Bi-elliptic transfer — when it wins over Hohmann: Question bank

Before you start, keep these earned symbols in mind:

$r_{1}$ = radius of the inner starting circular orbit; $r_{2}$ = radius of the outer target circular orbit.
$r_{b}$ = the far intermediate radius the bi-elliptic transfer flings you out to ( $r_{b} > r_{2}$ ).
$R = r_{2} / r_{1}$ = how many times bigger the target orbit is than the start.
$μ = GM$ = the gravitational parameter (planet's mass $\times$ big-G), which sets all speeds via the Vis-viva equation.
$Δ v$ = the total "speed budget" a maneuver spends — the currency of fuel.

(Note on callouts: the boxed headers below like [!formula] and [!intuition] are just labelled boxes — if your reader shows them as plain text, read the bold word as the box's title. The physics is in the text either way.)

The picture behind every question

The whole maneuver lives in one diagram. Study it before answering — most traps below are about which point a burn happens at and which way it pushes.

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

Look at the figure: the small blue circle is the start ( $r_{1}$ ), the green circle is the target ( $r_{2}$ ), and the two dashed ellipses both stretch out to the far point $r_{b}$ on the left. Burn 1 (orange) happens at $r_{1}$ and lifts the far end to $r_{b}$ . Burn 2 (blue) happens way out at $r_{b}$ and lifts the near end from $r_{1}$ up to $r_{2}$ . Burn 3 (red) happens back at $r_{2}$ and is a brake.

Why the semi-major axis is the average of the two apsidal radii

Look at the ellipse figure. The long axis (the major axis) runs in a straight line from the nearest point (periapsis, distance $r_{p}$ from the planet) through the planet's focus to the farthest point (apoapsis, distance $r_{a}$ ). That whole straight segment has length $r_{p} + r_{a}$ . The letter $a$ is the semi-major axis — half the major axis — so $a = \frac{r _{p} + r _{a}}{2},$ which is literally the arithmetic mean of the two endpoints' distances. That's all it is: $a$ is the midpoint distance of the long axis. So for ellipse 1 (periapsis $r_{1}$ , apoapsis $r_{b}$ ) we get $a_{1} = (r_{1} + r_{b}) /2$ , and for ellipse 2 (periapsis $r_{2}$ , apoapsis $r_{b}$ ) we get $a_{2} = (r_{2} + r_{b}) /2$ .

Where the magic numbers 11.94 and 15.58 actually come from

The cleanest way to see them is the bi-parabolic limit: push $r_{b} \to \infty$ . Then each ellipse becomes a parabola (an escape coast), burn 2 shrinks to zero, and the two remaining burns become just "jump onto escape speed" and "come off escape speed": $Δ v_{bipar} = burn 1: circular \to escape at r_{1} (\frac{2 μ}{r _{1}} - \frac{μ}{r _{1}}) + burn 3: escape \to circular at r_{2} (\frac{2 μ}{r _{2}} - \frac{μ}{r _{2}}) .$ Set this equal to $Δ v_{H}$ , divide through by $μ / r_{1}$ , and everything becomes a function of $R = r_{2} / r_{1}$ alone (the physics is scale-free). Solving that single equation numerically gives $R = 15.58$ : at exactly this ratio the ideal bi-parabolic ties Hohmann, and above it bi-elliptic wins. The lower number $11.94$ comes from the same comparison but keeping $r_{b}$ finite and asking "is there any $r_{b}$ that beats Hohmann?" — below $11.94$ the answer is no for every $r_{b}$ . The second figure shows both curves crossing between these two markers:

The orange Hohmann curve and blue bi-parabolic curve cross at $R = 15.58$ ; the blue curve stays below for larger $R$ . The two vertical gray markers are the thresholds $11.94$ (finite- $r_{b}$ onset) and $15.58$ (guaranteed win) every question below refers to.

