Concept

Trajectory optimisation — where it appears

The numerical search for the sequence of manoeuvres meeting a mission's constraints at least cost. It is what turns a candidate flyby sequence into a flyable trajectory, and it works against a full ephemeris rather than the circular coplanar model that generated the candidate.

Named by 2 essays across 2 fields — each of them below, with the objects they name alongside it.

5 transfers through the same two points in the same 1400 days. Time of flight against semi-major axis for every transfer through two points 135° apart at 1 and 1.524 AU, with the revolution count running from 0 to 2. Each count contributes two branches, and for N ≥ 1 the pair folds: the time has a minimum at a = 1.2426 AU for one revolution — only 2.2 per cent above the minimum-energy value of 1.2161, which is why the horizontal axis is the excess over that value and logarithmic — so a flight time above it is met twice and below it not at all. Reading the crossings of the 1400-day line off the drawn curves gives 5 of them — 0 revs high, 1 rev low, 1 rev high, 2 revs low, 2 revs high — which is 2N + 1 with N = 2, and the count is a property of the time rather than of the geometry. That is the practical content: a root-finder started from a single guess returns one of these 5 and gives no sign that the other 4 exist, and the cheapest of them is often not the one nearest the guess.

One time of flight and five ways round

Lambert's theorem says two positions and an interval fix the transfer. Allow the transfer to complete whole revolutions and that stops being true: the flight time folds, one number admits five arcs, and the cheapest of them is usually not the one a solver started nearest to.

orbits · Lambert's problem

Along a contour is free; across one costs, and 6 resonances sit on this one. Perihelion against aphelion, in units of Jupiter's orbit, with contours of constant v∞ — which is the Tisserand parameter through v∞² = (3 − T)v_p². Every contour crosses the line r_p = r_a = 1, because a spacecraft has to be at the planet's orbit to have an encounter there at all, and every point on one contour is reachable from every other point on it with no propellant: a flyby rotates the v∞ vector without changing its length, which moves the pump angle and slides the spacecraft along the curve. A burn is the only thing that moves it between curves, and that is the whole economy of a gravity-assist tour. The diagonals are resonant orbits — 1:1, 3:2, 2:1, 5:2, 3:1, 4:1 with the planet — and each is a straight line of slope −1 because a resonance fixes the semi-major axis and r_p + r_a = 2a. They matter because a spacecraft on one comes back to the same place at the same time as the planet, which is what makes a second encounter possible without waiting for a chance alignment. The marked points are where the v∞ = 0.3 contour meets each: a tour is a walk along the highlighted curve from one to the next, and the arithmetic that decides whether it can be walked is the turn one flyby delivers. At 1.35 Jupiter radii and 3.9 km s⁻¹ that turn is 163°, so the longest step drawn here needs 0.1 encounters — which is why a real tour has dozens of flybys and why the ones with the largest steps are the ones that need a deep-space manoeuvre in between.

The same planet, three times

A flyby cannot change the encounter speed, only its direction, so a tour has to be designed in the space of what is conserved. One pass moves a spacecraft along a single curve and no further than the planet can bend it — and reaching a distant target means walking that curve, returning to the same planet again and again.

spaceflight · Gravity assist

Named alongside it

The objects these essays reach for when they reach for this one.

Bifurcation Broken plane manoeuvre Crank angle A deep-space manoeuvre Departure energy Flyby sequence Lambert's problem Launch energy Minimum-energy transfer Multi-revolution transfer Parabolic limit Porkchop plot

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