Lambert's problem — connecting two positions in given time: Question bank

Before the traps, we earn every symbol used below — none of these appear in a question until they are defined here.

Definition The symbols this page uses

$r_{1}, r_{2}$ = the two position vectors from the central body to the start and end points; $r_{1}, r_{2}$ are their lengths.
$c$ = chord = the straight-line distance $∣ r_{2} - r_{1} ∣$ between the two points (a ruler laid between them, not the flight path).
$r_{1} + r_{2}$ = sum of the two radial distances.
$s = \frac{1}{2} (r_{1} + r_{2} + c)$ = semiperimeter of the triangle whose corners are the central body, point 1, and point 2.
$a$ = semi-major axis of the transfer orbit (its "size"); $a_{m i n} = s /2$ is the smallest ellipse that can reach both points.
$e$ = eccentricity of the orbit — a pure number telling how stretched the ellipse is ( $e = 0$ is a circle, $e \to 1$ is a parabola).
$Δ θ$ = transfer angle, the angle swept from $r_{1}$ to $r_{2}$ in the direction of motion (see the sign convention below).
$Δ t$ = time of flight; $N$ = number of complete revolutions before arrival.

Definition Mean anomaly

$M$ , eccentric anomaly $E$ , and the auxiliary angles $α, β$ The mean anomaly $M$ is a fictitious angle that grows uniformly with time (it is what a clock would read on the orbit). The eccentric anomaly $E$ is a geometric angle that locates a body on its ellipse measured from the ellipse's centre (not the focus). Kepler's equation links them: $M = E - e sin E$ , where $e$ is the eccentricity defined above. Here $E_{1}, E_{2}$ are the eccentric anomaly's values at the start and end points. When Kepler's law is written at both ends and subtracted, the two eccentric anomalies fold into two auxiliary angles $α, β$ defined by $sin \frac{α}{2} = \frac{s}{2 a}, sin \frac{β}{2} = \frac{s - c}{2 a} .$ Geometrically, $α$ packages the "full-triangle" size $s$ and $β$ packages the "chord-shortened" size $s - c$ , each divided by the orbit size $a$ . They are the pure-angle handles that turn Lambert into a one-unknown ( $a$ ) equation.

Figure — Lambert's problem — connecting two positions in given time

Figure 1 — the space triangle. Two teal radius arrows

r_{1}, r_{2}

reach out from the central body (black dot at the focus) to the two mission points. The straight burnt-orange line between them is the chord

c

; the plum dashed curve bulging outward is the actual flight arc the spacecraft flies. The small black arc near the focus marks the transfer angle

Δ θ

. Use this to feel the difference between chord (a ruler) and arc (the trajectory) referenced in TF4, SE1, and EC-items.

Figure 2 — why $a \geq s /2$ . The horizontal axis is the orbit size

a

; the curves are

sin (α /2) = s /2 a

(orange) and

sin (β /2) = (s - c) /2 a

(teal). The dotted horizontal line at height 1 is the ceiling every sine must obey. The plum dashed vertical line marks

a_{m i n} = s /2

, where the orange curve touches 1 (

α = π

); the shaded plum band to its left is the forbidden region where the formula would demand a sine bigger than 1. This is the picture behind SE2, WHY3, and EC7.

True or false — justify

TF1. "Two points and a fixed flight time always pin down exactly one orbit."

False — for each

a > a_{m i n}

there are generally two ellipses (focus on either side of the chord), plus a short-way/long-way choice and multi-rev branches, so many orbits can share the same

Δ t

TF2. "The time of flight depends on the eccentricity

e

of the transfer."

False in the sense that matters — Lambert's theorem says

Δ t

depends only on

a

c

, and

r_{1} + r_{2}

; the eccentricity

e

never appears as an independent input, it drops out in the trig collapse.

TF3. "Making the transfer orbit bigger (larger

a

) always makes the trip take longer."

False — near

a_{m i n}

the time is not monotonic in

a

; the two solution branches bracket a minimum-time point, so increasing

a

can shorten or lengthen the trip depending on the branch.

TF4. "The chord

c

is the distance the spacecraft actually travels."

