Astronomy & Celestial Mechanics
the sky, computed
Concept

Root finding — where it appears

The numerical task of finding where a function equals zero, usually by stepping from a guess using the function's slope. How quickly and reliably it converges depends as much on the variable the function is written in as on the method.

Named by 2 essays across one field — 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 Lambert's time of flight in the semi-major axis and in Lancaster and Blanchard's variable. The non-dimensional time of flight T = t√(2μ/s³) of the transfer between radii 1 and 1.524 through 135 degrees, against two variables that label the same family of orbits. Left: against the semi-major axis in units of the minimum-energy ellipse's. The family is two curves — the fast branch, which passes the empty focus on one side, and the slow branch, on the other — joined at a = aₘ with T = 1.566, where both have a vertical tangent: the time changes infinitely fast with the semi-major axis there, and every a above aₘ corresponds to two times. Right: against x, the variable of Lancaster and Blanchard, in which a = aₘ/(1 − x²). The same family is one smooth, decreasing curve through T = 1.566 at x = 0 to the parabolic time 0.662 at x = 1, with no tangent vertical and every time given by exactly one x. Nothing about the orbits differs between the panels; only the label on them does.

The variable that makes Lambert's problem easy

Written in the semi-major axis, the time-of-flight equation has two branches joined by a vertical tangent, and a solver near the join struggles. Written in a variable Lancaster and Blanchard chose in 1969, it is one smooth curve — and every trajectory library now solves it that way.

orbits · Lambert's problem

Named alongside it

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

Lambert's problemMinimum-energy transferPorkchop plotTime of flightBifurcationChange of variableConditioningDeparture energyMulti-revolution transferNewton's methodParabolic limitSemi-major axis

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