Astronomy & Celestial Mechanics
the sky, computed
Concept

Minimum-energy transfer — where it appears

The transfer orbit of least semi-major axis, and therefore least orbital energy, that passes through two given points. It is where the two branches of Lambert's solution meet, and it is the limit every multi-revolution transfer crowds up against.

Named by 3 essays across one field — each of them below, with the objects they name alongside it.

Flight time against semi-major axis, for a fixed 135° sweep. Lambert's theorem drawn: the time to fly between two points 1 and 1.524 AU out and 135° apart, against the semi-major axis of the orbit that does it. Nothing else about the orbit enters — not its eccentricity, not where its periapsis is, not how it is oriented — which is the content of the theorem and the reason a two-point transfer is a one-dimensional search rather than a six-dimensional one. Two branches: the lower one is the ellipse whose arc stays short of apoapsis, falling towards the parabolic floor at 103.2 days as a grows without limit; the upper one is the ellipse of the same size whose arc runs through apoapsis, rising without limit. They meet at a = s/2 = 1.2161 AU, 244.2 days, which is the minimum-energy transfer and the slowest ellipse available — every faster one is bigger. Each branch is monotone, checked point by point across the drawn range, so a horizontal line cuts each at most once: for a given pair of points and a given time there is exactly one ellipse, and at 260 days it is a = 1.2189 AU on the upper branch. The freedom a mission designer has is not in this picture: it is the choice of the two points, which is what a porkchop plot sweeps.

Two places and a clock decide the path

The time to fly between two points depends on the semi-major axis, the chord between them, and the sum of their distances — and on nothing else about the orbit. Not the eccentricity, not where periapsis is, not the orientation. Lambert's theorem is why an interplanetary launch date is the root of one equation.

orbits · Lambert's problem 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 problemPorkchop plotRoot findingSemi-major axisTime of flightBifurcationChange of variableCharacteristic energyConditioningConic sectionsDeparture energyHohmann transfer

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