Exercises — Bi-elliptic transfer — when it wins over Hohmann

• Three burns, two ellipses.
• Burn 1 at $r_1$ (raise apoapsis to $r_b$), Burn 2 at $r_b$ (raise the opposite apsis from $r_1$ up to $r_2$), Burn 3 at $r_2$ (circularize).
• Contrast with Hohmann transfer: two burns, one ellipse.

The intermediate radius must satisfy $r_b>r_2$ (recall $r_b$ is the overshoot apoapsis defined above). Here $r_b=15<20=r_2$, which is geometrically impossible: the first ellipse's apoapsis $r_b$ has to be beyond the target so the second ellipse can "coast back down" to $r_2$. If $r_b\le r_2$ it is not a bi-elliptic transfer at all (at $r_b=r_2$ it degenerates to Hohmann).

• $R<11.94$: Hohmann always wins — no choice of $r_b$ lets bi-elliptic beat it.
• $R>15.58$: bi-elliptic wins provided $r_b$ is chosen large enough.
• $11.94<R<15.58$: it depends on $r_b$ (grey zone).

• $v_{c1}=\sqrt{\mu /r_1}=\sqrt{1/1}=1$.
• $v_{c2}=\sqrt{\mu /r_2}=\sqrt{1/20}=0.223607$. Why this formula? It is vis-viva with $a=r$ (a circle), so the two energy terms leave only $v^2=\mu /r$. What it means: the outer orbit is slow — that slowness is the resource bi-elliptic exploits.

• $a_1=\frac{r_1+r_b}{2}=\frac{1+100}{2}=50.5$. Why: $r_1$ is this ellipse's periapsis, $r_b$ its apoapsis; average them.
• Speed on ellipse 1 at $r_1$: vis-viva with $r=r_1,a=a_1$: $\sqrt{2-\frac{1}{50.5}}=\sqrt{1.98020}=1.40719$. Why here: Burn 1 fires at $r_1$, so we evaluate the new orbit's speed at $r_1$.
• Subtract circular speed: $\Delta v_1=1.40719-1=0.40719$. Why subtract? We were already moving at $v_{c1}=1$ in a circle; the burn only pays for the extra speed needed to stretch the orbit outward — a pure prograde kick, same direction, so the cost is the gap.

• $a_2=60$. Why: ellipse 2 has apoapsis $r_b=100$ and periapsis $r_2=20$; average them.
• Speed on ellipse 2 at $r_b$ (vis-viva, $r=100,a=60$): $\sqrt{\frac{2}{100}-\frac{1}{60}}=\sqrt{0.02-0.016667}=\sqrt{0.003333}=0.057735$.
• Speed on ellipse 1 at $r_b$ ($r=100,a=50.5$): $\sqrt{\frac{2}{100}-\frac{1}{50.5}}=\sqrt{0.02-0.019802}=\sqrt{0.000198}=0.014079$.
• $\Delta v_2=0.057735-0.014079=0.043656$. Read the figure below: both speeds are prograde arrows at the far point, and both share the same dominant $\frac{2}{r_b}=0.02$ term — so the arrows are nearly equal length and you pay only the small orange gap between them, not their sum. This tiny far-out cost is the entire reason bi-elliptic can win.

• $a_2=60$ (from 2.3). Where and why Burn 3 fires: at $r_2=20$, which is the periapsis of ellipse 2, so the craft arrives moving faster than a circle there — vis-viva at $r=20,a=60$: $\sqrt{\frac{2}{20}-\frac{1}{60}}=\sqrt{0.083333}=0.288675$.
• Target circular speed: $v_{c2}=\sqrt{1/20}=0.223607$.
• $\Delta v_3=0.288675-0.223607=0.065068$ — a brake (retro): you are above circular speed, so you must slow down to circularize. Take the magnitude.
• Total: $0.40719+0.043656+0.065068=0.51591$.
• Intuition tie-back: the three burns are "climb out" (0.407) + "cheap far-out reshape" (0.044) + "gentle brake in" (0.065). The tiny middle term is the whole trick. Since $R=20>15.58$, bi-elliptic should win — and the parent's Hohmann $=0.5347>0.51591$ confirms it. ✅

