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

Lambert's theorem says the time depends on exactly three scalars: $a$, $c$, and $r_1+r_2$. So the correct set is (b), (c), (d). The eccentricity by itself (a) and the arc's individual shape (e) do not enter — that is the whole miracle: $e$ cancels out of the time law.

Chord. $\cos 90^{\circ }=0$, so $c=\sqrt{1^2+1.5^2-2(1)(1.5)(0)}=\sqrt{1+2.25}=\sqrt{3.25}=\mathrm{1.8028.}$ Semiperimeter. $s=\frac{1}{2}(1+1.5+1.8028)=\frac{1}{2}(4.3028)=\mathrm{2.1514.}$

$a_{\min }=\frac{s}{2}=\frac{2.1514}{2}=\mathrm{1.0757.}$ At $a=a_{\min }=s/2$, the ratio $\frac{s}{2a}=\frac{s}{s}=1$, so $\sin \frac{\alpha }{2}=\sqrt{1}=1\Rightarrow \frac{\alpha }{2}=\frac{\pi }{2}\Rightarrow \alpha =\pi .$ For $\beta$: $\frac{s-c}{2a}=\frac{2.1514-1.8028}{2(1.0757)}=\frac{0.3486}{2.1514}=0.1620$, so $\sin \frac{\beta }{2}=\sqrt{0.1620}=0.4025\Rightarrow \frac{\beta }{2}=0.4145\text{ rad}\Rightarrow \beta =0.8290\text{ rad}.$ So at the minimum-energy ellipse, $\alpha =\pi$ exactly — this is the landmark from the parent note.

Plug into Lambert's law with $\alpha =\pi$, $\beta =0.8290$:

• $\alpha -\sin \alpha =\pi -\sin \pi =\pi -0=\mathrm{3.1416.}$
• $\beta -\sin \beta =0.8290-\sin (0.8290)=0.8290-0.7368=\mathrm{0.0922.}$
• Bracket: $(\alpha -\sin \alpha )-(\beta -\sin \beta )=3.1416-0.0922=\mathrm{3.0494.}$
• $a^{3/2}=1.0757^{1.5}=\mathrm{1.1157.}$

$\sqrt{\mu }\Delta t=1.1157\times 3.0494=3.402\Rightarrow \Delta t_{\min }=\mathrm{3.402.}$ This is the fastest the minimum-energy ellipse flies this arc. Any other feasible transfer uses a larger $a$.

We need the root of $F(a)=\text{(Lambert time)}-4.0$. At $a_0=1.0757$: Lambert time $=3.402$, so $F(a_0)=3.402-4.0=-0.598$. Numerical slope. Take a nearby point $a=1.20$:

• $\frac{s}{2a}=\frac{2.1514}{2.40}=0.8964\Rightarrow \sin (\alpha /2)=0.9468\Rightarrow \alpha /2=1.2464\Rightarrow \alpha =2.4928$.
• $\alpha -\sin \alpha =2.4928-0.6003=\mathrm{1.8925.}$
• $\frac{s-c}{2a}=\frac{0.3486}{2.40}=0.14525\Rightarrow \sin (\beta /2)=0.38112\Rightarrow \beta /2=0.39125\Rightarrow \beta =0.7825$.
• $\beta -\sin \beta =0.7825-0.7053=\mathrm{0.0772.}$
• Bracket $=1.8925-0.0772=1.8153$; $a^{3/2}=1.3145$; time $=1.3145\times 1.8153=2.386$.
• $F(1.20)=2.386-4.0=-1.614$.

Wait — the time decreased as $a$ grew! This is the non-monotonic behavior warned about in the parent note. On this branch (the one continuing smoothly from $a_{\min }$ with $\alpha$ increasing past $\pi$), moving $a$ up in this range moves the time the "wrong" way. The Newton slope here is $F^{\prime }\approx \frac{-1.614-(-0.598)}{1.20-1.0757}=\frac{-1.016}{0.1243}=-\mathrm{8.17.}$ Newton step: $a_1=a_0-\frac{F(a_0)}{F^{\prime }}=1.0757-\frac{-0.598}{-8.17}=1.0757-0.0732=\mathrm{1.0025.}$ This pushes $a$ below $a_{\min }$ — which is infeasible ($a\ge s/2$). The lesson: near $a_{\min }$ the wrong branch of $\alpha$ makes naive Newton diverge. The correct increasing-time branch requires taking $\alpha \to 2\pi -\alpha$ (the second branch past $a_{\min }$), which we explore next.

