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Unit 6 · Lesson 6-3
Bi-elliptic Transfers and the Oberth Effect
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Bi-elliptic Transfers and the Oberth Effect

Three-impulse transfers, the 11.94 crossover ratio and the Oberth effect

A counter-intuitive transfer
You are in a circular orbit and you want to be in a much larger circular one. The Hohmann transfer is proven optimal. Two burns, minimum energy, a century of textbooks agree. Here is an alternative. Instead of transferring directly, burn much harder — hard enough to fling yourself out to an apogee far beyond your target. Not slightly beyond. Ten times beyond. A million kilometres, if you like. Out there, essentially at rest, make a tiny burn to raise your perigee up to the target radius. Then fall all the way back in, and circularise on arrival. Three burns instead of two. A journey of weeks or months instead of hours. An apogee absurdly far from anywhere you wanted to be. This is a worse idea in every respect except one. Past a radius ratio of about twelve, it uses less propellant. And the reason it works is a fact you already derived in Unit 1 and have been using ever since without noticing what it implies.
Watch first · presented by Shoaib Aziz
The numbers in this film are the labs' own: from 7,000 km to 140,000 km (a ratio of 20) Hohmann costs 2.87 + 1.17 = 4.035 km/s in 27.5 hours, while a bi-elliptic transfer out to 700,000 km costs 3.07 + 0.329 + 0.491 = 3.893 km/s, saving 142 m/s but taking 27.8 days; Hohmann always wins below a ratio of 11.94 and bi-elliptic wins above 15.58 if it goes far enough; 0.5 km/s at GTO perigee lifts apogee to almost 127,000 km but the same burn at apogee lifts perigee only to about 13,000 km, 5.53 times less energy; and going out 1,000 times the target radius saves 210 m/s at the cost of 185 years.
The mechanism is the Oberth effect, and it is worth stating precisely because it governs far more than this one manoeuvre.
Energy is ε=v2/2−μ/r\varepsilon = v^2/2 - \mu/r. Add a small Δv\Delta v along the velocity and the energy changes by
Δε=(v+Δv)22−v22=v Δv+Δv22≈v Δv\Delta\varepsilon = \frac{(v+\Delta v)^2}{2} - \frac{v^2}{2} = v\,\Delta v + \frac{\Delta v^2}{2} \approx v\,\Delta v

The energy bought by a burn is proportional to the speed you already had. A fixed Δv is worth more when you are moving faster — the Oberth effect, named for Hermann Oberth, who described it in 1929.

