Flying further away in order to turn

The first rung of this anchor established that a plane change costs 2v sin(Δi/2) and is therefore proportional to the speed at which it is done. The second took that seriously and divided a rotation between the two burns of a transfer, finding an interior optimum worth 25 metres a second.

Both took the available speeds as given: whatever the vehicle happens to be doing at the two ends of a transfer it was flying anyway. There is a third option, and it is the one that sounds absurd.

A vehicle can make a slow place. Climbing to a very high apoapsis costs a burn, and the burn’s cost does not grow with the angle being turned, while the rotation’s cost falls as the speed there falls. Two costs, one fixed and one shrinking, and a crossover.

The manoeuvre is not a curiosity. It is flown, under a different name, by most geostationary satellites launched from high latitudes, and the arithmetic that decides when it is worth flying is the same arithmetic that decides everything else on this anchor: where the vehicle is when the rotation happens.

Why the crossover is an angle #

The bi-elliptic transfer as a way of changing radius has a crossover too, and it is a radius ratio: 11.94, derived once in 1959, above which flying out to infinity and back beats a Hohmann. That number appears nowhere here, and the reason is worth being clear about.

For a pure radius change, both routes’ costs are set by the same two circular speeds, so the comparison depends only on their ratio. For a plane change the rotation introduces a second variable — the angle — whose cost scales differently between the two routes. The three-burn route’s rotation happens at almost no speed, so its cost is almost independent of the angle; the single burn’s rotation happens at orbital speed, so its cost grows as sin(Δi/2). One curve is nearly flat and the other rises, and where they cross is decided by the angle.

For a rotation at fixed radius the arithmetic closes exactly. The three-burn route, with the apoapsis taken to infinity, costs the escape burn out and the escape burn back: 2(√2 − 1)v = 0.8284v, with the rotation itself free because the speed at infinity is zero. The single burn costs 2v sin(Δi/2). Setting those equal gives

$$\sin(\Delta i/2) = \sqrt{2} - 1, \qquad \Delta i = 2\arcsin(\sqrt2 - 1) = 48.94°$$

and, as with the angle at which turning costs more than leaving, the speed cancels. That crossover is a property of the geometry and holds at any altitude around any body.

What the far apoapsis is worth, and what it costs #

The saving is bounded and the time is not, which is what makes this manoeuvre rare rather than standard.

At a fixed radius the best the three-burn route can ever do is 0.8284 v, whatever the angle, because the rotation becomes free. Against a single burn’s 2v sin(Δi/2), that caps the saving at 41 per cent for a 90° turn and 59 per cent for a full reversal. Those are large fractions of a large number — for a low Earth orbit at 7.67 km/s, a 90° rotation costs 10.8 km/s directly and 6.4 by the long route.

That is the practically useful reading of the whole rung. The limiting bi-elliptic is a mathematical bound; a real one flies to a few tens of radii, captures ninety-five per cent of the available saving, and takes weeks rather than years.

The version that is actually flown #

The pure rotation above is a clean problem and not a common mission. The common one is a rotation combined with a radius change — a satellite going from an inclined low orbit to an equatorial geostationary one — and there the numbers are different and the manoeuvre is real.

A crossover at 36.6° matters because real launch sites produce real inclinations in that neighbourhood. Kennedy delivers 28.5°, Baikonur 51.6°, and a mission launched from the second into geostationary orbit is on the wrong side of the line — which is why supersynchronous transfers are used from high-latitude sites and not from equatorial ones.

The intermediate radius is not a free parameter in practice #

Three things bound the far apoapsis from above, and none of them is in the arithmetic.

The first is the Hill sphere. A vehicle further than about 1.5 million kilometres from Earth is not bound to Earth in any useful sense — the Sun’s tidal influence exceeds Earth’s own — and a two-body calculation of the return leg stops applying. That is about 235 starting radii from a low orbit, so the limiting bi-elliptic drawn above at 200 is already at the edge of where its own model works.

The second is the Moon. At 60 Earth radii, anything on a trajectory beyond that is subject to lunar perturbations of tens of metres a second per pass, which is comparable to the entire second burn.

The third is time, and it is the binding one for every real mission.

The saving, in kilometres a second #

The units on the figures are the starting orbit’s circular speed, which keeps them general and hides how large the numbers are. Converting them for a low Earth orbit is worth doing once.

