Gauss's method for Lambert's problem

WHAT is Lambert's problem?

WHY it matters: every interplanetary transfer, rendezvous, and orbit-determination-from-two-observations reduces to this. It's the workhorse of mission design (porkchop plots come from solving Lambert millions of times).

The geometric quantities we need

Let $\theta = \Delta\nu$ be the transfer angle (change in true anomaly between the two points).

Deriving Gauss's key ratio $y$

Kepler's 2nd law says the radius vector sweeps equal areas in equal times, so $\Delta t = \frac{2\,(\text{area of elliptic sector swept})}{h}, \qquad h=\sqrt{\mu p}.$

The two radii and the chord also enclose a triangle of area $A_{\triangle} = \tfrac12 r_1 r_2 \sin\theta.$

Gauss expressed the sector-to-triangle ratio through the semi-latus rectum $p$ . Working through the sector area for an ellipse gives the two fundamental relations:

Why two equations? One ( $y^2=\frac{m}{\ell+x}$ ) is geometry ↔ auxiliary variable; the other ( $y^2(y-1)=\dots$ ) is time ↔ auxiliary variable via Kepler's equation. The true orbit satisfies both simultaneously → solve the coupled pair.

HOW to solve — the iteration

Worked example The Gauss iteration algorithm

Compute $\cos\theta,\ \ell,\ m$ from the given data.
Guess $y=1$ (sector≈triangle).
Compute $x=\dfrac{m}{y^2}-\ell$ . (Why? Invert equation 1.)
Compute $X(x)$ from the series.
Update $y = 1 + (\ell + x)\big(x - \tfrac12 + X(x)\big)$ . (Substitute $y^2=m/(\ell+x)$ into equation 2 and divide by $y^2$ : $y-1 = \dfrac{m}{y^2}\big(x-\tfrac12+X\big) = (\ell+x)\big(x-\tfrac12+X\big)$ .)
Repeat 3–5 until $y$ converges.

Why it converges fast: for typical transfers $y$ is close to 1, and the map is a contraction — usually 3–5 iterations to $10^{-8}$ .

Once $y$ is known, recover the orbit:

Figure — Gauss's method for Lambert's problem

Worked example

Worked example A short-way transfer

Given (canonical units, $\mu=1$ ): $r_1=2$ , $r_2=3$ , $\theta=40^\circ$ , $\Delta t=1.8$ .

Step 1 — geometry. $\cos40^\circ=0.766$ , $\cos(20^\circ)=0.9397$ . $\ell=\frac{2+3}{4\sqrt6(0.9397)}-\tfrac12 = \frac{5}{9.206}-0.5 = 0.0433.$ Why? pure formula plug-in from geometry constants.

Step 2 — time constant. $m=\frac{1\cdot 1.8^2}{(2\sqrt6\cdot0.9397)^3}=\frac{3.24}{(4.603)^3}=\frac{3.24}{97.5}=0.03323.$

Step 3 — iterate. Start $y=1$ :

$x=m/y^2-\ell=0.03323-0.0433=-0.0101$
$X(-0.0101)\approx \tfrac43(1+1.2(-0.0101))=1.317$
$y=1+(\ell+x)\big(x-\tfrac12+X\big)=1+(0.0332)(-0.0101-0.5+1.317)=1+(0.0332)(0.807)=1.0268$

Next: $x=0.03323/1.0268^2-0.0433=0.0315-0.0433=-0.0118$ ; recompute $X$ and update → $y$ settles near $1.027$ . Converged in ~3 steps.

Step 4 — recover. Use converged $y$ to get $p$ , then $f,g,\dot g$ , then $\mathbf{v}_1,\mathbf{v}_2$ . Why trust it? Because both Gauss equations are now satisfied, geometry AND time-of-flight are simultaneously consistent — that IS the connecting orbit.

Steel-manned mistakes

Active recall

Recall Test yourself (hide the answers)

What are the inputs and outputs of Lambert's problem?
What does the ratio $y$ physically compare?
Why does Gauss's method need two equations?
Where does the method fail, and why?
How do you get $\mathbf v_1$ once $y$ is found?

Recall Feynman: explain to a 12-year-old

Imagine a stone thrown so it lands on a far hill in exactly 3 seconds. If I tell you where it started, where it landed, and how long it took, there's exactly one curved path (for a chosen curve direction) that fits. Gauss's trick: he compares the "pie slice" the planet sweeps to the flat "triangle" between the two spots. He guesses their ratio is about 1, checks whether that guess would take the right amount of time, and nudges the guess until the clock matches. When it matches, he's found the path — and from the path he knows how fast it was going.