Figure 6.7
the same 0.5 km/s burnat perigee — v = 10.07 km/sΔε = 5.161 km²/s² · apogee → 127×10³ kmat apogee — v = 1.62 km/sΔε = 0.934 km²/s² · perigee → 13.1×10³ km5.53× more energyΔε ≈ v·Δv — the energy youbuy is proportional to thespeed you already had.
The same 1 km/s burn applied at perigee and at apogee of a GTO. At perigee, where the spacecraft is doing 10.07 km/s, the resulting apogee climbs enormously. The identical burn at apogee, at 1.62 km/s, barely moves perigee. Same propellant, same engine, five and a half times the energy.
💡Where does the extra energy come from?
This is the question that makes people suspicious of the Oberth effect, and the answer is genuinely satisfying. Energy is conserved, so the extra cannot come from nowhere. It comes from the propellant. Consider the exhaust. In the spacecraft's frame it leaves at vev_e regardless of where the burn happens. But in the inertial frame, its speed is the vector sum of vev_e and the spacecraft's own velocity. Burn at perigee, moving fast, and the exhaust is thrown backwards so hard that it ends up nearly stationary in the inertial frame — or even moving backwards. Its kinetic energy has been almost entirely stripped away. Burn at apogee, moving slowly, and the exhaust retains a great deal of kinetic energy that simply flies away with it, wasted. So the Oberth effect is not free energy. It is a more efficient transfer of energy from propellant to vehicle — extracting more of the chemical energy you already paid for. Nothing is violated, and the spacecraft simply does a better job of robbing its own exhaust.
There ain't no such thing as a free lunch.
– Robert A. Heinlein, The Moon Is a Harsh Mistress, 1966
Now the bi-elliptic transfer. Three burns, two ellipses, one intermediate apogee radius rbr_b that you get to choose:
BurnWhereWhat it doesCost driver
1On the inner circle at r₁Raise apogee to the far point r_bLarge — near escape speed if r_b is huge
2At the far apogee r_bRaise perigee from r₁ to r₂Tiny — you are barely moving out there
3On arrival at r₂Circularise (a retrograde burn)Small — you arrive too fast, so you slow down
Δv1=μ(2r1−1a1)−μr1,a1=r1+rb2Δv2=μ(2rb−1a2)−μ(2rb−1a1),a2=r2+rb2Δv3=μ(2r2−1a2)−μr2\begin{aligned}\Delta v_{1} &= \sqrt{\mu\left(\tfrac{2}{r_1}-\tfrac{1}{a_1}\right)} - \sqrt{\tfrac{\mu}{r_1}}, & a_{1} &= \tfrac{r_1+r_b}{2}\\[4pt] \Delta v_{2} &= \sqrt{\mu\left(\tfrac{2}{r_b}-\tfrac{1}{a_2}\right)} - \sqrt{\mu\left(\tfrac{2}{r_b}-\tfrac{1}{a_1}\right)}, & a_{2} &= \tfrac{r_2+r_b}{2}\\[4pt] \Delta v_{3} &= \sqrt{\mu\left(\tfrac{2}{r_2}-\tfrac{1}{a_2}\right)} - \sqrt{\tfrac{\mu}{r_2}}\end{aligned}

The three burns. Note Δv₃ comes out negative — it is a braking burn, and you take its magnitude. Propellant does not care about direction.

