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Unit 9 · Lesson 9-1
Lambert's Problem: Formulation and Geometry
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Lambert's Problem: Formulation and Geometry

The boundary-value problem, the transfer triangle, Lambert's theorem and Lagrange's time-of-flight equation

Palermo, 1 January 1801
Giuseppe Piazzi finds a faint moving point of light in Taurus. He follows it for six weeks, then loses it in the glare of the Sun. The object — Ceres, the first asteroid — is expected to reappear late in the year, somewhere in a patch of sky many degrees wide, and nobody knows where to point a telescope. A 24-year-old mathematician in Brunswick, Carl Friedrich Gauss, takes Piazzi's handful of positions and asks a question that turns out to be the central problem of this unit: given where a body was at two different times, which orbit connects those two positions in exactly that interval? His prediction sends the astronomers to the right spot. On 31 December 1801 Ceres is recovered, almost exactly where Gauss said it would be. Every unit so far has run time forwards: here is the state, integrate. This one runs it the other way. You know two positions and the clock between them; the velocity is the unknown. That is Lambert's problem, and it is underneath every rendezvous, every interception and every launch to another planet.
Watch first · presented by Shoaib Aziz
The numbers in this film are the labs' own: an interceptor 7,000 km from Earth's centre aims at a point 8,000 km out a quarter turn ahead, so the chord is 10,630 km and the minimum-energy ellipse has a = 6,407.5 km (29 km more than Earth's radius) and takes 41.19 min, while the parabola needs 16.78 min; 20 minutes gives a = 14,344 km, e = 0.537 and a 2.34 km/s burn, and 30 minutes gives a = 6,929 km, e = 0.16 and 1.21 km/s, an ellipse whose perigee is 558 km below the ground but never on the arc that is flown; the fast branch at 1,000 times the minimum size gives 16.79 min, and 41.19 + 43.88 = 85.07 min, exactly one period.
Pauca sed matura. (Few, but ripe.)
– The motto of Carl Friedrich Gauss, who published little and only when it was finished
Statement. Given two position vectors r⃗1\vec{r}_1, r⃗2\vec{r}_2 relative to an attracting centre with gravitational parameter μ\mu, a time of flight Δt\Delta t, a direction of motion (prograde or retrograde) and a number of complete revolutions NN, find the conic that passes through r⃗1\vec{r}_1 at t1t_1 and r⃗2\vec{r}_2 at t1+Δtt_1 + \Delta t. The output is the velocity at each end, v⃗1\vec{v}_1 and v⃗2\vec{v}_2.
Compare with propagation (in the unit on Kepler's equation and orbit propagation): there, r⃗1\vec{r}_1 and v⃗1\vec{v}_1 were given and r⃗2\vec{r}_2 was computed. That is an initial-value problem and always has exactly one answer. Lambert's problem is a two-point boundary-value problem: half the state is given at each end. It can have no solution, one, or several — and finding them requires iteration.
💡The transfer triangle
Three points fix everything: the focus FF and the two endpoints P1P_1, P2P_2. They form a triangle with sides r1r_1, r2r_2 and the chord c=∣r⃗2−r⃗1∣=r12+r22−2r1r2cos⁡Δθc = |\vec{r}_2 - \vec{r}_1| = \sqrt{r_1^2 + r_2^2 - 2 r_1 r_2 \cos\Delta\theta} and its semiperimeter s=(r1+r2+c)/2s = (r_1 + r_2 + c)/2. The transfer angle Δθ\Delta\theta is the angle swept at the focus. From the vectors alone you get cos⁡Δθ=r^1⋅r^2\cos\Delta\theta = \hat{r}_1\cdot\hat{r}_2, which cannot tell 90°90° from 270°270°. The direction of motion settles it: for a prograde transfer, if (r⃗1×r⃗2)z≥0(\vec{r}_1 \times \vec{r}_2)_z \ge 0 the short way is the right way; otherwise the motion goes the long way round and Δθ→360°−Δθ\Delta\theta \to 360° - \Delta\theta. Transfers with Δθ<180°\Delta\theta < 180° are short-way (Type I in interplanetary work); those above are long-way (Type II).
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Where can the second focus be? For an ellipse, the distances from any point to the two foci add up to 2a2a. So the empty focus F∗F^* must sit at distance 2a−r12a - r_1 from P1P_1 and 2a−r22a - r_2 from P2P_2: on the intersection of two circles. Two circles meet in two points, one on each side of the chord — hence, for any admissible aa, two different ellipses pass through both endpoints.
As aa shrinks the circles shrink, and at one value they just touch, on the chord itself. That is the minimum-energy ellipse. The circles touch when (2a−r1)+(2a−r2)=c(2a - r_1) + (2a - r_2) = c, i.e.
am=r1+r2+c4=s2a_m = \frac{r_1 + r_2 + c}{4} = \frac{s}{2}

Below a_m the two circles do not meet and no ellipse can contain both points. The minimum-energy ellipse has the smallest |ε| = μ/2a of any bound transfer — but, careful, not necessarily the smallest Δv: Δv depends on the velocities you start and end with, not on the transfer energy alone.

