Question bank — Gauss's method for Lambert's problem

False as stated — only unique once you also fix the "short way vs long way" branch and the number of full revolutions; geometry + $\Delta t$ alone leave a directional ambiguity.

True — a thin sector and its triangle enclose almost the same area, so sector/triangle $\approx 1$; this is exactly why the iteration starts at $y=1$.

False — $c^2=r_1^2+r_2^2-2r_1{r}_2\cos \theta$ is pure geometry of the two points; it is fixed before any orbit is chosen.

False — it enters as $\Delta t^2$, echoing Kepler's third law (time-squared scaling); treating it linearly gives a wrong $m$ and nonsense $y$.

False — one links geometry to the auxiliary variable, the other links time to it via Kepler's equation; the true orbit is where both hold at once, so you need both.

False — for hyperbolic or fast transfers the "eccentric anomaly" analogue makes $x<0$ (as in the worked example where $x\approx -0.01$); the series $X(x)$ is still valid there.

False — as $\theta \to 180^{\circ }$, $\cos (\theta /2)\to 0$, so $\ell$ and $m$ blow up and the iteration fails; near-antipodal transfers need Battin/Izzo instead.

False — the Lagrange f and g functions give ${\mathbf{v}}_1,{\mathbf{v}}_2$ in closed form from $p$; that is the whole point of using $f,g$.

The error: $\arccos$ only returns $0$–$180^{\circ }$, so it cannot distinguish the short way from the long way ($\theta$ vs $360^{\circ }-\theta$); you must use the orbit-normal / prograde flag to pick the branch.

Wrong: $m\propto \Delta t^2$, so doubling $\Delta t$ quadruples $m$, not doubles it.

Reversed — it is sector ÷ triangle; the curved sector is the bigger region, so $y\ge 1$ for elliptic short transfers.

A poor seed: $y=0$ makes $x=m/y^2-\ell$ diverge (division by zero) and there is no physical zero-area sector; the correct, contraction-friendly seed is $y=1$.

Error: $\cos (\theta /2)$ goes negative past $180^{\circ }$, flipping the signs inside $\ell$ and $m$; you must track that sign, not assume it stays positive.

Error: at $\theta =0$, $g\to 0$ and ${\mathbf{r}}_2\parallel {\mathbf{r}}_1$, so the division is singular — the two points coincide directionally and the problem is degenerate.

Wrong: the $p$ formula contains $x$, which is only known after $y$ converges; geometry alone under-determines $p$ because it fixes the shape but not the energy/time.

The triangle is computable from the given data alone ($\frac{1}{2}{r}_1{r}_2\sin \theta$), giving a fixed known reference; the sector is the unknown swept area, so their ratio isolates exactly the one thing the time-of-flight constrains.

Kepler's second law says equal areas are swept in equal times ($\Delta t=2(\text{sector area})/h$, with $h=\sqrt{\mu p}$ the steady area-rate), which is what converts a time input into an area condition — the bridge between the clock and the geometry.

For ordinary short-way arcs $y$ sits very near $1$ and the update map's slope there is well below $1$ in magnitude, so each step shrinks the error by a large factor — 3–5 steps reach $10^{-8}$.

That is Kepler's-third-law scaling (period${}^2\propto$ size${}^3$); packaging time this way makes $m$ dimensionless and puts the whole time dependence into one number.

Because ${\mathbf{r}}_2=f{\mathbf{r}}_1+g{\mathbf{v}}_1$ is a linear relation, so inverting it recovers ${\mathbf{v}}_1$ algebraically — no second pass through Kepler's equation.

They come from term-by-term expansion of the elliptic sector-area integral in powers of $x$; each fraction is the next coefficient in that convergent series, valid because $|x|$ stays small.

A porkchop plot sweeps thousands of departure/arrival date pairs, and each pair is one Lambert solve; a fast, few-iteration method like Gauss makes scanning that whole grid cheap.

Both diverge because $\cos (\theta /2)\to 0$ sits in their denominators; the method loses accuracy and fails to converge, so switch to a Universal variable formulation solver.

The sector area equals the triangle area — a degenerate flat (straight-line) transfer where the swept region collapses onto the triangle; real ellipses give $y$ slightly above $1$.

The chord and $\sin \theta$ vanish, $g\to 0$, and the "orbit" is undetermined — the endpoints are radially aligned, a degenerate case Gauss's formulas cannot resolve.

Perfectly fine — the geometry constant $\ell$ simplifies ($r_1+r_2=2r_1$) and the method runs normally; equal radii just mean a symmetric transfer, not a singularity.

Basic Gauss assumes a single arc; multi-revolution transfers introduce extra solution branches, and the single-branch iteration will not find them — you must handle each revolution count separately, as Izzo Lambert solver does.

The eccentric-anomaly analogue and $x$ turn negative, but the series $X(x)$ still converges for small $|x|$, so Gauss can handle mildly hyperbolic cases; strongly hyperbolic ones favour the Universal variable formulation.

That is normal and not an error — it simply means $E_2-E_1$ is small or the arc bends the "other way"; keep iterating and $y$ still settles (there it converged near $1.027$).