Flashcards

What does Lambert's problem take as input and produce as output?

Input:

\mathbf r_1,\mathbf r_2

and time-of-flight

\Delta t

. Output: the connecting Kepler orbit, i.e.

\mathbf v_1,\mathbf v_2

In Gauss's method, what does the ratio

y

represent?

The ratio of the swept elliptic sector area to the triangle area between the two radii and the chord.

Why does Gauss's method use two coupled equations?

One links geometry to the auxiliary variable (

y^2=m/(\ell+x)

); the other links time-of-flight to it (

y^2(y-1)=m(x-\tfrac12+X)

). The true orbit satisfies both.

What is the correct iteration update for

y

y = 1 + (\ell+x)(x-\tfrac12+X(x))

, obtained by dividing the 2nd Gauss equation by

y^2

and using the 1st.

What is the chord length formula for the transfer?

c^2=r_1^2+r_2^2-2r_1r_2\cos\theta

(law of cosines).

How does

\Delta t

enter Gauss's constant

m

\Delta t^2

m=\mu\Delta t^2/(2\sqrt{r_1r_2}\cos(\theta/2))^3

(Kepler-3 scaling).

Where does Gauss's method break down?

Near

\theta=180°

(antipodal), since

\cos(\theta/2)\to0

makes

\ell,m

blow up and iteration fails.

Once

y

(hence

p

) is known, how are velocities recovered?

Via Lagrange coefficients:

\mathbf v_1=(\mathbf r_2-f\mathbf r_1)/g

\mathbf v_2=(\dot g\mathbf r_2-\mathbf r_1)/g

What is a good starting guess for the iteration and why?

y=1

, because for modest transfer angles the sector ≈ triangle, and

y

stays near 1.

Connections

Lambert's problem — parent problem
Kepler's equation — supplies the time↔eccentric-anomaly link
Lagrange f and g functions — used to extract velocities
Universal variable formulation — robust alternative near $180°$
Battin's method and Izzo Lambert solver — modern replacements
Kepler's second law — equal-area basis of the sector/triangle ratio
Porkchop plot — application: mass Lambert solves for launch windows

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Lambert's problem simple si baat hai: tumhe do positions pata hain — kahaan se object chala ( $\mathbf r_1$ ) aur kahaan pahuncha ( $\mathbf r_2$ ) — aur kitne time mein pahuncha ( $\Delta t$ ). Sawaal: konsi orbit in dono ko jodti hai? Gauss ne iska ek bahut clever tarika nikala. Usne kaha, do radii aur unke beech ki chord se ek triangle banta hai, aur planet jo area sweep karta hai woh ek sector (pie slice) hota hai. Inn dono ke ratio ko — sector ÷ triangle — usne $y$ naam diya.

Trick yeh hai: shuru mein maan lo $y=1$ (matlab sector aur triangle almost barabar). Is guess se ek auxiliary variable $x$ nikaalo, phir ek chhoti si series $X(x)$ se naya $y$ calculate karo. Sahi update formula hai $y = 1 + (\ell+x)\,(x-\tfrac12+X(x))$ — yeh dusri Gauss equation ko $y^2$ se divide karke aur pehli equation daal ke aata hai. Yeh loop 3-4 baar chalao aur $y$ settle ho jaata hai. Ismein do equations hain — ek geometry batati hai, doosri time-of-flight ( $\Delta t^2$ ke through, Kepler ke third law jaisa). Dono jab ek saath satisfy ho jaayein, tab tumhe sahi orbit mil gayi.

$y$ mil gaya toh $p$ (semi-latus rectum) nikaalo, phir Lagrange $f, g$ functions use karke seedha $\mathbf v_1$ aur $\mathbf v_2$ nikaal lo — dobara Kepler solve karne ki zaroorat nahi. Yeh method mission design mein bahut kaam aata hai: Mars ya kisi bhi planet ka transfer, launch window (porkchop plot) — sab ismein Lambert solve karke banta hai.

Ek warning: agar transfer angle $\theta$ 180° ke paas ho (dono points bilkul opposite side), toh $\cos(\theta/2)$ zero ho jaata hai aur method fat jaata hai. Aise case mein universal-variable ya Izzo/Battin solver use karo. Aur haan, $\Delta t$ hamesha square karke aata hai ( $m\propto\Delta t^2$ ) — yeh mat bhoolna.

WHAT is Lambert's problem?

The geometric quantities we need

Deriving Gauss's key ratio $y$

HOW to solve — the iteration

Worked example

Steel-manned mistakes

Active recall

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Orbital Mechanics & Astrodynamics

Connections