One square root that raises the orbit and turns it

A low-thrust spiral between two circular orbits costs $|v_1 - v_2|$, with no thrust level in it and no transfer time. That is the whole cost of raising an orbit continuously, and it is a fifth more than the two-impulse transfer on the same pair of circles.

Almost no mission only raises an orbit. A launch from a site at latitude $\phi$ delivers a payload into an orbit inclined at least $\phi$, and a geostationary satellite needs an inclination of zero, so the plane has to be turned as well as the orbit raised. Edelbaum solved that case in 1961, and the answer is one line:

$$\Delta v = \sqrt{v_1^2 + v_2^2 - 2v_1v_2\cos\!\left(\tfrac{\pi}{2}\Delta i\right)},$$

with $\Delta i$ in radians. It is the law of cosines with a factor of $\pi/2$ inside the angle, and everything interesting about low-thrust steering is in that factor.

Where the half-pi comes from #

Set $v_1 = v_2 = v$ and the expression describes a pure turn at fixed radius:

$$\Delta v = 2v\sin\!\left(\tfrac{\pi}{4}\Delta i\right),$$

against the impulsive $2v\sin(\Delta i/2)$ — the third side of an isosceles triangle with two sides of length $v$ and an apex angle $\Delta i$.

For a small angle both sines linearise and the ratio is exactly $(\pi/4)/(1/2) = \pi/2$. A continuous plane change costs a factor of 1.571 more than an impulsive one, which is the opposite of what the phrase “low thrust turns for free” suggests.

The reason is geometric and is worth seeing. An impulse rotates the velocity vector once, through the whole angle, and the cost is the chord of that rotation. Continuous steering rotates it in infinitesimal steps while the thrust is also fighting to keep the orbit circular — the optimal steering law points the thrust out of plane by an angle that varies around each revolution, and the out-of-plane component is only part of the thrust at any moment.

Integrated over the whole manoeuvre, the wasted fraction comes to exactly $\pi/2$ in the small-angle limit. It is not a loss of efficiency in any mechanism; it is what the integral of a varying steering angle comes to.

The crossing at 140 degrees #

The crossover is worth pausing on because it is a clean piece of arithmetic with a physical reading.

Setting $2v\sin(\pi\Delta i/4) = 2v\sin(\Delta i/2)$ and using $\sin x = \sin(\pi - x)$ gives $\tfrac{\pi}{4}\Delta i = \pi - \tfrac{1}{2}\Delta i$, so

$$\Delta i = \frac{\pi}{\pi/4 + 1/2} = 2.444\ \mathrm{rad} = 140.0°.$$

Below that angle the impulsive turn is cheaper; above it the continuous one is. The physical reading is that an impulse has to stop the vehicle and start it again when the turn approaches a reversal — the chord of a 180° rotation is $2v$, the whole speed twice over — while a continuous turn never has to, because it can spiral outward, turn cheaply at the top where the speed is low, and spiral back in.

The crossing also explains a piece of folklore. It is sometimes said that low-thrust vehicles are good at large plane changes, and the statement is true in the narrow sense that they are relatively better at large ones than at small ones. It is usually offered as a reason to fly a 30° change electrically, where the factor is still 1.55 against the impulsive value and the statement does not apply.

No mission has ever needed an angle past the crossover. Plane changes flown in practice run from a few degrees to about thirty, and a full reversal has no use — so the result is nearly always quoted as its first half, and the second half is a curiosity.

It is a curiosity with one application. A spacecraft escaping the ecliptic to observe the Sun’s poles needs an inclination change of about ninety degrees relative to the ecliptic, and that manoeuvre is at the expensive end of both curves. The mission that has done it used a gravity assist at Jupiter instead, which costs nothing at all and is the reason neither curve was consulted.

What the closed form is worth #

The value of Edelbaum’s expression is not that it is more accurate than an integration — it is less — but that it makes a trade visible that an integration does not.

The expression is the law of cosines, so $\Delta v$, $v_1$, $v_2$ and $\tfrac{\pi}{2}\Delta i$ form a triangle. The transfer’s cost is the third side of a triangle whose other two sides are the circular speeds and whose apex angle is the scaled plane change.

That gives an immediate result nobody derives from an integration: the plane change is cheapest when it is made where the vehicle is slowest, and in a spiral outward the vehicle is slowest at the end. The optimal steering therefore does most of the turning late, and the optimal-control solution — which is what the expression is a closed form of — puts the out-of-plane angle near zero at the start and near its maximum at the finish.

An impulsive mission does the same thing for the same reason, by putting the plane change into the apoapsis burn. The two strategies agree about where to turn and disagree about how much it costs to do so.

Reading the triangle for the whole mission #

Because the expression is a law of cosines, the three quantities in it can be read off a triangle, and doing so answers a question mission designers ask constantly.