Circular speed at 400 km is 7.669 km/s. A 90° rotation there costs 2 × 7.669 × sin(45°) = 10.85 km/s directly — more than reaching orbit in the first place, and more than any satellite has ever carried. By the limiting three-burn route it costs 0.8284 × 7.669 = 6.35 km/s, which is still enormous and is now merely impossible rather than absurd.

A 60° rotation costs 7.67 directly and 6.35 the long way. A 30° rotation costs 3.97 directly and 6.35 the long way, so the long route loses — as the crossover at 48.9° requires.

What those numbers say is that a large plane change in low orbit is not something that gets done. The reason the manoeuvre appears in this collection at all is the combined case, where the vehicle was going to raise its apoapsis anyway and the rotation rides along, and there the sums come down to the hundreds of metres a second that a satellite can actually carry.

What the three burns are #

The route is worth naming burn by burn, because the middle one is not what an intuition about “turning at apogee” expects.

The first burn raises the apoapsis from the starting circular orbit to the intermediate radius. It is a pure prograde burn in the original plane, and it is nearly an escape burn — 0.414 v for the limiting case.

The second happens at the far apoapsis and does two things at once: it rotates the plane, and it adjusts the periapsis from the starting radius to the target radius. Both are done as one vector sum, and at a large apoapsis the vehicle is moving so slowly that the whole burn is a few tens of metres a second. This is where the manoeuvre earns its keep.

The third circularises at the target. It is again a pure prograde or retrograde burn in the final plane.

Where the time goes #

The flight time is the reason this manoeuvre is rare, and it is worth quantifying because the scaling is severe.

The semi-major axis of the outbound ellipse is (1 + B)/2 in starting radii, so by the harmonic law the round trip takes ((1 + B)/2)^(3/2) starting periods for each leg. At B = 20 that is 68 periods; at B = 200 it is 2,015; at B = 1,000 it is 22,394. For a low Earth orbit with a 90-minute period, those are four days, four months and four years.

Four days is a mission. Four years is not, for anything with a schedule, and it is also four years of radiation dose, four years of station-keeping, and four years during which a launch window for the next thing has come and gone. The bound the arithmetic puts on the saving is 41 per cent; the bound the schedule puts on the manoeuvre is a few tens of radii.

What the picture cannot show #

The burns are impulsive, the field is a point mass, and the Moon is not in it. Each omission is worth a sentence and the last of them is the largest.

Impulsive burns at a very high apoapsis are a better approximation than usual, because the vehicle is moving slowly and a burn of a few tens of metres a second at 1.6 per cent of orbital speed takes a small fraction of the local orbital period. It is the first burn, near periapsis at full orbital speed, that suffers the usual finite-burn losses, and those are the same for both routes.

The point-mass field is not innocuous at 200 starting radii. That is 1.3 million kilometres from Earth, well beyond the Moon and comparable to the Earth’s Hill sphere at 1.5 million — so a vehicle there is barely bound, its trajectory is perturbed by the Sun at the tens-of-metres-a-second level, and the two-body arithmetic above stops being the right description. The manoeuvre and the three-body dynamics meet, and what comes out of the meeting is better rather than worse: a vehicle at the edge of the Hill sphere can be turned by the Sun’s perturbation for free, which is the low-energy-transfer trick and is a different subject.

The Moon is the other omission and it is an opportunity rather than a limitation. A lunar flyby is a way of getting a large plane change at no propellant cost at all, and it is flown: the Hiten spacecraft and several stranded geostationary satellites have been recovered by routing them past the Moon to change their inclination. Against that, everything on this page is a way of doing badly what a gravity assist does for nothing — where a suitable body exists.

The number this rung shares with the one before it #

Two exact angles have now appeared on this anchor, both with the speed cancelling, and it is worth putting them side by side because they are the same computation twice.

Turning costs as much as leaving at Δi = 2 arcsin((√2 − 1)/2) = 23.9°. The bi-elliptic beats the single burn at Δi = 2 arcsin(√2 − 1) = 48.9°. The second is the first with the argument doubled, and the reason is transparent once both are written out: the first compares a rotation with one escape burn and the second compares it with two.

