how lopsided the orbit is. At 0 the "orbit" is a perfect circle and nothing ever changes speed. Pull it up and the path stretches into an ellipse, with the star sitting off to one side — at a focus, not the center — which is itself part of why the near and far sides look so different. Push past 1 and something categorical happens: the path stops closing. Below 1 the planet swings out and is always pulled back, forever, on schedule. At exactly 1 it's thrown out with precisely enough speed to never quite stop receding — the exact edge between falling back and getting away. Past 1 it has speed to spare: one pass by the star and it's gone for good.

The rule, exactly. In units where the gravitational pull and the orbit's width at the star are both set to 1, the path is the classic conic section r(θ) = 1 ⁄ (1 + e·cos θ) and the angle advances at exactly the rate that keeps the swept area constant — dθ⁄dt = h ⁄ r², h = 1 — so equal-area sweeping isn't assumed, it's the literal update rule. For e<1 the orbit closes with period T = 2π·a^1.5 (Kepler's third law, a = 1⁄(1−e²)). Verified in node (improve/verify/54-kepler-orbits.js): the integrated period matches that closed form to within 0.2% at four eccentricities; a periapsis-side time-window and an apoapsis-side time-window of equal duration sweep areas agreeing to within 1% of each other and of the predicted h·Δt⁄2, despite one covering 26× the angle of the other; specific angular momentum measured independently from the simulated trajectory's own finite-difference velocity holds within 0.3% of 1 throughout. Counter-example: a "fake" orbit that traces the identical ellipse but sweeps angle at a constant rate instead — same shape, wrong time-law — sweeps 31× more area on its far side than its near side in equal time, so the equal-area law is a genuine consequence of the 1⁄r² rule, not just of having an elliptical curve. Escaping orbits (e≥1) never turn back: theta approaches its limiting angle asymptotically and distance from the star grows past 20× its closest approach without bound, over the same span in which every e<1 orbit already completed a full lap.

← the cabinet · Thought Toys — a cabinet of explorable explanations. Exhibit 54.