• $a_1=\frac{1+60}{2}=30.5$, $a_2=\frac{20+60}{2}=40$. Why: same "average the apsides" rule with the new $r_b=60$.
• Burn 1 (at $r_1$, enter ellipse 1): $\sqrt{2-\frac{1}{30.5}}-1=\sqrt{1.967213}-1=1.402574-1=0.402574$.
• Burn 2 (at $r_b=60$, ellipse 1→2): $\sqrt{\frac{2}{60}-\frac{1}{40}}-\sqrt{\frac{2}{60}-\frac{1}{30.5}}=\sqrt{0.008333}-\sqrt{0.000546}=0.091287-0.023372=0.067915$.
• Burn 3 (brake at $r_2=20$): $\sqrt{\frac{2}{20}-\frac{1}{40}}-0.223607=\sqrt{0.075}-0.223607=0.273861-0.223607=0.050254$.
• Total: $0.402574+0.067915+0.050254=0.520743$.
• Interpretation: compared to $r_b=100$'s $0.51591$, this is slightly worse. The far point at $r_b=60$ is not as slow, so Burn 2 is bigger (0.068 vs 0.044). Larger $r_b$ pushed the total down — the trend continues toward the bi-parabolic floor.

• Burn 2 ↓ (physical reason): it fires at $r=r_b$. In vis-viva the term $\frac{2}{r_b}\to 0$ as $r_b\to \infty$ for both ellipse speeds, so both roots shrink toward zero and their difference vanishes. Farther out = slower everything = cheaper reshape. This is the far-out "calm zone" of the parent's merry-go-round analogy.
• Burn 1 ↑ (physical reason): at $r_1$ the ellipse-1 speed is $\sqrt{2/r_1-1/a_1}$ with $a_1=(r_1+r_b)/2$. As $r_b\to \infty$, $1/a_1\to 0$, so this speed $\to \sqrt{2/r_1}=$ escape speed. Flinging farther out demands a bigger initial kick.
• Net: going $r_b:60\to 100$, the Burn-2 saving (plus the Burn-3 shift) outweighed the Burn-1 rise, so the total dropped $0.520743\to 0.51591$. The two effects are a tug-of-war that only stops improving at the bi-parabolic limit.

• $a_1=50.5,a_2=(6+100)/2=53$.
• Burn 1: $\sqrt{2-\frac{1}{50.5}}-1=1.407190-1=0.407190$ — identical to Ex 2.2, because Burn 1 depends only on $r_1$ and $r_b$, not on $r_2$.
• Burn 2: $\sqrt{\frac{2}{100}-\frac{1}{53}}-\sqrt{\frac{2}{100}-\frac{1}{50.5}}=\sqrt{0.001132}-\sqrt{0.000198}=0.033651-0.014079=0.019572$.
• Burn 3 (brake at $r_2=6$): $\sqrt{\frac{2}{6}-\frac{1}{53}}-\sqrt{\frac{1}{6}}=\sqrt{0.314570}-0.408248=0.560866-0.408248=0.152618$.
• Total: $0.407190+0.019572+0.152618=0.579380$.
• On the parent's $0.6034$: that value used the rounded intermediate roots (e.g. $\sqrt{0.3144}$, $a_2$ printed as $53$) and accumulated those roundings; carrying full precision the honest total is $0.5794$. The $\approx 0.024$ gap is not trivial rounding of the final answer — it is the compounded effect of rounding each square root to 4 digits before subtracting nearly-equal numbers (catastrophic cancellation in Burn 2 especially). Either way the conclusion stands: $0.5794>0.4925$, Hohmann wins.
• Culprit: Burn 1 ($0.407$). At small $R$ you still pay the full escape-like cost of flinging out to $r_b=100$, but the target sits at only $r_2=6$ — that huge climb is mostly wasted, and the meagre Burn-2 saving cannot repay it. $R=6<11.94$ → Hohmann wins.

Hohmann, $a_H=(1+13)/2=7$:

• Burn A: $\sqrt{2-\frac{1}{7}}-1=\sqrt{1.857143}-1=1.362770-1=0.362770$. Why: vis-viva at $r_1$ on the Hohmann ellipse, minus $v_{c1}$.
• Burn B: $v_{c2}-\sqrt{\frac{2}{13}-\frac{1}{7}}=\sqrt{1/13}-\sqrt{0.153846-0.142857}=0.277350-0.104828=0.172522$. Why: Hohmann arrives at $r_2$ below circular speed (it's apoapsis), so this burn adds speed.
• Hohmann total $=0.362770+0.172522=0.535292$.