False — the craft flies the curved arc;

c

is only the straight-line gap

∣ r_{2} - r_{1} ∣

. Geometry enters Lambert purely through this straight chord, which is exactly the surprising content of Lambert's theorem.

TF5. "If I swap the short way for the long way but keep

r_{1}

r_{2}

fixed, the chord

c

changes."

False —

c = ∣ r_{2} - r_{1} ∣

is set entirely by the two endpoints, so it is identical for short and long way; what changes is the swept arc, handled by flipping the sign of the auxiliary angle

β

(per its sign convention above).

TF6. "

a_{m i n} = s /2

is the orbit of least time of flight."

False —

a_{m i n} = s /2

is the minimum-energy orbit (smallest possible

a

), not the minimum-time orbit; those are different special points on the

Δ t

versus

a

curve.

TF7. "A propagator like Kepler's equation could solve Lambert's problem directly."

False — a propagator solves the initial value problem (needs position and velocity); Lambert is a boundary value problem (two positions and a time, velocity unknown), so it must be inverted, not just propagated.

TF8. "Lambert's problem outputs the orbit's shape; the velocities are a separate mission-design step."

False — recovering

v_{1}, v_{2}

via the Lagrange

f, g

functions is part of solving Lambert; those velocities (and hence the

Δ v

) are the whole point of solving it.

TF9. "For a fixed geometry there is a shortest possible flight time, below which no ellipse works."

True — the parabolic escape limit sets a floor; ask for a shorter

Δ t

than the parabola's and you must use a hyperbolic transfer (

a < 0

), not an ellipse.

Spot the error

SE1. "

c = r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} cos Δ θ

, and since

cos

can be negative,

c

can exceed

r_{1} + r_{2}

Error —

c

can never exceed

r_{1} + r_{2}

; that's the triangle inequality. The maximum

c = r_{1} + r_{2}

is reached only in the degenerate case where

r_{1}

and

r_{2}

point in exactly opposite directions.

SE2. "I got

sin (α /2) = s /2 a = 1.2

, so

α /2 \approx 7 4^{\circ}

Error — a sine can never exceed 1, so

s /2 a = 1.2

is impossible; it means the chosen

a

is below

a_{m i n} = s /2

, i.e. a nonphysical ellipse. Increase

a

SE3. "For the long way (

Δ θ = 27 0^{\circ}

) I used the same

+ β

as the short way."

Error — long way needs

β \to - β

(per the sign convention); the sign of

β

encodes short-vs-long, and keeping it positive gives a badly wrong

Δ t

for the same

a

SE4. "Newton's method converged, so I have the answer."

Error — Lambert has multiple branches; Newton can land on the wrong ellipse or wrong revolution number without a bracketed guess (start from

a_{m i n}

-based bounds and specify

N

SE5. "The transfer angle is

Δ θ = arccos (\frac{r _{1} \cdot r _{2}}{r _{1} r _{2}})

, done."

Error —

arccos

only returns

0

π

, so it can't tell short way from long way; you must resolve the direction with the orbit normal / prograde-retrograde choice to know whether

Δ θ

2 π - Δ θ

applies.

SE6. "Since

α, β

appear symmetrically in

(α - sin α) - (β - sin β)

, swapping them just flips a sign, no big deal."

Error —

α

is tied to

s

and

β

s - c

via different arguments; they are not interchangeable, and swapping them describes a different (wrong) geometry.

SE7. "

a_{m i n} = s /2

where

s = \frac{1}{2} (r_{1} + r_{2} + c)

, so

a_{m i n} = r_{1} + r_{2} + c

Error — arithmetic slip:

a_{m i n} = s /2 = \frac{1}{4} (r_{1} + r_{2} + c)

, not

r_{1} + r_{2} + c

Why questions

WHY1. Why does the time of flight collapse to depend on only

a

c

, and

r_{1} + r_{2}

Because when Kepler's equation is written at both endpoints and subtracted, the eccentricity

e

and the individual eccentric anomalies

E_{1}, E_{2}

(defined above) combine into the two auxiliary angles

α, β

, which themselves depend only on

s

and

s - c

over

a

— that's Lambert's theorem's algebraic miracle.

