The engine is chosen by the calendar

Choosing a chemical propellant is choosing a molecular weight, and the choice is made from a short list. Hydrogen and oxygen give about 450 seconds of specific impulse; hypergolics give 320; a solid gives 280. A designer picks one and lives with it.

An electric engine has no such list. Its exhaust speed is set by the voltage across the accelerating grid, and the voltage is an adjustable number — the same hardware can be operated anywhere over a factor of three. So the specific impulse becomes a design variable, and the question of what value to choose is a real optimisation rather than a menu.

The answer is not “as high as possible”, and the reason is that a faster exhaust needs more power for the same thrust.

The one line the whole trade is in #

Write the vehicle as three masses: payload, power plant, propellant. The rocket equation gives the propellant fraction from the mission’s velocity budget and the exhaust speed:

$$\frac{m_p}{m_0} = 1 - e^{-\Delta v/c}.$$

The jet power is $P = \tfrac12\dot m c^2$, so for a propellant mass expended over a thrusting time $t$,

$$P = \frac{m_p c^2}{2t}.$$

And the power plant’s mass is $\alpha P$, with $\alpha$ the specific mass in kilograms per watt — a property of solar arrays and radiators and power conditioning, and the one number that decides whether any of this works.

Putting them together and dividing by the initial mass,

$$\lambda(c) = e^{-\Delta v/c} - \frac{\alpha c^2}{2t}\left(1 - e^{-\Delta v/c}\right).$$

The first term rises with $c$ and the second falls with it as $c^2$, so there is an interior maximum. A slow exhaust burns propellant; a fast one needs a power plant heavier than the propellant it saved.

Where the optimum actually sits #

Differentiating gives no closed form, so the maximum is found numerically. What it depends on is the interesting part.

It depends on $\Delta v$, as expected. It depends on $\alpha$, as expected — a lighter power plant permits a faster exhaust.

And it depends on $t$, which is not obvious at all. The thrusting time appears nowhere in the rocket equation and nowhere in the trajectory, and it enters here because the same energy delivered over a longer interval needs less power, and less power means a lighter plant.

So a mission with two years to fly wants a different engine from one with six months, even if both have the same velocity budget and the same destination. The exhaust speed that is optimal at 200 days is slow enough to be within reach of the best chemical engines, and the one optimal at 2,000 days is four times higher.

Reading the two limits #

The shape of the curve is clearer from its ends than from its peak, and both ends have simple descriptions.

At small $c$ the exponential dominates. $e^{-\Delta v/c} \to 0$, so the payload fraction goes to zero from below — the vehicle is almost entirely propellant, and the power plant, though light, is subtracted from nothing. This is the regime a chemical engine would be in if it had to pay for a power plant, and it is why it does not have one.

At large $c$ the power plant dominates. Expanding for $c \gg \Delta v$ gives $\lambda \approx 1 - \Delta v/c - \alpha c\Delta v/2t$, which falls linearly in $c$ without limit. The propellant saved by doubling the exhaust speed is halved; the power plant needed is doubled; and the second eventually wins by an unbounded margin.

The optimum is where the two derivatives cancel, and its position scales as roughly $(\Delta v\, t/\alpha)^{1/3}$ — a cube root, which is why the answer is so insensitive. A factor of eight in any of the three inputs moves the best exhaust speed by a factor of two, and a factor of eight is more than any of them is uncertain by.

That insensitivity is the reason the trade is usually presented as a rule of thumb rather than an optimisation. The rule is that the optimum exhaust speed is roughly half the mission’s $\Delta v$ for a typical schedule, and it is right to within a factor of 1.5 across most of the space missions have occupied.

Why there is no such trade for a chemical stage #

The comparison with chemical propulsion is worth making explicit, because the absence of this optimisation there is structural rather than a matter of having fewer choices.

A chemical engine is energy-limited rather than power-limited. The energy comes from the propellant itself, so the power available scales with the mass flow, and the “power plant” — the combustion chamber — does not get heavier when the exhaust speed rises. The $\alpha c^2/2t$ term simply does not exist.

The consequence is that for a chemical vehicle the payload fraction rises monotonically with exhaust speed, and the optimum is at the highest $c$ chemistry allows. There is nothing to trade.

So the two propulsion families are optimised by different arguments entirely. A chemical designer maximises $c$ subject to what will burn; an electric designer maximises $\lambda$ subject to what the power system weighs, and the answer is an interior point.

The term that is not a constant #

The specific mass $\alpha$ is treated above as a property of the vehicle, and it is a property of the vehicle and where the vehicle is.