Bi-elliptic, $r_b=100$: $a_1=50.5,a_2=(13+100)/2=56.5$:

• Burn 1: $0.407190$.
• Burn 2: $\sqrt{\frac{2}{100}-\frac{1}{56.5}}-\sqrt{\frac{2}{100}-\frac{1}{50.5}}=\sqrt{0.002301}-\sqrt{0.000198}=0.047968-0.014079=0.033889$.
• Burn 3 (brake): $\sqrt{\frac{2}{13}-\frac{1}{56.5}}-\sqrt{1/13}=\sqrt{0.136147}-0.277350=0.368981-0.277350=0.091631$.
• Bi-elliptic total $=0.407190+0.033889+0.091631=0.532710$.

Result: $0.532710<0.535292$ → bi-elliptic wins, but barely ($\approx 0.5\%$). This is the grey zone: the outcome hinges on $r_b$ and the margin is razor-thin. A planner weighs that sliver against a far longer flight time.

Take limits term by term (with $r_1=1,r_2=R$), each with its physical meaning.

• Burn 1 (the climb): $a_1=\frac{1+r_b}{2}\to \infty$, so $\sqrt{2-1/a_1}\to \sqrt{2}$. Thus $\Delta v_1\to \sqrt{2}-1$ — you reach escape speed at $r_1$ and no more.
• Burn 2 (far-out reshape): both speeds at $r_b$ scale like $\sqrt{2/r_b}\to 0$; their difference $\to 0$. So $\Delta v_2\to 0$ — free plane/apsis reshape at infinity.
• Burn 3 (brake in): $a_2=\frac{R+r_b}{2}\to \infty$, so $\sqrt{2/R-1/a_2}\to \sqrt{2/R}$. Thus $\Delta v_3\to \sqrt{2/R}-\sqrt{1/R}=\frac{\sqrt{2}-1}{\sqrt{R}}$ — the same escape-vs-circular gap, scaled down by the slower speeds at $r_2$. $\boxed{\Delta v_{\text{bipar}}=(\sqrt{2}-1)\left(1+\frac{1}{\sqrt{R}}\right)}$ At $R=20$: $(\sqrt{2}-1)(1+1/\sqrt{20})=0.414214\times 1.223607=0.506841$. Compare Ex 3.1's finite value $0.51591$ — the bi-parabolic limit is lower (it is the floor), confirming larger $r_b$ keeps helping toward this bound.

Hohmann in normalized units, $a_H=\frac{1+R}{2}$ (derived the same way as Ex 4.2, now symbolic): $\Delta v_H=\left(\sqrt{2-\frac{2}{1+R}}-1\right)+\left(\frac{1}{\sqrt{R}}-\sqrt{\frac{2}{R}-\frac{2}{1+R}}\right)$ At $R=11.94$: $a_H=6.47$.

• Burn A: $\sqrt{2-2/12.94}-1=\sqrt{1.845440}-1=0.358470$.
• Burn B: $1/\sqrt{11.94}-\sqrt{2/11.94-2/12.94}=0.289396-\sqrt{0.012944}=0.289396-0.113772=0.175624$.
• $\Delta v_H=0.534094$.
• Bi-parabolic: $(\sqrt{2}-1)(1+1/\sqrt{11.94})=0.414214\times 1.289396=0.534089$.
• $|0.534094-0.534089|\approx 5\times 10^{-6}$ → equal to rounding. ✅ This equality is precisely why $11.94$ is the lower magic number: below it the bi-parabolic (best-possible bi-elliptic) can no longer match Hohmann.

At $R=10$ (below threshold):

• Hohmann: $a_H=5.5$; Burn A $=\sqrt{2-2/11}-1=0.348469$; Burn B $=1/\sqrt{10}-\sqrt{0.2-2/11}=0.316228-0.134840=0.181388$; total $=0.529857$.
• Bi-parabolic: $0.414214\times (1+1/\sqrt{10})=0.414214\times 1.316228=0.545225$.
• $0.529857<0.545225$ → Hohmann strictly wins, as promised for $R<11.94$. ✅

A pure plane change of angle $\Delta i$ at speed $v$ costs $\Delta v\approx 2v\sin (\Delta i/2)$ — it is proportional to the speed at the burn point (see Plane change maneuvers). At $r_1$ the spacecraft is fast ($v_{c1}$ large), so rotating the orbit plane is expensive. At $r_b$ the speed is tiny (vis-viva: $v\to 0$ as $r_b\to \infty$), so the same angular rotation costs almost nothing.

• The rule: rotate where you are slow. Fling out to $r_b$, do the plane change (and periapsis raise) at that lazy far point, then coast back. When the required plane change is large, this can save more than the extra climbs cost — a direct cousin of the size-change crossover, and the inverse of the Oberth effect (which says speed changes are cheap where you are fast).