A rotation split between two burns

The first rung of this anchor established the one thing everybody knows about plane changes: they cost 2v sin(Δi/2), the cost is proportional to the speed, and the answer to “where should the turn be done” is therefore “wherever the vehicle is slowest”. For a transfer to geostationary orbit that is the apogee of the transfer ellipse, where the vehicle moves at 1.6 km/s instead of 7.7.

That answer is nearly right and it is not the optimum, and the gap between them is a piece of second-order reasoning that recurs across this collection. The optimum is not at either end of the range, and the reason it cannot be is available before any arithmetic is done.

Why the optimum cannot be at an end #

The argument is a comparison of two derivatives at zero.

Take a transfer whose perigee burn changes the speed from v₁ to v₂ in the same plane. Adding s degrees of rotation to that burn makes it a vector sum rather than a difference: the cost becomes √(v₁² + v₂² − 2v₁v₂ cos s), and the two speeds are nearly equal in direction, so for small s the cost rises from |v₂ − v₁| by a term proportional to s². It is quadratic. The first tenth of a degree is almost free.

Meanwhile the apogee burn is doing the remaining Δi − s degrees, and its cost is 2v sin((Δi − s)/2), which for small changes falls linearly in s. Linear against quadratic: at s = 0 the total derivative is strictly negative, so moving some of the turn to perigee always helps, at every angle and in every transfer. The optimum is interior by construction rather than by numerical accident.

It is worth stating what makes the perigee term quadratic, because it is not a general property of plane changes. At perigee the vehicle is not merely turning; it is accelerating in the same plane, and the rotation is applied to a burn that already exists. Two vectors of similar magnitude at a small angle differ in length by second order in the angle — the chord of a shallow triangle — and that is where the s² comes from. A rotation applied where no burn is happening costs 2v sin(s/2), which is linear in s for small s, and would give no interior optimum at all.

There is a way of seeing the same thing without derivatives, and it is worth having because it explains the sign as well as the existence. The perigee burn is already several kilometres a second — it is what raises the apogee — and rotating a large vector by a small angle changes its length hardly at all while changing its direction by exactly that angle. The vehicle therefore arrives at apogee already tilted, for nearly nothing. What it paid was the difference between the chord and the arc of a shallow triangle, and that difference is what the s² is.

Turned around: a plane change is expensive where it is the whole manoeuvre and cheap where it is a small perturbation on a manoeuvre that is happening for another reason. That statement covers the entire subject of this ladder, and the rest of it is working out where such manoeuvres exist.

The sizes, in order #

Three savings are available on the same transfer and they are three orders of magnitude apart, which is the practical content of this rung.

Doing the turn at apogee rather than in the low orbit saves about three kilometres a second, because the same burn is worth more where the vehicle is slower and the speed ratio here is 21 per cent. Combining the turn with the circularisation rather than doing them in sequence saves 1.146 km/s more, because two sides of a triangle exceed the third. Splitting a couple of degrees of the rotation into the perigee burn saves a further 25 metres a second.

The last of those is a tenth of a per cent of a 24-tonne geostationary satellite’s launch mass, which is about twenty kilograms of payload — worth taking and not worth an essay on its own. What makes it worth this one is the shape of the argument, which is that an optimum sitting at a boundary is usually a sign that a second-order term has been dropped.

How the optimum moves #

The split is a couple of degrees at every angle drawn, and it does not scale with the turn in the way an intuition about “a small fraction of it” would suggest.

The turnover is the kind of feature a closed-form approximation would miss and a search finds without being told. It is also the reason the optimum here is found by ternary search on the total rather than by differentiating: the total is a sum of two square roots of trigonometric expressions, and its derivative is not an expression anybody wants to set to zero.

There is a second reason the optimum is small, and it is about where the vehicle is rather than about the algebra. The perigee burn happens at 7.7 km/s and the apogee burn at 1.6, so a degree removed from apogee saves 2 × 1.618 × sin(0.5°) = 28 m/s while a degree added at perigee costs a quadratic term whose coefficient carries the 7.7. The ratio of the two coefficients is what fixes the optimum at a couple of degrees rather than at a couple of tens, and it is the same speed ratio that made the apogee turn cheap in the first place. The high speed at perigee is what makes the correction second order and what makes its coefficient large.