A solar-electric system’s power falls as the inverse square of the distance from the Sun, so its effective $\alpha$ rises as the square. An array sized for 10 kW at the Earth delivers 4.3 kW at Mars and 370 W at Jupiter, so the same vehicle has a specific mass 27 times worse at Jupiter than at the Earth.

That is why the optimisation above cannot be done once for a mission that changes its distance, and why interplanetary electric trajectories are optimised with the power as a function of position rather than as a constant. The engine’s operating point is then varied along the trajectory — throttled down and its voltage lowered as the vehicle recedes — which is exactly the dial this essay is about, turned continuously.

For missions beyond about Jupiter a solar array becomes impractical and the alternative is a radioisotope or fission source, whose $\alpha$ is worse near the Sun and unchanged far from it. The crossover between the two is around 4 to 5 astronomical units, and it is decided by this arithmetic.

What the thrusting time really is #

One subtlety deserves separating, because it is where this analysis connects back to the trajectory.

The $t$ in the expression is the thrusting time, not the mission duration. A vehicle that coasts for part of its transfer has a thrusting time shorter than its flight time, and coasting is often optimal — an interplanetary transfer thrusts near the beginning and near the end and coasts in between.

So the optimisation is coupled to the trajectory: the trajectory decides how much of the flight is under thrust, the thrusting time decides the best exhaust speed, and the exhaust speed changes the acceleration and therefore the trajectory. The three have to be solved together, and every serious low-thrust mission design is that joint optimisation rather than the one-variable version drawn here.

What the one-variable version supplies is the structure — that there is an optimum, that it rises with mission duration, and that the curve near it is flat — and those three statements survive the coupling.

The mission that made it concrete #

The trade was first worked out in the 1950s, decades before anything flew on it, and the first spacecraft to be designed around it rather than merely to carry an ion engine was Deep Space 1 in 1998.

Its successor made the argument visible. Dawn carried a single ion propulsion system through a decade, rendezvoused with two main-belt bodies, and entered orbit about each of them — a mission with a velocity budget above 11 km/s, which no chemical vehicle of its launch mass could have flown at all. Its thrusters ran for over five years of cumulative thrusting.

Its specific impulse was not a constant. It was throttled across a range from about 1,900 to 3,100 seconds through the mission, set at each point by the power available at that heliocentric distance and by the thrust needed. That throttling is precisely this optimisation carried out continuously, with $\alpha$ varying as the inverse square of distance.

The trade also explains a choice that looks odd from outside. A vehicle designed for the inner solar system runs its engine near the top of its range, where the exhaust is fast and the thrust small, because power is plentiful; the same engine at Jupiter’s distance would be run at the bottom, where the exhaust is slow and each watt produces more thrust. The hardware is identical and the operating point is not.

The same figure makes a point about how flat the optima are. Over the whole 120-to-560-day range the best exhaust speed moves by less than a factor of two, and the payload fraction at any fixed engine within that range varies by a few per cent. A satellite operator choosing a transfer duration is trading revenue against a payload penalty of single-digit per cent, which is exactly the kind of trade that gets made on commercial rather than technical grounds.

Three exact relations and one empirical number #

Nothing here is measured; all of it follows from three relations, and each is exact within its own idealisation.

The rocket equation is exact for a constant exhaust speed.

The jet power relation $P = \tfrac12\dot m c^2$ is the kinetic energy of the exhaust per unit time and is also exact, and the real electrical power is larger by the inverse of the thruster’s efficiency, which for a gridded ion engine is 60 to 70 per cent and rises with voltage. Folding that efficiency in shifts the optimum slightly upward, because a faster exhaust is also a more efficient one.

The specific mass $\alpha$ is the empirical term, and it is where the numbers come from rather than from physics. The 25 kg/kW used in the first figure is a conservative whole-system value including arrays, structure, power processing and thermal control; the 8 kg/kW is what the best current systems achieve.

The optimum is proportional to $\alpha^{-1/3}$ and to $t^{1/3}$, roughly, which is why a factor of three in either moves the answer by only 1.4 — and why the disagreement between published values of $\alpha$ does not produce a disagreement about which engine to fly.

An engine that cannot be operated anywhere on the axis #

The engine cannot be operated anywhere on the axis. A gridded ion thruster has a practical range of perhaps 2,000 to 5,000 seconds, a Hall thruster 1,500 to 3,000, and reaching 11,000 requires a different device entirely. The optimum computed for a 2,000-day mission is outside the range of the hardware that exists, which is a real finding and is the reason those missions fly below their optimum.