WHY2. Why introduce the auxiliary angles

α, β

instead of solving in

E_{1}, E_{2}

directly?

Because

α, β

are exactly the geometry-folded combinations of

E_{1}, E_{2}

that survive the subtraction; they turn a two-endpoint problem into a clean function of one unknown

a

, which is what makes the root-find one-dimensional.

WHY3. Why is

a \geq s /2

a hard lower bound?

Because

sin (α /2) = s /2 a

and a sine can't exceed 1, forcing

2 a \geq s

; equality gives the thinnest allowed ellipse, the minimum-energy orbit.

WHY4. Why must we choose prograde vs retrograde (short vs long way) before solving?

Because the transfer angle

Δ θ

— and therefore the sign of

β

— is not fixed by the two positions alone; the same endpoints admit two arcs, and each gives a different time for the same

a

WHY5. Why do the Lagrange

f, g

functions let us recover velocity from just the two positions?

Because

r_{2} = f r_{1} + g v_{1}

(see the definition above) expresses the future state as a linear combination of initial position and velocity; once

a

(hence

f, g, \overset{g}{˙}

) is known and both positions are known, we algebraically solve for

v_{1}

then

v_{2}

WHY6. Why can't we just pick

a = a_{m i n}

and be done for any mission?

Because

a_{m i n}

fixes a specific flight time (the minimum-energy time); a mission with a different required

Δ t

needs a larger

a

, chosen so Lambert's equation equals that

Δ t

, not the minimum-energy one.

WHY7. Why is Lambert's problem the backbone of launch-window and rendezvous planning?

Because every such task is "be here now, be there at a set later time"; Lambert converts that boundary condition into the required departure velocity and hence

Δ v

, which is exactly what fuel budgets and porkchop plots need.

Edge cases

EC1. What happens to

c

and

s

when

Δ θ \to 0

(the two points coincide in direction, same radius)?

The chord

c \to 0

and

s \to r_{1}

; the "space triangle" degenerates to a line, and Lambert becomes ill-posed — you need a full revolution (

N \geq 1

) for any finite time.

EC2. What happens at

Δ θ = π

(points diametrically opposite)?

r_{1}

and

r_{2}

are exactly antiparallel the transfer plane is undefined, so short/long way is ambiguous and the orbit normal can't be fixed from the positions alone; the chord reaches its maximum,

c = r_{1} + r_{2}

EC3. What if the requested

Δ t

is smaller than any elliptic transfer allows?

No ellipse works; you cross into a hyperbolic transfer with

a < 0

, and the elliptic formula (with real

α, β

) must be replaced by its hyperbolic analog using

sinh

EC4. What is the parabolic case, and where does it sit in the

a < 0

a > s /2

classification?

The parabola is the exact boundary

a \to \infty

(zero orbital energy); it is the single fastest bound-adjacent transfer for the geometry. Times longer than the parabola's are elliptic (

s /2 \leq a < \infty

); times shorter are hyperbolic (

a < 0

). So the parabola is the hinge separating the two families.

EC5. What if

r_{1} = r_{2}

exactly?

Perfectly valid — the geometry is symmetric,

α, β

still follow from

s

and

s - c

; it corresponds to same-altitude transfers (the Hohmann-like family lives near here when

Δ θ = π

EC6. What does asking for

N \geq 1

(multi-revolution) do to the number of solutions?

It multiplies the branches — for each revolution count there can be two more solutions (a "low" and "high"

a

), so long-duration transfers may have many valid orbits, and you must specify

N

and branch.

EC7. What is the meaning of a solution with

α = π

That's the minimum-energy orbit

a = s /2

; the auxiliary angle

α

has hit its maximum for that geometry, marking the boundary between the two

a

-branches for a fixed short/long way.

Recall One-sentence summary

Lambert traps almost always come from forgetting a branch, a sign, or a lower bound — multiple orbits per geometry, $β$ 's sign for direction, and $a \geq s /2$ — plus confusing chord with arc and time-with- $a$ monotonicity.

See also: Kepler's Equation and Time of Flight · Lagrange Coefficients (f and g functions) · Universal Variable Formulation · Hohmann Transfer · Porkchop Plots and Launch Windows · Orbital Rendezvous and Targeting