What a percentage of a budget means here #

A saving of 25 metres a second sounds negligible against 4.2 kilometres a second, and on a launch vehicle it would be. On a satellite it is not, and the difference is worth explaining because it is why geostationary operators care about this arithmetic and launch operators do not.

A geostationary satellite spends its own propellant on the last part of the transfer and on station-keeping for fifteen years afterwards. Its dry mass is fixed by what it does; its wet mass is fixed by what the launcher can lift; and the difference between them is the propellant available for both jobs. Every metre a second saved on the transfer is a metre a second available for station-keeping, and station-keeping consumes about 50 m/s a year. Twenty-five metres a second is therefore six months of operational life, on an asset worth a few hundred million.

That conversion — Δv into years — is the currency this end of the subject actually uses, and it is why an optimisation worth a tenth of a per cent of a budget is worth doing.

The angle at which turning costs more than leaving #

One number on the cost figure deserves separating out, because it is the most surprising thing on this ladder and it is easy to read past.

A plane change of Δi from a circular orbit costs 2v sin(Δi/2). Escaping from the same orbit costs (√2 − 1)v, since escape speed is √2 times circular. Setting those equal gives sin(Δi/2) = (√2 − 1)/2, so Δi = 23.9° — and the speed cancels. That angle is a property of the geometry alone: at any altitude, around any body, turning an orbit by more than about twenty-four degrees costs more than leaving the body altogether.

Kennedy’s latitude is 28.5°, which is past it. A satellite launched due east from Florida and then rotated into the equatorial plane in low orbit would spend more velocity on the rotation than on escaping Earth, which is why nobody does it and why every geostationary mission carries its inclination up to apogee and removes it there. The escape speed and the circular speed differ by a factor of √2 everywhere, so the comparison is the same for a satellite of Mars or of the Moon, and the answer is the same 23.9°.

What the picture cannot show #

Three simplifications run through every figure here, and the first is the one that matters.

The transfer is impulsive. Both burns are treated as instantaneous changes of velocity at a point, and a real apogee burn on a geostationary transfer lasts several minutes during which the vehicle moves through several degrees of true anomaly. Spreading a burn over an arc costs a few per cent in finite-burn losses, because the thrust is no longer aligned with the ideal direction throughout, and the loss grows with the plane-change component — so a real optimum sits slightly differently from the one computed. The direction of the correction is toward doing less at apogee, since that is the burn that is longest.

The orbits are coplanar circles joined by a coplanar ellipse, with the rotation applied at one or both ends. A real geostationary transfer is not that: the ascending node has to be at the equator crossing, the argument of perigee matters, and the oblateness of the Earth regresses the node during the transfer by a small amount that has to be planned for.

And the two burns are treated as available at will. A real transfer may spend several revolutions in the transfer ellipse before the apogee burn, for tracking, checkout and thermal reasons, and a supersynchronous transfer — raising the apogee above geostationary and coming back down — is a bi-elliptic manoeuvre in disguise that beats everything here at large angles. That is the next rung.

An optimum that is worth less than the error in the inputs #

One more thing about the 25 metres a second deserves saying, because it bears on when an optimisation of this kind should be believed.

The number is computed for a specific transfer between two specific circular orbits at a specific inclination, with impulsive burns and a spherical Earth. Every one of those assumptions is wrong by more than 25 m/s in its effect on the total. Finite-burn losses on a real apogee burn are 20 to 60 m/s. The launcher’s dispersion in delivered transfer orbit is worth tens of metres a second of correction. The oblateness of the Earth changes the node during the transfer by enough to matter.

So a designer taking the 2.23° optimum literally is optimising a quantity smaller than the uncertainties around it. What the calculation is actually good for is the sign: it says that some of the rotation belongs at perigee, which is not obvious, and that the amount is a couple of degrees rather than a couple of tens. Both of those survive every correction listed; the third decimal place does not.