Draw $v_1$ and $v_2$ from a common vertex with $\tfrac{\pi}{2}\Delta i$ between them. The transfer’s cost is the side joining their tips. That means:

When the two speeds are very different, the angle matters little. A transfer from low orbit to a far one has $v_1 \gg v_2$, so the triangle is long and thin and rotating one short side changes the long side hardly at all. A transfer to the Moon’s distance with a 28.5° plane change costs barely more than one without.

When the two speeds are comparable, the angle dominates. A transfer between two orbits of similar size is a nearly isosceles triangle, and then the third side is almost entirely the angle.

So the plane change’s cost is not a fixed quantity to be added to a mission’s budget — it is a term that depends on how large a raise it is combined with. A plane change is cheapest when bundled with the largest available raise, which is the same principle as the impulsive rule of turning at apoapsis and is arrived at by a different route.

Why low thrust is flown anyway #

Everything above says that a continuous transfer is dearer in velocity change, and it is flown for almost every commercial geostationary satellite launched since about 2015. The reconciliation is the exponential in the rocket equation.

The propellant fraction is $1 - e^{-\Delta v/c}$, and $c$ for a gridded ion engine is around 30 km/s against 3 km/s for a chemical stage. So a chemical transfer at 4.26 km/s needs a propellant fraction of $1 - e^{-1.42} = 0.76$, and an electric one at 5.95 km/s needs $1 - e^{-0.20} = 0.18$.

A forty per cent penalty in $\Delta v$ buys a factor of four in propellant mass, which is why the dearer manoeuvre is the cheaper mission. The exponential is over $\Delta v/c$ and the electric engine moves the denominator by a factor of ten while the trajectory moves the numerator by a factor of 1.4.

That comparison is also why the plane change specifically is where the two approaches are closest. The plane change costs the electric transfer 1.300 km/s against the chemical’s 0.363, a ratio of 3.6, which is the worst ratio anywhere in the manoeuvre — but it is 1.3 km/s on a budget where the exponential is being divided by thirty.

The form the expression takes when the radii are equal #

There is a case where Edelbaum’s result becomes something other than a transfer, and it is used more often than the transfer is.

A satellite in low Earth orbit with electric propulsion can be commanded to change its inclination without changing its altitude, and the cost is the pure-turn expression above. At 7.7 km/s a single degree costs $2 \times 7.7 \times \sin(0.785°) = 0.211$ km/s, against the impulsive $2\times7.7\times\sin(0.5°) = 0.134$.

A degree of inclination in low orbit costs 0.211 km/s, and a degree at geostationary costs 0.084, because the cost scales with the local circular speed and low orbit is where the vehicle is fastest. That is the arithmetic behind a standing rule of mission design: the plane is chosen at launch, and changed afterwards only under duress and preferably from as far out as possible — which is exactly why a plane change is flown from an apoapsis raised for the purpose.

The rule has one common exception. A constellation that needs its planes spread in right ascension can get the spreading for free from the oblateness of the Earth, which precesses each orbit’s node at a rate depending on its altitude and inclination. Placing satellites at slightly different altitudes and waiting produces any relative node spacing at no propellant cost at all, and the waiting is measured in months.

What happens when the spiral goes inward #

The expression is symmetric in $v_1$ and $v_2$, so it covers descent as readily as ascent, and one application of that is worth stating.

A spacecraft spiralling down — from an interplanetary arrival hyperbola’s capture orbit to a low orbit about a planet — ends where the circular speed is highest, so the useful bundling runs the other way: the plane change should be made early, at the slow end, before the descent.

That is the manoeuvre every electric mission arriving at a planet has to fly, and it is where the drawn asymmetry between the two curves is most pronounced. An impulsive arrival can insert directly into the desired plane by aiming the approach; a low-thrust arrival cannot, because it has no impulse to aim, and the plane has to be corrected afterwards at whatever speed the capture orbit gives it.

So the same expression describes two manoeuvres with opposite best practices, and which end is the cheap one is decided by which end is slower. The corresponding impulsive rule says the same thing and is easier to state because an impulse happens at a point.

Nothing here is measured, and that is what makes it checkable #

Nothing in this essay is measured; all of it is derived, and the derivations are checkable in a way an observation is not.

Edelbaum’s expression is exact for the optimal steering law under the assumption that the orbit remains circular throughout — which is the approximation, and it is good when the thrust acceleration is small compared with the local gravity. For an ion engine at $10^{-4}$ of local gravity it is very good indeed.

The impulsive comparison is vis-viva arithmetic with no approximation in it.

The check the figures make is the one that matters: setting $\Delta i = 0$ must reproduce $|v_1 - v_2|$ exactly, since with no plane change the expression has to become the spiral’s own cost. It does, to the last digit, and a version with the $\pi/2$ in the wrong place would not.

How the steering law actually behaves #

The closed form is an integral of a steering law, and the law is worth describing because it is the part a spacecraft executes.

