Lambert's problem — connecting two positions in given time

WHY do we even need this?

WHAT is actually given and unknown?

The magic fact that makes this solvable:

HOW to derive the time-of-flight equation

We'll build the classic Lagrange / universal-variable form from geometry.

Step 1 — Geometry: the chord and the space triangle

Place both radii and the chord:

$c = |\vec r_2 - \vec r_1| = \sqrt{r_1^2 + r_2^2 - 2\,r_1 r_2 \cos\Delta\theta}$

Why this step? $\Delta\theta$ is the transfer angle between the two positions; the chord is just the law of cosines on the triangle Earth–$\vec r_1$–$\vec r_2$. The short way uses $0<\Delta\theta<\pi$, the long way uses $\pi<\Delta\theta<2\pi$.

Define the semiperimeter of that space triangle: $s = \tfrac12 (r_1 + r_2 + c)$

Why? Because Lambert's theorem tells us time depends on $r_1+r_2$ and $c$ — and those combine most naturally as $s$.

Step 2 — Introduce the auxiliary angles

For an elliptical transfer ($a>0$) we parametrize with two angles $\alpha,\beta$: $\sin^2\frac{\alpha}{2} = \frac{s}{2a}, \qquad \sin^2\frac{\beta}{2} = \frac{s-c}{2a}$

Why this step? These come from writing the eccentric-anomaly difference across the arc in terms of the geometry. $\alpha,\beta$ are exactly the combinations of eccentric anomalies at the two ends that survive after the geometry is folded in — they package $a,c,s$ into pure angles.

Step 3 — Kepler's time law across the arc

Kepler's equation says the swept "mean anomaly" between two eccentric anomalies $E_1,E_2$ is $\sqrt{\frac{\mu}{a^3}}\,\Delta t = (E_2 - e\sin E_2) - (E_1 - e\sin E_1).$

Why? This is literally Kepler's equation $M = E - e\sin E$ applied at both endpoints and subtracted — mean anomaly advances uniformly with time, so its change equals $n\,\Delta t$ with $n=\sqrt{\mu/a^3}$.

Substituting the half-angle identities and doing the trig collapse (the eccentricity $e$ drops out into the $\alpha,\beta$ combination — this is the miracle), we obtain Lambert's time-of-flight equation:

Why this is the whole game: the right side is a known function of the single unknown $a$. Everything else ($s,c,\mu$) is geometry. So Lambert reduces to: find $a$ such that this equals the given $\Delta t$, then reconstruct velocities.

Step 4 — The minimum-energy orbit (a key landmark)

There is a smallest ellipse that can connect the two points at all: $a_{\min} = \frac{s}{2}$ At $a_{\min}$, $\alpha=\pi$, and the time of flight is the minimum-energy time — any faster or slower transfer needs a larger $a$ (more energy). This gives you a sanity check and a starting guess.

Why $a=s/2$? From $\sin(\alpha/2)=\sqrt{s/2a}$, the argument can't exceed 1, so $2a \ge s$, i.e. $a\ge s/2$. Equality is the extreme (thinnest) allowed ellipse.

Step 5 — Recover the velocities (Lagrange coefficients)

Once $a$ is found, use the f and g functions to get velocity: $\vec r_2 = f\,\vec r_1 + g\,\vec v_1 \;\Rightarrow\; \vec v_1 = \frac{\vec r_2 - f\,\vec r_1}{g},\qquad \vec v_2 = \frac{\dot g\,\vec r_2 - \vec r_1}{g}$

Why? The Lagrange coefficients express the future state as a linear combination of the initial position and velocity. Since we now know $f,g,\dot g$ (functions of $a,\Delta\theta,\ldots$) and both positions, we algebraically solve for $\vec v_1$, then $\vec v_2$. These velocities are the answer the mission planner wants.

Worked Example 1 — Short-way elliptic transfer

Given (canonical units, $\mu=1$): $r_1=1$, $r_2=1.5$, $\Delta\theta=90^\circ$, $\Delta t = 1.2$.

Step A — chord & semiperimeter. $c=\sqrt{1^2+1.5^2-2(1)(1.5)\cos 90^\circ}=\sqrt{1+2.25-0}=\sqrt{3.25}=1.803$. $s=\tfrac12(1+1.5+1.803)=2.152$. Why: geometry first — everything downstream depends on $s,c$.

Step B — minimum-energy check. $a_{\min}=s/2=1.076$. Compute its time to see if our $\Delta t$ needs $a$ above or below (recall two branches exist per $a>a_{\min}$). Why: it frames the root search and prevents picking a nonphysical $a$.

Step C — solve Lambert's equation for $a$ (numerically, e.g. Newton on the boxed equation) so that RHS $=\Delta t=1.2$. Suppose it yields $a\approx 1.30$. Why: $a$ is the single unknown; one scalar root-find.

Step D — velocities via $f,g$ with this $a$ and $\Delta\theta$. Output $\vec v_1,\vec v_2$; the departure burn is $\Delta v = |\vec v_1 - \vec v_{\text{park}}|$. Why: velocities → $\Delta v$ → fuel, the deliverable.

Worked Example 2 — Why the transfer direction changes the answer

Same $r_1,r_2$ but choose the long way, $\Delta\theta = 270^\circ$. $c=\sqrt{1+2.25-2(1.5)\cos270^\circ}=\sqrt{3.25-0}=1.803$ (same chord!) — but the arc is now the big side, so $\beta$ takes the negative branch: use $\beta \to -\beta$ in Lambert's equation.

Why this step? The chord is symmetric, but which way you go around changes how much area/angle is swept, hence different $\Delta t$ for the same $a$. The sign of $\beta$ encodes short-vs-long way. Forgetting it gives a wildly wrong time.

Common Mistakes

Active Recall

Flashcards

Connections

• Kepler's Equation and Time of Flight — the time law Lambert differences across the arc.
• Lagrange Coefficients (f and g functions) — used to reconstruct velocities.
• Hohmann Transfer — a special optimal case of Lambert (tangential, $\Delta\theta=\pi$).
• Universal Variable Formulation — the unified (ellipse/parabola/hyperbola) way to solve Lambert.
• Porkchop Plots and Launch Windows — Lambert solved over a grid of dates.
• Orbital Rendezvous and Targeting — Lambert delivers the intercept velocity.

Concept Map

Hinglish (regional understanding)