Nothing on the plot is a lifetime. An ion engine has a finite operating life set by grid erosion, typically tens of thousands of hours, and a 2,000-day thrusting time is at or beyond it. Missions of that duration fly multiple thrusters and operate them in sequence, which is mass the model does not carry.

And the propellant’s own properties are absent. Xenon is dense and stores well and is scarce and expensive; krypton is cheaper and less dense and needs a larger tank. The tank is part of the propellant mass in this model and its scaling with the choice is not.

What a mission gives up to fly this way #

The payload fraction on the vertical axis is not the whole of the comparison, and the omission is the same one the combined raise and turn had to note: time.

A vehicle thrusting for 2,000 days is a vehicle not doing anything else for 2,000 days, and for a commercial satellite that is revenue forgone; for a science mission it is an operations budget and a delayed result; for a crewed one it is out of the question entirely.

So the real optimisation has a schedule on one axis and a mass on the other, and the curve drawn here is a slice through it at fixed schedule. Moving along the family of curves — choosing a longer mission to get a better payload fraction — is the actual decision, and it is made against a cost per day rather than against physics.

The three curves in the first figure are therefore not three cases; they are three points on the decision the designer is making, and the fact that the best engine differs between them is the reason the propulsion and the schedule cannot be chosen separately.

There is one class of mission where the schedule costs nothing, and it is where electric propulsion has been most transformative: a spacecraft already in orbit with nothing to do. Station-keeping, orbit raising after a launch shortfall, and end-of-life disposal all have arbitrary schedules, and for all three the exponential’s advantage is collected at no cost at all. The orbit that has to be paid for every year is the clearest case: a geostationary satellite spends about 50 m/s a year holding its slot, indefinitely, and there is no schedule pressure on any of it.

The shape of the answer #

The recurring structure is a quantity whose benefit and whose cost have different powers of the same variable, so that their difference has an interior maximum.

Here the benefit of a faster exhaust is exponential in $1/c$ and the cost is quadratic in $c$, and an exponential beaten by a quadratic produces a peak. Both terms are monotone and the answer is not, which is the general reason interior optima exist and is worth recognising when the same shape appears elsewhere.

It appears in the choice of how many stages to fly, where each added stage buys a smaller exponential and costs a fixed structural mass — and the split between two stages is not an equal one for the same reason the optimum here is not at either end. It appears in the efficiency with which a halo turns baryons into stars, where two suppressions bound a peak from opposite sides. And it appears in the optimum aperture of a telescope against its cost, which is the same arithmetic in money.

What distinguishes this one is that the variable is continuously adjustable in flight, so the optimum is not a design decision made once but a set point that can be moved. Very few optimisations in engineering have that property.

A second dial that is not usually turned #

The exhaust speed is the adjustable parameter this essay is about. There is a second, and it is adjustable in a way that has been exploited only recently.

The thrust at fixed power is $2P/c$, so lowering the exhaust speed raises the thrust — an engine throttled down in specific impulse produces more force from the same watts. That is the trade a vehicle makes when it needs to get somewhere quickly rather than efficiently, and it is why the operating point can be moved within a single mission.

The consequence for the trajectory is that the acceleration is not a constant, which breaks the assumption behind the closed-form transfer cost — though not badly, since that expression has no thrust level in it at all. What it changes is the duration, and the duration is one of the inputs to this optimisation.

So the two dials are coupled through $t$, and turning one changes the optimum setting of the other. A vehicle that throttles down to get more thrust finishes sooner, which shortens $t$, which lowers the optimum exhaust speed, which is the direction it was already moving. The feedback is stabilising rather than otherwise, which is a small mercy and is not obvious in advance.

Still open: how light a power plant can be made #

Every quantity in the trade is settled except $\alpha$, and $\alpha$ is the one that decides everything.

At 25 kg/kW electric propulsion is competitive for geostationary transfer and for slow interplanetary missions. At 5 kg/kW it becomes competitive for fast ones. At 1 kg/kW — which nothing approaches and which a nuclear-electric system might eventually reach — the optimum exhaust speed rises past 20,000 seconds and the payload fractions transform.

So the future of the whole family is a materials and power-systems question rather than a propulsion one, and the trajectory analysis above will not change when it is answered. The curves keep their shape; their peaks move.

From here: the one vehicle this arithmetic does not price #

Every vehicle considered so far carries its reaction mass and pays the exponential. There is one that does not, and it changes the problem rather than optimising it: a sail has no propellant, no rocket equation and no $\Delta v$ budget at all, and what limits it is a structure rather than a mass ratio.