True or false — justify

TF1. A bi-elliptic transfer always uses fewer burns than a Hohmann transfer.

False — it uses more: three burns and two ellipses versus Hohmann's two burns and one ellipse. Its advantage (when it exists) is lower total

Δ v

, never fewer impulses.

TF2. Because orbital speeds are tiny far out, the middle burn is always the cheapest of the three.

True in the sense that it is the smallest of the three: at large

r_{b}

both speeds in

Δ v_{2}

are

μ (2/ r_{b} - 1/ a)

, and since

2/ r_{b}

shrinks toward zero while

1/ a_{1} \approx 1/ a_{2}

, the two square roots collapse to nearly the same tiny value and their difference vanishes — but "cheap middle burn" does not mean the whole transfer is cheap.

TF3. Bi-elliptic transfer beats Hohmann whenever the target orbit is larger than the start.

False — a larger target is necessary but nowhere near sufficient. Below

R = 11.94

Hohmann always wins; the detour only pays off for very large ratios.

TF4. If

R = 20

, bi-elliptic is guaranteed to beat Hohmann for any choice of

r_{b}

False —

R = 20 > 15.58

guarantees a win exists for a suitable $r_{b}$ , but a badly chosen (too small)

r_{b}

can still lose. The threshold promises that some

r_{b}

works, not every

r_{b}

TF5. Burn 3 (circularizing at

r_{2}

) is a prograde acceleration that "finishes the climb".

False — at

r_{2}

you are at periapsis of the second ellipse, moving faster than the circular speed there, so burn 3 is a retro-burn (brake). You take its magnitude in the

Δ v

sum.

TF6. Taking

r_{b} \to \infty

(the bi-parabolic limit) gives the mathematically minimal

Δ v

for any

R

Partly true and dangerous: the bi-parabolic limit gives the best case used to derive the thresholds, but it also demands infinite transfer time, and it only wins for

R

above the crossover in the first place. See Bi-parabolic transfer.

TF7. For the same start and target, bi-elliptic and Hohmann take roughly the same time.

False — bi-elliptic swings out to

r_{b} ≫ r_{2}

, so its coast legs can take months to years versus Hohmann's fixed half-period. Time is the hidden price of the fuel saving.

TF8. Both maneuvers assume the two orbits are coplanar (in the same flat plane).

True — the whole

Δ v

comparison here is for coplanar circular orbits. Adding an inclination change is a separate cost handled by Plane change maneuvers.

TF9. The saving in a bi-elliptic transfer comes from a smaller first burn than Hohmann's first burn.

False — burn 1 is actually larger than Hohmann's first burn (you aim at

r_{b} > r_{2}

, a bigger ellipse). The net saving comes from the tiny middle burn plus a gentler final circularization, which together can outweigh the heavier burn 1.

TF10. The threshold

R = 15.58

is derived from the bi-parabolic (

r_{b} \to \infty

) case.

True — setting

Δ v_{bipar} = Δ v_{H}

and solving for

R

gives exactly

15.58

; that's the ratio where even the ideal bi-elliptic only ties Hohmann, so any larger

R

guarantees a strict win.

Spot the error

SE1. "

a_{1} = (r_{1} + r_{b}) /2

because the semi-major axis is the sum of the two apsidal radii."

Error: it is the average (arithmetic mean), not the sum —

a_{1} = (r_{1} + r_{b}) /2

. The major axis (length

r_{1} + r_{b}

) is the whole long line;

a

is half of it.

SE2. "The first ellipse connects

r_{1}

r_{2}

, the second connects

r_{2}

r_{b}

Error: ellipse 1 connects

r_{1}

(periapsis) to

r_{b}

(apoapsis), and ellipse 2 connects

r_{2}

(periapsis) to

r_{b}

(apoapsis). Both ellipses share the far apsis

r_{b}

; that shared point is where burn 2 happens.

SE3. "Since burn 2 raises periapsis, and periapsis is the low point, burn 2 must happen at low altitude."