💡Why the bi-elliptic transfer can be cheaper
The bi-elliptic transfer makes a specific trade, and it is worth naming. It pays more on burn 1. Going to a much higher apogee than necessary costs extra up front — though by less than you would think, because the required speed at r1r_1 asymptotes to escape speed rather than growing without bound. All the way to infinity costs only 2\sqrt{2} times circular speed. It gets burns 2 and 3 almost free. Out at rbr_b the spacecraft is barely moving, so a tiny nudge swings perigee across an enormous distance. That is the Oberth effect running in reverse: where speed is low, a burn changes the orbit's shape dramatically for very little Δv. And it dodges the expensive middle. The Hohmann transfer's cost peaks at a radius ratio near 15.6 precisely because the intermediate regime is where both burns are substantial. The bi-elliptic route steps over that region entirely. So it is not a trick. It is the recognition that in a 1/r1/r potential, plane and shape changes are cheap where you are slow, and energy changes are cheap where you are fast — and if you can arrange to make each kind of change in the right place, you win.
Figure 6.8
Hohmann4.035 km/s · 1.15 dbi-elliptic3.893 km/s · 28 dsaves 142 m/s — 3.5%costs 27 extra daysRatio r₂/r₁ = 20, with anintermediate apogee fivetimes further out than thetarget itself.
The two routes, drawn to scale for a radius ratio of 20. Hohmann is the compact ellipse touching both circles. Bi-elliptic swings far off the right of the page and comes back. It is dramatically longer, obviously more complicated, and slightly cheaper.
The crossover is a clean, classical result, and it has two numbers rather than one — which is where most summaries get it wrong.
Radius ratio r₂/r₁Verdict
Below 11.94Hohmann always wins, for every possible choice of r_b
11.94 to 15.58Depends on r_b — bi-elliptic wins only if the intermediate apogee is large enough
Above 15.58Bi-elliptic always wins, for a large enough r_b
As r_b → ∞The bi-elliptic cost approaches its limit, and the saving is at most a few percent
Figure 6.9
510152025300.40.50.6radius ratio r₂ / r₁total Δv ÷ v_circ(r₁)11.9415.58Hohmannr_b = 5 r₂r_b = 20 r₂r_b → ∞Below 11.94 Hohmannalways wins.Above 15.58 bi-ellipticalways can.Note the vertical scale:the saving is a few percent.
Total Δv against radius ratio for Hohmann and for bi-elliptic transfers with several intermediate radii. The curves cross in a fan between 11.94 and 15.58. Note the vertical scale: the saving above the crossover is single-digit percent, and it is bought with weeks or months of extra flight time.
📐Worked Example: LEO to a very high orbit
From r1=7000r_1 = 7000 km to r2=140,000r_2 = 140{,}000 km — a ratio of 20, comfortably past both crossovers. Hohmann. at=73,500a_t = 73{,}500 km. vc1=7.546v_{c1} = 7.546, vt1=10.415v_{t1} = 10.415, so Δv1=2.869\Delta v_1 = 2.869 km/s. vc2=1.687v_{c2} = 1.687, vt2=0.521v_{t2} = 0.521, so Δv2=1.167\Delta v_2 = 1.167 km/s. Total = 4.035 km/s, transfer time 27.5 hours. Bi-elliptic with rb=700,000r_b = 700{,}000 km (five times the target radius): a1=353,500a_1 = 353{,}500 km, a2=420,000a_2 = 420{,}000 km. Δv1=10.619−7.546=3.073\Delta v_1 = 10.619 - 7.546 = 3.073 km/s — more than the Hohmann first burn, as promised. Δv2\Delta v_2 out at 700,000 km: 0.436−0.106=0.3300.436 - 0.106 = 0.330 km/s. Three hundred metres per second, to swing perigee across 133,000 km. Δv3=1.687−2.178\Delta v_3 = 1.687 - 2.178, magnitude 0.491 km/s — a braking burn, because you arrive too fast. Total = 3.893 km/s. The verdict. Bi-elliptic saves 142 m/s, about 3.5%. And it costs roughly 28 days of flight time instead of 27.5 hours — a factor of twenty-four. So would you fly it? For a crewed mission, obviously not. For a satellite with a hard delivery date, no. For a deep-space probe already spending years in transit, where 142 m/s might be the difference between carrying an extra instrument and not — quite possibly yes. The answer is a mission-design judgement, and the arithmetic above is what informs it.
🛰Where the Oberth effect really earns its keep
The bi-elliptic transfer is mostly a theoretical curiosity. The Oberth effect underneath it is anything but. Every interplanetary departure is an Oberth manoeuvre. A probe leaving for Jupiter does not burn from a high parking orbit. It drops to a low perigee — as deep in Earth's well as it dares — and burns there, where it is moving fastest. The saving over a high-altitude departure is measured in kilometres per second. Powered flybys take it further. Arrive at a planet on a hyperbolic trajectory, and at closest approach you are moving faster than anywhere else on the entire trajectory. Burn there and the Oberth multiplier is enormous. Proposals for reaching the outer solar system quickly rely on a powered perihelion pass close to the Sun, where speeds exceed 200 km/s and a modest burn buys a colossal amount of energy. And it constrains where you cannot burn. A spacecraft that has already escaped, coasting far from any mass, is in the worst possible place to change its trajectory. Which is why interplanetary trajectory correction manoeuvres are tiny, carefully rationed, and scheduled near planets whenever possible.
One practical variant deserves a mention, because it is flown far more often than the bi-elliptic transfer ever is. A one-tangent burn keeps the first burn tangential but makes the second one at an angle, arriving at the target orbit before apogee. It is more expensive than Hohmann and considerably faster — and that trade, fuel against time, is what most real transfer design actually consists of.
↓But there is a problem
Every transfer in this unit so far — Hohmann, bi-elliptic, one-tangent — has changed the size and shape of the orbit. Not one of them has changed the plane. And every one of them has been cheap, in the sense that a few km/s moves you across tens of thousands of kilometres. Now try to change the plane. A satellite in LEO wants to shift its orbital plane by one single degree. Not to a different altitude. Not to a different shape. The same orbit, tilted by one degree — a change so small you could not see it in any diagram in this course. That costs more than raising your altitude by a hundred kilometres. And getting from a Cape Canaveral launch to the equatorial GEO belt requires 28.5 degrees of it.