At every instant the thrust is tilted out of the orbital plane by an angle $\beta$, and the optimal $\beta$ is not constant. It is zero at the nodes — where an out-of-plane push does nothing to the inclination — and maximum at the points a quarter of a revolution away, where the whole of it goes into rotating the plane. So the thrust vector yaws back and forth once per revolution, and the sign flips each half-orbit.

Averaged over one revolution, the fraction of the thrust that goes into turning is $2/\pi$ of $\sin\beta$, and $2/\pi$ is where the $\pi/2$ in the closed form comes from. The factor is not a loss to anything physical; it is the mean of a sine over a half-period.

The magnitude of $\beta$ then rises through the transfer, from near zero at the start to its largest at the end, for the reason given above: turning is cheapest where the vehicle is slowest. Edelbaum’s solution gives that schedule in closed form too, and a spacecraft flying the manoeuvre follows a yaw profile computed from it and updated as the orbit is measured.

A real vehicle also has to keep its solar arrays pointed at the Sun while yawing the thrust vector twice a revolution, which is a constraint the trajectory optimisation does not contain and the attitude control system does.

Time is absent from every curve #

Time is absent from every curve. Edelbaum’s expression has no thrust level in it and therefore no duration, exactly as the spiral’s did, so a 5.95 km/s manoeuvre at $10^{-4}$ g takes about seven months and the plot says nothing about it. For a commercial satellite that is seven months of revenue forgone, and it is the real cost of electric propulsion rather than the $\Delta v$.

The radiation belts are not in it either. A slow spiral through the Van Allen belts exposes the solar arrays to a dose a fast chemical transfer avoids, and the array degradation over a seven-month transit is a design driver. A mission planner comparing the two options is comparing a mass, a schedule and a lifetime, and only the first is on the axis.

And the drawn impulsive transfer is the optimal two-burn one, which is not always what is flown. A supersynchronous transfer — going beyond geostationary and coming back — turns more cheaply still, and a three-burn version beats the two-burn one for large plane changes, so the impulsive curve drawn is an upper bound on what impulses can do.

A result that is a triangle #

The thing worth carrying is that a manoeuvre with two independent requirements has a cost that is the third side of a triangle rather than the sum of two costs.

An impulsive plane change combined with a circularisation burn has the same property — the two are vector-added rather than added, which is why doing them together at apoapsis costs 0.363 km/s where doing them separately would cost 1.514.

Edelbaum’s result is the continuous version of the same statement, with a scaled angle. In both cases the saving comes from the fact that a velocity change is a vector, and two requirements pointing in different directions are met by one vector rather than two.

The scaling factor is the difference between the two, and it is the price of never being able to make the whole change at once. A single impulse applies the full rotation in one place; continuous steering applies it everywhere, and everywhere is not the best place.

The comparison that is not a comparison #

One habit worth resisting is reading the two curves as a choice between two options for one vehicle. They are not.

A vehicle with an ion engine cannot fly the impulsive trajectory — it has no impulse — and a vehicle with a chemical stage cannot fly the continuous one, because it would exhaust its propellant in minutes. Each curve belongs to a different vehicle, and the curves cross nothing because they are not alternatives at a fixed mass.

The comparison that a designer actually makes is between two whole vehicles, each with its own dry mass, propellant fraction, power system and schedule, and the $\Delta v$ is one input to each. Putting the two on one axis makes the trajectory difference visible and makes the vehicle difference invisible, which is the usual trade a figure of this kind carries.

That said, the hybrid exists and is flown. A chemical stage performs the raise and part of the turn, and electric propulsion finishes the plane change and the station-keeping over the following months. The budget then splits between the two curves, at a point chosen to trade launch mass against time in transit, and it is the arrangement most large geostationary satellites now use.

Still open: whether the closed form is the optimum #

Edelbaum’s expression is derived for a particular steering law under the circular-orbit assumption, and the assumption is what makes it closed-form. The true optimum — with the thrust direction free at every instant and the orbit allowed to become eccentric — comes from Pontryagin’s principle and has no closed form at all.

Numerical solutions find improvements of a few per cent for large plane changes, obtained by letting the orbit become slightly eccentric so that the turning can be concentrated near apoapsis, which is exactly the impulsive strategy reappearing inside the continuous solution.

Whether those few per cent are worth having is not a question the trajectory answers. An eccentric spiral spends longer in the radiation belts, needs a more complicated steering profile, and saves a quantity that the rocket equation’s exponential has already divided by thirty. The same tension between a theoretically optimal manoeuvre and a flyable one recurs throughout this field, and the theoretical optimum usually loses.

From here: the quantity taken as given so far #

Everything here has taken the exhaust speed as given, at whatever the engine happens to produce. It is not given: an electric engine’s exhaust speed is a design choice, and the choice is not free, because the power needed per newton of thrust rises in proportion to it.

Putting the engine’s own specification on the axis finds that the best exhaust speed for a mission depends on how long there is to fly it, and not on how far it is going.