Error: burn 2 happens at apoapsis (

r_{b}

, the high point). A tangential burn at apoapsis changes the opposite apsis — here it lifts the periapsis from

r_{1}

up to

r_{2}

. That's why it's cheap: it's done where speed is smallest.

SE4. "In the crossover rule,

R

is the ratio

r_{b} / r_{1}

Error:

R = r_{2} / r_{1}

, the ratio of target to start radius —

r_{b}

(the detour radius) is a separate free parameter you choose. Confusing them wrecks the 11.94 / 15.58 thresholds.

SE5. "We can drop the absolute values in the

Δ v

sum since all three burns speed the craft up."

Error: burn 3 is a brake (deceleration), so its raw signed value is negative. Fuel cost is

Δ v = ∣ v_{after} - v_{before} ∣

for every burn, so each term needs its own bars before summing.

SE6. "

v_{c} (r) = μ / r

comes from setting

r = a

in vis-viva, so it also gives ellipse speeds."

Error: setting

a = r

is valid only on a circular orbit where

r

is constant. On an ellipse

r

varies while

a

is fixed, so

v_{c} = μ / r

does not apply — you must use full vis-viva.

SE7. "Because the middle burn is nearly free, a bi-elliptic transfer is basically a free plane change."

Error: the middle burn being small is real, but you pay for it by climbing twice deep into the gravity well (a heavier burn 1 and no chance to reuse Hohmann's efficiency). The "free lunch" only survives the accounting when

R

is large.

SE8. "The 15.58 threshold comes from solving the bi-elliptic total at some fixed finite

r_{b}

Error: 15.58 comes from the bi-parabolic limit

r_{b} \to \infty

(the best possible case). The finite-

r_{b}

comparison is what produces the lower number 11.94, the onset of the gray zone.

Why questions

WHY1. Why does the second burn shrink as

r_{b}

grows?

Because both speeds in

Δ v_{2} = ∣ μ (2/ r_{b} - 1/ a_{2}) - μ (2/ r_{b} - 1/ a_{1}) ∣

are pulled toward the same tiny value when

r_{b}

is huge — the

2/ r_{b}

contribution goes to zero and

a_{1}, a_{2}

become nearly equal — so the difference collapses.

WHY2. Why does going farther than needed ever save fuel instead of always wasting it?

Reshaping an orbit (changing

a

) is cheapest where speeds are smallest, i.e. far out. The bi-elliptic parks its expensive "raise periapsis" job at

r_{b}

where velocities are minuscule — an inverted Oberth effect logic (burns matter most where speed is high, so the reshaping burn is placed where speed is low).

WHY3. Why is there a whole gray zone (

11.94 < R < 15.58

) instead of one clean threshold?

Because for these ratios the winner depends on the chosen

r_{b}

: small

r_{b}

favours Hohmann, large

r_{b}

favours bi-elliptic. Only below 11.94 does no

r_{b}

help, and only above 15.58 does the win become guaranteed even in the ideal limit. See Transfer time vs delta-v tradeoffs.

WHY4. Why can't we just always pick

r_{b} = \infty

to guarantee the best

Δ v

Because infinite

r_{b}

means the craft coasts out forever — infinite transfer time — and it only wins at all when

R

already exceeds the crossover. Real missions choose a finite

r_{b}

trading fuel against time.

WHY5. Why does the crossover analysis use the bi-parabolic limit to state the magic numbers?

The bi-parabolic case (

r_{b} \to \infty

) is the theoretical best a bi-elliptic can do, so comparing it to Hohmann tells you when bi-elliptic can ever win. The 15.58 threshold is exactly where this best case ties Hohmann. A Bi-parabolic transfer replaces each ellipse with a parabola (escape-speed coast), the

r_{b} \to \infty

ideal.

WHY6. Why is burn 1 of the bi-elliptic larger than the first Hohmann burn, given they start identically?