That distinction — between an optimisation that says where to look and one that says what to do — is worth carrying, because a great many published optima in this subject are the first kind reported as the second.

The habit #

An optimum found at the end of its allowed range is a result that should be checked rather than reported, and the check is to look for a term of different order.

The standard answer here — do the whole rotation where the vehicle is slowest — is an optimisation over where, and it correctly finds the end of that range. It becomes a boundary answer only when the question is widened to how much where, and the widened problem has an interior answer because the two ends are not the same kind of manoeuvre: one is a rotation applied to an existing burn and the other is a rotation applied to nothing.

The same shape appears elsewhere in this collection whenever a resource is divided between two mechanisms whose costs scale differently. The division of a launcher’s velocity budget between two stages is interior for the same structural reason, with the ends being vehicles that do not close rather than manoeuvres that cost nothing. And the split of an observing programme between more targets and longer exposures is interior because the noise contributions scale differently in each.

In every case the diagnostic is the same. Compute the derivative at the boundary; if it is not zero, the boundary is not the answer.

The manoeuvre this is a special case of #

Everything above is a two-burn transfer with the rotation split between the burns, and the general problem it belongs to has a name and a much larger literature.

Given a start state, an end state and no constraint on the number of burns, the minimum-fuel impulsive transfer is a question in optimal control, and the governing result is Lawden’s primer vector theory: there is a vector function along the trajectory whose magnitude decides where burns should occur, and a transfer is optimal only if that magnitude reaches one exactly at each burn and stays below one everywhere else. The theory does not say what the answer is; it supplies a test that a candidate answer must pass, and the test is what tells a designer whether adding a burn would help.

Applied here, it says exactly what the second-order argument said: the two-burn solution with all the rotation at apogee fails the test, because the primer magnitude exceeds one somewhere near perigee, which is the signal that a burn there would reduce the total. Adding it, and optimising the split, restores the condition.

The reason this is worth a paragraph rather than a rung of its own is that the general theory rarely produces a closed form. What it produces is a diagnostic, and the diagnostics in this subject — the primer vector, the derivative at a boundary, the crossing of a linear term and a quadratic one — say where to look rather than what will be found.

What is actually flown #

It is worth saying what a real geostationary transfer does, because it is neither the textbook answer nor the optimum computed here.

A launcher from Kourou delivers a satellite to a transfer orbit at about 6° of inclination — the site’s latitude of 5.2° plus a little. The satellite’s own apogee engine then performs a combined manoeuvre: it circularises and removes the inclination in one burn, or in several burns at successive apogees if the engine is small. The perigee split is not usually flown as such, because the launcher rather than the satellite performs the perigee burn and the two are contracted separately; what the launcher does instead is deliver a transfer orbit whose apogee is above geostationary, which achieves the same end by a different route.

Most modern geostationary satellites do something else again. Electric propulsion at 3,000 seconds of specific impulse makes the transfer’s Δv nearly irrelevant to the mass budget and makes its duration the constraint, so the vehicle spirals out over three to six months with the plane change spread continuously along the way. On that trajectory none of the arithmetic here applies: there are no burns to split, the rotation is done a fraction of a degree at a time wherever the vehicle happens to be, and the relevant optimisation is Edelbaum’s rather than this one.

That is the honest position of this rung. It is exact for the vehicle it describes, that vehicle is what flew for forty years, and it is being replaced by one the analysis does not cover.

Where this ladder goes next #

Everything above takes the transfer’s apogee as given: geostationary radius, because that is where the satellite is going. The speed at that apogee is what makes the turn cheap, and it is 1.6 km/s.

There is nothing stopping a vehicle from flying higher. A transfer to an apogee well above the target, with the rotation done at that far apogee and a third burn to come back down, buys a slower place to turn at the cost of two extra burns and a great deal of time. The next rung is where that trade changes hands, and the answer is an angle rather than a distance: below about thirty-nine degrees the single combined burn wins, and above it flying further away is cheaper.

Beyond it: the launch azimuth, where a plane change is paid at the worst speed there is and a site’s latitude is a floor no trajectory removes; the sun-synchronous inclination near 98°, where the nodal regression is made to match the year and an inclination does 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.