So the ladder has a structure. Below 23.9° a rotation in low orbit is cheaper than escaping, and everything can be done locally. Between 23.9° and 48.9° a rotation costs more than escaping but less than escaping and coming back, so the right answer is to go somewhere slower without going all the way. Above 48.9° the round trip to infinity is cheaper than turning at home, and the further out the better.

Three regimes, two boundaries, and both boundaries are properties of √2 — which is the ratio between escape and circular speed, the one number in orbital mechanics that is the same everywhere.

Why the classical result is about infinity and the useful one is not #

There is a pattern in this subject worth naming, because it appears in both bi-elliptic results and in several others.

The clean answer — 11.94 for the radius change, 48.9° for the rotation — is always the limiting case, with the intermediate apoapsis taken to infinity. That is what makes it exact and quotable, and it is what makes it useless as a design point, because an infinite apoapsis takes infinite time.

What a designer needs is the family of answers at finite apoapsis, and those are not clean: 48.2° at twenty radii, 48.9° at a thousand, with the crossover creeping toward the limit and the saving creeping toward its bound. The figures here draw the family rather than the limit for that reason, and the interesting property of the family turns out to be how fast it converges — ninety-five per cent of the available saving at twenty radii, which is a four-day round trip rather than a four-year one.

That convergence is the reason the manoeuvre is practical at all, and it is invisible in the classical statement. A result derived at infinity says nothing about whether the approach is quick or slow, and in this case it happens to be quick. The bi-elliptic radius change has the same structure with the opposite answer: its limiting saving is small and its convergence is slow, so the limit and the practice are both discouraging.

The habit #

The structure worth extracting is that a cost proportional to a state variable can be reduced by changing the state, provided the change is cheaper than the saving — and that the trade is decided by which cost grows with the thing being bought.

Here the rotation’s cost is proportional to speed and the speed can be lowered by climbing. The climb’s cost is fixed; the rotation’s saving grows with the angle. So the answer is an angle, and above it the manoeuvre is worth doing however far it has to go.

The same reasoning applied to a radius change gives a much weaker result, because there the climb’s cost and the saving both scale with the same speeds and the trade is nearly balanced everywhere — which is why 11.94 is a large ratio and 48.9° is a modest angle. And applied to the Oberth effect it runs in reverse: a burn that adds energy is worth more where the vehicle is fast, so the cheapest place to accelerate is the most expensive place to turn, and a mission that wants to do both has to choose.

What it is worth on a real satellite #

The supersynchronous transfer is the version of this manoeuvre with a schedule, and it is worth pricing on a specific case because the numbers are the reason it is flown.

A satellite launched from Baikonur reaches a transfer orbit at 51.6° of inclination. Removing that at geostationary apogee, combined with the circularisation, costs about 1.83 km/s. Raising the apogee to twice geostationary radius first, doing the rotation there where the speed is a third lower, and then lowering the perigee-side of the orbit back down costs about 1.72 — a saving of 110 metres a second for an extra week of transfer and one extra burn.

Against fifteen years of station-keeping at 50 m/s a year, 110 metres a second is more than two years of operational life. That is why it is flown, and it is also why the manoeuvre is never taken to the limiting apoapsis: the marginal saving falls quickly past a factor of two or three in radius while the flight time and the radiation dose through the Van Allen belts do not.

The belts are worth mentioning because they are a cost the arithmetic here has no term for. A supersynchronous transfer crosses the trapped-particle regions twice per revolution rather than once, and the total dose over a multi-week transfer is a real constraint on the solar arrays and the electronics. A manoeuvre that is free in propellant can be expensive in something else.

Where this ladder goes next #

Everything so far has been about a rotation performed in orbit, where the vehicle is moving at kilometres a second and the cost is measured against a transfer budget. The most expensive plane change anybody performs is not in orbit at all.

The next rung is the launch azimuth. A vehicle leaving the surface reaches an inclination fixed by its latitude and the direction it flies, and it cannot reach an inclination below its own latitude by any azimuth whatever. Getting there afterwards means a plane change at low-orbit speed, which is the most expensive place available — and the site’s latitude is therefore a floor that no trajectory removes and no engineering reduces.

Beyond it: nodal regression under the Earth’s oblateness, and the inclination near 98° chosen so the regression matches the year, which is an inclination doing work rather than costing money; and the plane change as a term in a Lambert solution, where the two ends differ in plane and the split between them is one more interior optimum.