Both start at

r_{1}

on the same circular orbit, but bi-elliptic's ellipse reaches all the way to

r_{b} > r_{2}

, a bigger, more energetic orbit (larger

a_{1}

), so it demands a bigger speed jump to enter. This bigger burn 1 is the "climb tax" you hope the cheap middle burn repays.

WHY7. Why does the comparison depend only on the ratio

R

and the choice of

r_{b} / r_{1}

, not on absolute sizes or

μ

Vis-viva scales cleanly: rewrite everything in units where

μ = 1

and

r_{1} = 1

, and all

Δ v

values become functions of

R

and

r_{b}

alone. The physics is scale-free, which is why one universal threshold table exists — see Semi-major axis and orbital energy.

WHY8. Why is

a = (r_{p} + r_{a}) /2

literally just "the middle of the long axis"?

The major axis is a single straight line from periapsis (

r_{p}

from the focus) to apoapsis (

r_{a}

from the focus), total length

r_{p} + r_{a}

; the semi-major axis

a

is defined as half that line, i.e. the arithmetic mean of the two endpoint distances.

Edge cases

EC1. What happens to a bi-elliptic transfer as

r_{b} \to r_{2}

(the smallest allowed detour)?

It degenerates: the "detour" barely exceeds the target, ellipse 2 shrinks toward a circle, and the maneuver collapses toward an ordinary Hohmann-like two-burn transfer with no saving. The requirement is strictly

r_{b} > r_{2}

EC2. What if

R = 1

exactly (

r_{1} = r_{2}

, same-size orbits)?

No transfer is needed at all — target and start coincide. Any bi-elliptic detour is pure wasted

Δ v

(two climbs to nowhere). This is the extreme of the "small

R

= waste" regime.

EC3. What if the target orbit is smaller than the start (

r_{2} < r_{1}

, i.e.

R < 1

The maneuver runs in reverse (you're deorbiting/lowering), and the same vis-viva accounting applies with roles swapped. The 11.94 / 15.58 thresholds are stated for raising orbits; for lowering, use the ratio the correct way round (

R = larger / smaller

EC4. At exactly

R = 11.94

, who wins?

This is the boundary where Hohmann's guaranteed advantage ends — at

11.94

the two are essentially tied for the best finite

r_{b}

, so neither strictly beats the other; it is the onset of the gray zone, not a clean flip.

EC5. At exactly

R = 15.58

(the upper bi-parabolic threshold), who wins?

This is the point where even the ideal case flips: at

R = 15.58

the bi-parabolic (

r_{b} \to \infty

) transfer exactly ties Hohmann, and for any

R

above it bi-elliptic strictly wins with a large enough

r_{b}

. At the threshold itself they are tied in the limit; a finite

r_{b}

still needs

R

a touch higher to guarantee a win.

EC6. In the

r_{b} \to \infty

limit, does the total time also go to infinity?

Yes — the coast out to and back from an ever-more-distant apoapsis takes ever longer, so the bi-parabolic limit is a fuel ideal that is never achievable in finite mission time.

EC7. Does the bi-elliptic advantage survive if we must also change orbital plane?

It can grow: a plane change is far cheaper at the low speeds near

r_{b}

, so combining the inclination change with the far burn (Plane change maneuvers) can push the crossover in bi-elliptic's favour even below

R = 11.94

EC8. What is the smallest number of ellipses and burns a bi-elliptic transfer can have?

Exactly two ellipses and three burns by definition — remove one ellipse and you're back to a Hohmann transfer, so this is the irreducible structure of the maneuver.

Recall One-line survival summary

Bi-elliptic = 3 burns, 2 ellipses, cheap middle burn far out, expensive climb tax; wins only for large $R$ (>15.58 guaranteed, <11.94 never, tied at the thresholds in the limit), and always costs far more time. Every burn: $Δ v = ∣ v_{after} - v_{before} ∣$ .