Bi-elliptic transfer — when it wins over Hohmann

WHAT is a bi-elliptic transfer?

WHY would a detour ever save fuel?

The key physics is the vis-viva equation — the speed on any orbit:

$v^2 = \mu\left(\frac{2}{r} - \frac{1}{a}\right)$

HOW to derive the total $\Delta v$ (from scratch)

We derive everything from vis-viva. Define $\mu = GM$ .

Speeds on circular orbits. Setting $a = r$ in vis-viva: $v_{c}(r) = \sqrt{\frac{\mu}{r}}$ So $v_{c1}=\sqrt{\mu/r_1}$ , $v_{c2}=\sqrt{\mu/r_2}$ .

Ellipse 1 connects $r_1$ (periapsis) and $r_b$ (apoapsis): $a_1 = \frac{r_1 + r_b}{2}$

Why? Semi-major axis is the average of the two apsidal radii.

Ellipse 2 connects $r_2$ (periapsis) and $r_b$ (apoapsis): $a_2 = \frac{r_2 + r_b}{2}$

Now apply vis-viva at each burn point.

Burn 1 — at $r_1$ , jump from circular ( $a=r_1$ ) to ellipse 1 ( $a=a_1$ ): $\Delta v_1 = \sqrt{\mu\left(\frac{2}{r_1}-\frac{1}{a_1}\right)} - \sqrt{\frac{\mu}{r_1}}$

Burn 2 — at $r_b$ , jump from ellipse 1 to ellipse 2 (both raise the far apsis's opposite side): $\Delta v_2 = \sqrt{\mu\left(\frac{2}{r_b}-\frac{1}{a_2}\right)} - \sqrt{\mu\left(\frac{2}{r_b}-\frac{1}{a_1}\right)}$

Why this is small: at large $r_b$ , the $\frac{2}{r_b}$ term dominates both roots and they nearly cancel.

Burn 3 — at $r_2$ , drop from ellipse 2 to circular ( $a=r_2$ ). This is a braking burn, so we take the magnitude: $\Delta v_3 = \sqrt{\mu\left(\frac{2}{r_2}-\frac{1}{a_2}\right)} - \sqrt{\frac{\mu}{r_2}}$

The crossover rule (the 80/20 takeaway)

Let $R = r_2/r_1$ . In the bi-parabolic limit ( $r_b\to\infty$ ) one can show the classic result:

If $R < 11.94$ → Hohmann always wins.
If $R > 15.58$ → bi-elliptic (with large enough $r_b$ ) always wins.
Between $11.94$ and $15.58$ → depends on the chosen $r_b$ .

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

Worked examples

Worked example Example 1 — LEO to a huge orbit (

$R = 20$ ) Take $\mu=1$ , $r_1=1$ , $r_2=20$ , and choose $r_b = 100$ .

Circular speeds: $v_{c1}=\sqrt{1/1}=1$ , $v_{c2}=\sqrt{1/20}=0.2236$ . Why? $v_c=\sqrt{\mu/r}$ .

Ellipse axes: $a_1=(1+100)/2=50.5$ , $a_2=(20+100)/2=60$ . Why? average of apsidal radii.

Burn 1: $\sqrt{2/1-1/50.5}-1 = \sqrt{1.9802}-1 = 0.4072$ . Why? vis-viva at $r_1$ minus circular speed.

Burn 2: $\sqrt{2/100-1/60}-\sqrt{2/100-1/50.5}=\sqrt{0.003333}-\sqrt{0.000198}=0.0577-0.0141=0.0437$ . Why? far-out burn — notice how tiny it is.

Burn 3 (brake): $\sqrt{2/20-1/60}-\sqrt{1/20}=\sqrt{0.08333}-0.2236=0.2887-0.2236=0.0651$ .

Total bi-elliptic: $0.4072+0.0437+0.0651 = 0.516$ .

Hohmann: $a_H=(1+20)/2=10.5$ . Burn A $=\sqrt{2/1-1/10.5}-1=\sqrt{1.9048}-1=0.3801$ . Burn B $=\sqrt{1/20}-\sqrt{2/20-1/10.5}=0.2236-\sqrt{0.00476}=0.2236-0.0690=0.1546$ . Total Hohmann $=0.5347$ .

Result: bi-elliptic $0.516 < 0.535$ Hohmann → bi-elliptic wins ✅ (since $R=20>15.58$ ).