For a fixed $a>a_{\min }$, the number $x=\sqrt{s/2a}$ satisfies $0<x<1$. The equation $\sin (\alpha /2)=x$ has two solutions for $\alpha /2$ in $(0,\pi )$: $\frac{\alpha }{2}=\arcsin x\text{or}\frac{\alpha }{2}=\pi -\arcsin x,$ giving $\alpha =2\arcsin x$ (the "$\alpha <\pi$" branch) or $\alpha =2\pi -2\arcsin x$ (the "$\alpha >\pi$" branch).

• $\alpha <\pi$ branch: the swept eccentric-anomaly arc is small → shorter time.
• $\alpha >\pi$ branch: the arc wraps the far side → longer time.

Physically these are two genuinely different ellipses (focus on opposite sides of the chord relative to the swept region) that both connect ${\vec{r}}_1,{\vec{r}}_2$ with the same $a$. At exactly $a=a_{\min }$ the two branches merge at $\alpha =\pi$. This is why "one time → one orbit" is false, and why a solver must know which branch it wants before iterating.

(i) Chord. $\cos 270^{\circ }=0$, identical to $\cos 90^{\circ }$, so $c=\sqrt{1+2.25-2(1.5)(0)}=\sqrt{3.25}=\mathrm{1.8028.}$ Same chord, same $s=2.1514$, same $a_{\min }=1.0757$. The chord is symmetric in going around either way — only the arc differs. (ii) Modification. For the long way we flip the sign of $\beta$: use $\beta \to -\beta$, so the bracket becomes $(\alpha -\sin \alpha )-(-\beta -\sin (-\beta ))=(\alpha -\sin \alpha )+(\beta -\sin \beta ).$ (iii) Long-way min-energy time. With $\alpha =\pi$ and $\beta =0.8290$ (from 2.1):

• $\alpha -\sin \alpha =3.1416$.
• $+(\beta -\sin \beta )=+0.0922$.
• Bracket $=3.1416+0.0922=3.2338$.
• $\times a^{3/2}=1.1157$: $\Delta t_{\text{long,min}}=1.1157\times 3.2338=3.608$.

Comparison: short way gave $3.402$, long way gives $3.608$. The long way is slower at the same $a$ — sensible, since it sweeps a bigger arc. The sign of $\beta$ is exactly what encodes this difference.

(i) Hohmann $a$. A Hohmann ellipse touches both circles, so its major axis is $r_1+r_2$: $a_H=\frac{r_1+r_2}{2}=\frac{1+1.5}{2}=\mathrm{1.25.}$ (ii) Geometry at $\Delta \theta =180^{\circ }$. $\cos 180^{\circ }=-1$: $c=\sqrt{1+2.25-2(1)(1.5)(-1)}=\sqrt{1+2.25+3}=\sqrt{6.25}=\mathrm{2.5.}$ This is just $r_1+r_2$ — makes sense, the two points are diametrically opposite so the chord is the whole diameter. $s=\frac{1}{2}(1+1.5+2.5)=\frac{1}{2}(5)=2.5,a_{\min }=\frac{s}{2}=\mathrm{1.25.}$ So $a_H=1.25=a_{\min }$: the Hohmann transfer is exactly the minimum-energy Lambert solution for a half-revolution between these radii. This is the beautiful unification — Hohmann is not a separate theory, it's the $\Delta \theta =\pi$, $a=a_{\min }$ corner of Lambert.