💡Lambert's theorem (1761)
The flight time between two points on a conic depends on only three quantities: the semi-major axis aa, the chord cc, and the sum of the radii r1+r2r_1 + r_2 (equivalently ss and cc). Not the eccentricity. Not the orientation. Not where the endpoints sit relative to perigee. Two triangles with the same r1+r2r_1 + r_2 and the same chord — however differently shaped — produce the same flight time for the same aa. Johann Heinrich Lambert stated it in 1761 while working on comet orbits, with a geometric argument. Lagrange gave the analytic proof in 1778. Its practical meaning is enormous: a search over five orbital elements collapses into one scalar equation with one unknown.
Lagrange's form of the theorem is explicit. Define two auxiliary angles from the triangle:
sin⁡α2=s2a,sin⁡β2=s−c2a\sin\frac{\alpha}{2} = \sqrt{\frac{s}{2a}}, \qquad \sin\frac{\beta}{2} = \sqrt{\frac{s-c}{2a}}

Principal values α₀, β₀ in [0, π]. The branch rules below decide which multiple of them applies.

μ Δt=a3/2[(α−sin⁡α)−(β−sin⁡β)]\sqrt{\mu}\,\Delta t = a^{3/2}\Big[(\alpha - \sin\alpha) - (\beta - \sin\beta)\Big]

Lagrange's time-of-flight equation. Short way: β = β₀; long way (Δθ > 180°): β = −β₀. Fast arc (empty focus beyond the chord, Δt below the minimum-energy time): α = α₀; slow arc (Δt above it): α = 2π − α₀.

💡Four branches, one curve
Fix the two points and plot Δt\Delta t against aa. On each side (short or long way) the curve has two arms meeting at ama_m: the fast arm falls from the minimum-energy time towards the parabolic limit as a→∞a \to \infty; the slow arm climbs without bound. A requested Δt\Delta t therefore lands on exactly one arm of each side, and a bracketing root-finder on aa finds it. A useful check: on the minimum-energy ellipse, the short-way time plus the long-way time equals one full period. The two arcs together make the whole ellipse.
The parabolic boundary. As a→∞a \to \infty the fast arm approaches the parabola through both points. Its flight time was found by Euler in 1743, for comets:
Δtp=132μ[s3/2∓(s−c)3/2]\Delta t_p = \frac{1}{3}\sqrt{\frac{2}{\mu}}\left[s^{3/2} \mp (s-c)^{3/2}\right]

Minus sign for Δθ ≤ 180°, plus for Δθ > 180°. If the requested time is shorter than Δt_p, no ellipse can do it: the transfer is a hyperbola. You can classify the conic before solving anything.

Figure 9.2
chord cr₁r₂same two points, five arcs30 mine = 0.2736 mine = 0.1644 mine = 0.2756 mine = 0.4372 mine = 0.55Every arc joins the same twopoints. Each takes a differenttime — and costs a differentamount to fly.
A fan of transfer arcs between the same two points. Each has a different semi-major axis and a different flight time. Short times bow the arc outward into high-energy paths; long times fold it back on itself.
From a to velocities. Once aa (and therefore α\alpha, β\beta) is known, the semi-latus rectum follows from the triangle:
p=4a(s−r1)(s−r2)c2sin⁡2 ⁣α+β2p = \frac{4a(s - r_1)(s - r_2)}{c^2}\sin^2\!\frac{\alpha + \beta}{2}
and the Lagrange coefficients — the same ff and gg that propagate a state in the unit on Kepler's equation and orbit propagation — now written with pp and the transfer angle:
f=1−r2p(1−cos⁡Δθ),g=r1r2sin⁡Δθμp,g˙=1−r1p(1−cos⁡Δθ)f = 1 - \frac{r_2}{p}(1-\cos\Delta\theta),\quad g = \frac{r_1 r_2 \sin\Delta\theta}{\sqrt{\mu p}},\quad \dot g = 1 - \frac{r_1}{p}(1-\cos\Delta\theta)

Then v₁ = (r₂ − f r₁)/g and v₂ = (ġ r₂ − r₁)/g. Note the division by sin Δθ hidden in g: at exactly 180° it collapses, and that is not a numerical accident.