Worked example Example 2 — GEO-like ratio (

$R = 6$ ) $r_1=1$ , $r_2=6$ , try $r_b=100$ . Hohmann ( $a_H=3.5$ ): $\sqrt{2-1/3.5}-1 = 0.3025$ ; second $=\sqrt{1/6}-\sqrt{2/6-1/3.5}=0.4082-\sqrt{0.0476}=0.4082-0.2182=0.1900$ . Total $=0.4925$ . Bi-elliptic ( $a_1=50.5,a_2=53$ ): Burn1 $=0.4072$ ; Burn2 $=\sqrt{0.003333}-\sqrt{0.000198}=0.0437$ ; Burn3 $=\sqrt{2/6-1/53}-\sqrt{1/6}=\sqrt{0.3144}-0.4082=0.5607-0.4082=0.1525$ . Total $=0.6034$ . Result: Hohmann $0.4925 <$ bi-elliptic $0.6034$ → Hohmann wins ✅ (since $R=6<11.94$ ). Detour was pure waste here.

Common mistakes

Feynman check

Recall Explain to a 12-year-old

Imagine you're on a merry-go-round and want to jump onto a much bigger, slower one far away. Turning yourself around is hard when you're spinning fast. So instead you first throw yourself waaay out to where everything drifts slow and lazy — out there you can gently nudge yourself to line up with the big ride, because nothing's whizzing by. Then you coast back and hop on. Going the long way sounds silly, but the gentle nudge in the calm far-away zone can save more energy than it costs — but only if the big ride is really, really far out (about 12× bigger). Otherwise, just take the straight two-jump path.

Flashcards

How many burns does a bi-elliptic transfer use, vs Hohmann?

Bi-elliptic uses 3 burns and 2 ellipses; Hohmann uses 2 burns and 1 ellipse.

Below what radius ratio

R=r_2/r_1

does Hohmann always beat bi-elliptic?

R < 11.94

Above what radius ratio is bi-elliptic (with suitable

r_b

) always better?

R > 15.58

Why is the middle (second) burn of a bi-elliptic transfer so cheap?

It happens at large radius

r_b

where orbital speeds are tiny, so the velocity change to raise periapsis is small (vis-viva:

2/r_b

term dominates and nearly cancels).

Is burn 3 (circularization at

r_2

) a prograde or retro burn?

Retro (brake) — at

r_2

you're at periapsis of ellipse 2, moving faster than circular speed.

What is the vis-viva equation?

v^2 = \mu(2/r - 1/a)

What is the semi-major axis of the first bi-elliptic ellipse?

a_1 = (r_1 + r_b)/2

What is the bi-parabolic limit and its cost?

The

r_b\to\infty

case giving minimum bi-elliptic

\Delta v

, but at the cost of infinite transfer time.

Main hidden downside of bi-elliptic even when it saves fuel?

Much longer transfer time (larger orbits → longer periods).

Connections

Hohmann transfer — the two-burn baseline this competes with
Vis-viva equation — the single tool all $\Delta v$ 's come from
Oberth effect — burns are more/less effective depending on local speed
Semi-major axis and orbital energy — why $a$ sets the speed
Plane change maneuvers — often combined with bi-elliptic (cheap far out)
Transfer time vs delta-v tradeoffs — the time penalty
Bi-parabolic transfer — the limiting best case

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, orbit change karne ke do popular tareeke hain. Ek Hohmann — do burns, ek beech-wala ellipse, seedha inner circle se outer circle. Dusra bi-elliptic — teen burns aur do ellipse. Isme trick ye hai ki pehle tum apne aap ko target se bhi bahut door fling karte ho, phir wahan se neeche aake settle hote ho. Sunne me lagta hai bewakoofi — jab kaam nazdeek ka hai to itni door kyun jao? Lekin physics ka mazaa yahin hai.

Reason vis-viva equation me chhupa hai: $v^2=\mu(2/r-1/a)$ . Jab tum bahut door ( $r_b$ bada) hote ho, saari speeds bahut choti ho jaati hain. Us calm zone me apni periapsis uthana (middle burn) almost muft padta hai, kyunki dono velocity vectors chhote hote hain. Iska price? Do baar gravity well me deep climb karna padta hai. Isliye ye sirf tab faayda deta hai jab target orbit bahut bada ho.

Magic number yaad rakho: agar $R=r_2/r_1 < 11.94$ , Hohmann hamesha jeetega. Agar $R > 15.58$ , bi-elliptic (sahi $r_b$ ke saath) jeet jaata hai. Beech me depends karta hai. Aur ek chhoti si but important baat — burn 3 ek brake hai, acceleration nahi, kyunki $r_2$ pe tum circular se tez chal rahe hote ho.

Ek aur cheez: bi-elliptic fuel bachata hai lekin time bahut leta hai — bade orbit ka period lamba hota hai, to transfer months–years lag sakta hai. To exam me bhi aur real mission me bhi, fuel-win alag hai aur mission-win alag. Dono tarazu me tolo.

Bi-elliptic transfer — when it wins over Hohmann

WHAT is a bi-elliptic transfer?

WHY would a detour ever save fuel?

HOW to derive the total $\Delta v$ (from scratch)

The crossover rule (the 80/20 takeaway)

Worked examples

Common mistakes

Feynman check

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Orbital Mechanics & Astrodynamics

Connections