(i) $a$. Since $\Delta t=\Delta t_{\min }$, the only ellipse achieving it is $a=a_{\min }=1.0757$. (ii) Number of solutions. For $\Delta t>\Delta t_{\min }$ there are two branches (short-time and long-time $\alpha$). But at $\Delta t=\Delta t_{\min }$ the two branches have merged into one — this is the tangent bottom of the time-vs-$a$ curve. So there is exactly one zero-rev short-way solution. (iii) Planner insight. A merged (double) root means the solution is a minimum of the $\Delta t(a)$ curve: $\frac{d\Delta t}{da}=0$ there. Small changes in the required $\Delta t$ move $a$ only slightly, but it's a knife-edge — you cannot fly faster than $\Delta t_{\min }$ on any ellipse of this geometry at all. In a porkchop plot this shows up as the boundary of the feasible $\Delta v$ region: below the minimum-energy contour there is simply no solution. Planners keep $\Delta t$ comfortably above $\Delta t_{\min }$ to keep two branches (hence flexibility) available.

Bracket. $\Delta t=4.5>\Delta t_{\min }=3.402$, so a solution exists with $a>a_{\min }=1.0757$. Because $4.5$ exceeds the minimum time, we want the long-time behavior — the $\alpha >\pi$ branch, where increasing the arc lengthens the flight. On the $\alpha >\pi$ branch we use $\alpha =2\pi -2\arcsin \sqrt{s/2a}$. Trial $a=1.30$:

• $\sqrt{s/2a}=\sqrt{2.1514/2.60}=\sqrt{0.8275}=0.9097\Rightarrow \arcsin =1.1441\Rightarrow \alpha =2\pi -2.2882=3.9950$.
• $\alpha -\sin \alpha =3.9950-(-0.7469)=4.7419$.
• $\sqrt{(s-c)/2a}=\sqrt{0.3486/2.60}=\sqrt{0.13408}=0.36617\Rightarrow \beta /2=0.37504\Rightarrow \beta =0.75008$; $\beta -\sin \beta =0.75008-0.68164=0.06844$.
• Bracket $=4.7419-0.0684=4.6735$; $a^{3/2}=1.4822$; time $=1.4822\times 4.6735=6.928$. Too big.

Trial $a=1.10$:

• $\sqrt{s/2a}=\sqrt{2.1514/2.20}=\sqrt{0.97791}=0.98890\Rightarrow \arcsin =1.42280\Rightarrow \alpha =2\pi -2.84560=3.4376$.
• $\alpha -\sin \alpha =3.4376-(-0.29327)=3.7309$.
• $\sqrt{(s-c)/2a}=\sqrt{0.3486/2.20}=\sqrt{0.15845}=0.39806\Rightarrow \beta /2=0.40952\Rightarrow \beta =0.81904$; $\beta -\sin \beta =0.81904-0.73053=0.08851$.
• Bracket $=3.7309-0.0885=3.6424$; $a^{3/2}=1.15369$; time $=1.15369\times 3.6424=4.202$. Slightly small.

Target $4.5$ lies between $a=1.10$ (time $4.202$) and $a=1.30$ (time $6.928$). Interpolating: $a\approx 1.10+\frac{4.5-4.202}{6.928-4.202}(1.30-1.10)=1.10+\frac{0.298}{2.726}(0.20)=1.10+0.0219=\mathrm{1.122.}$ Refine at $a=1.122$:

• $\sqrt{s/2a}=\sqrt{2.1514/2.244}=\sqrt{0.95873}=0.97914\Rightarrow \arcsin =1.36494\Rightarrow \alpha =2\pi -2.72988=3.5533$.
• $\alpha -\sin \alpha =3.5533-(-0.40241)=3.9557$.
• $\sqrt{(s-c)/2a}=\sqrt{0.3486/2.244}=\sqrt{0.15535}=0.39415\Rightarrow \beta /2=0.40527\Rightarrow \beta =0.81054$; $\beta -\sin \beta =0.81054-0.72469=0.08585$.
• Bracket $=3.9557-0.0859=3.8698$; $a^{3/2}=1.18845$; time $=1.18845\times 3.8698=4.599$. Very close to $4.5$.

One more nudge downward gives $a\approx 1.117$. To 3 sig figs, $a\approx 1.12$ on the $\alpha >\pi$ (long-time) branch.