⚠Δθ = 180° and Δθ = 0°
If the endpoints and the focus are collinear, the three points no longer define a plane. Infinitely many orbit planes contain the transfer, the equations divide by sin⁡Δθ=0\sin\Delta\theta = 0, and the problem has no unique answer. This is why an interplanetary Hohmann transfer — exactly 180° — sits on a ridge of the porkchop plot that real missions avoid (the lesson on launch windows and porkchop plots), and why a transfer that returns to the same planet a whole number of years later is not a Lambert problem at all (Lesson 10-5).
📐Worked Example: an intercept across a quarter turn
An interceptor is at r⃗1=(7000,0,0)\vec{r}_1 = (7000, 0, 0) km; the target point is r⃗2=(0,8000,0)\vec{r}_2 = (0, 8000, 0) km. Prograde. Geometry. (r⃗1×r⃗2)z>0(\vec r_1\times\vec r_2)_z > 0, so Δθ=90°\Delta\theta = 90°. c=70002+80002=10,630.1c = \sqrt{7000^2 + 8000^2} = 10{,}630.1 km; s=12,815.1s = 12{,}815.1 km. Minimum energy. am=s/2=6407.5a_m = s/2 = 6407.5 km, only 29 km more than Earth's radius. At ama_m, α=π\alpha = \pi and β0=2arcsin⁡(s−c)/s=0.8486\beta_0 = 2\arcsin\sqrt{(s-c)/s} = 0.8486 rad, so Δtm=am3/μ (π−β0+sin⁡β0)=41.19\Delta t_m = \sqrt{a_m^3/\mu}\,(\pi - \beta_0 + \sin\beta_0) = 41.19 min. Parabolic limit. Δtp=16.78\Delta t_p = 16.78 min. Any request faster than that is a hyperbola. A 20-minute intercept lies on the fast arm. Solving gives a=14,344a = 14{,}344 km, e=0.537e = 0.537, departure speed 9.28 km/s (Δv1=2.34\Delta v_1 = 2.34 km/s from a circular orbit). This arc passes through perigee on the way — at 271 km altitude, so it is flyable. A 30-minute intercept: a=6929a = 6929 km, e=0.160e = 0.160, Δv1=1.21\Delta v_1 = 1.21 km/s. Its ellipse has perigee 558 km below the surface — but perigee is not on the arc (the radial velocity is positive at departure and negative at arrival), so the spacecraft never goes below 7000 km. Always check the arc, not the ellipse.
Figure 9.3
204060801001207.588.59time of flight (minutes)departure speed (km/s)cheapest — 42 minHurrying costs energy.So does dawdling: belowthe minimum the arc hasto be bent into a longer,more eccentric shape justto waste enough time.r₁ = 7 000 km, r₂ = 8 000 kmΔθ = 90°
Departure speed against time of flight for fixed endpoints. Going faster costs energy, which is unsurprising. Going slower also costs energy, which is not: past the minimum the arc has to be bent into a longer, more eccentric path to use up the time.
🕰Four centuries of one problem
Euler (1743) solved the parabolic case for comets. Lambert (1761) stated the theorem; Lagrange (1778) proved it. Gauss (1809, Theoria Motus) built the method that recovered Ceres. Then the problem went quiet until spaceflight made it urgent: universal variables (Bate, Mueller & White, 1971), the unified formulation of Lancaster and Blanchard (NASA, 1969), Battin's elegant solution (1984–87), Gooding's robust procedure (1990), Izzo's fast solver (2015) and Russell's interpolated solver covering 'every Lambert problem' (2019). The Apollo Guidance Computer carried a Lambert routine among its conic subroutines; open-source libraries such as poliastro and pykep ship Izzo's.
↓But there is a problem
You now have a working Lambert solver: root-find Lagrange's equation for aa, rebuild pp, apply ff and gg. But it carries baggage. It needs to know which of four branches to search before it starts. It needs a separate formula for hyperbolas. It loses precision near the parabola, where a→∞a \to \infty. And it says nothing about multi-revolution transfers. Flight software cannot afford a case analysis. It needs one equation, one unknown, smooth across every conic, that a single Newton iteration can attack. Such an equation exists — and it is built from functions you met when propagating orbits.