The stage that has to be thrown away

Rung one of this ladder put the difficulty of spaceflight in one place: a rocket carries its own reaction mass, so $\Delta v = v_e\ln(m_0/m_f)$ and the mass ratio is the exponential of the velocity change rather than a multiple of it. Read as a prohibition, that equation invites the obvious conclusion — the exponential is what stops a single rocket from reaching orbit.

It does not, and the arithmetic says so in one line. An exponential is unbounded but finite everywhere, and the value it takes at the orbital requirement is unremarkable: at a specific impulse of 350 seconds, $e^{9.4/3.432} = 15.5$, which a vehicle 93.5 per cent propellant by mass satisfies. The quantity that runs out first is not the propellant. It is the tank it is in.

What one stage can do, and it is 8.67 km/s #

A stage is not propellant. It is propellant plus the tankage, pumps, engines, plumbing, avionics and airframe that deliver it, and the ratio of that hardware to the stage’s own gross mass is the structural coefficient $\varepsilon$, which for real launch hardware sits between 0.06 and 0.12.

Empty the stage of payload entirely and the mass ratio it can reach is fixed: everything but the structure has gone, so $m_0/m_f = 1/\varepsilon$, and the velocity change is

$$\Delta v_{\max} = v_e\ln\frac{1}{\varepsilon}.$$

That is a ceiling no amount of propellant raises, because the propellant is already in the numerator. At $\varepsilon = 0.08$ and $I_{sp} = 350$ s it comes to 8.67 km/s. The requirement is about 9.4. A single stage of ordinary construction, carrying nothing whatever, falls short of orbit by 0.73 km/s.

The 9.4 is not negotiable, and only 7.7 of it is orbital #

The other half of the inequality is a number about the Earth rather than about rockets, established by the speed that does not come back, and it cannot be argued with. Neither speed is measured directly: both follow from one equation and one constant, with $GM$ inferred from the period and size of a tracked orbit rather than from any mass in kilograms. The gap between 7.67 and 9.4 is the loss budget — roughly 1.2 km/s of gravity loss, 0.1 to 0.3 of drag, some steering, less 0.41 for launching eastward from Cape Canaveral — reconstructed by integrating flown trajectories from telemetry and radar tracking. Remove those losses and a single ordinary stage clears orbital speed with 1.0 km/s in hand, so the wall is partly atmosphere and gravity rather than orbital mechanics.

Four lines, and the tank is the only thing in them #

Normalise the lift-off mass to 1 and let $\lambda$ be the payload fraction: the stage’s gross mass is $1 - \lambda$, of which $\varepsilon(1-\lambda)$ is structure, so the burnout mass is that structure plus the payload,

$$\frac{m_0}{m_f} = \frac{1}{\varepsilon(1-\lambda) + \lambda} = e^{\Delta v/v_e},$$

which rearranges to

$$\lambda = \frac{e^{-\Delta v/v_e} - \varepsilon}{1 - \varepsilon}.$$

Everything the argument needs is in that numerator: a difference between what the engine buys and what the structure costs, which changes sign. Setting $\lambda = 0$ gives $\varepsilon = e^{-\Delta v/v_e}$, the ceiling above — and past it a negative payload fraction, which is not a hard mission but an impossible one.

Seven hundred and thirty metres a second #

The most interesting thing about the inequality is how nearly it goes the other way. A shortfall of 0.73 km/s in 9.4 is under 8 per cent, which in this subject is a rounding error in a loss budget — and it is why single-stage-to-orbit was a live engineering question for four decades rather than settled arithmetic.

Both terms are within ordinary reach. Take $\varepsilon$ from 0.08 to 0.06 — a fifth off the dry mass — and the ceiling rises to 9.66 km/s, clearing the requirement with 0.50 per cent of the vehicle as payload. Or improve the engine instead: at a hydrogen–oxygen specific impulse of 450 seconds the ceiling is $4.413 \times \ln 12.5 = 11.15$ km/s, a margin of 1.75 in hand rather than 0.73 short.

So the equation does not forbid a single-stage launcher, and it is worth being exact about what does. Hydrogen’s density is 71 kg m⁻³ against kerosene’s 810, so the tank that raises $I_{sp}$ to 450 holds eleven times the volume per unit mass and drives $\varepsilon$ back up: the two improvements are not independent. Add an engine that must work at sea level and in vacuum, and half a per cent of payload from which every gram of re-entry hardware is deducted, and the honest statement is that the equation permits single-stage-to-orbit and nothing else does.

A product of fractions, not a sum of stages #

Staging is usually introduced as an efficiency measure. On the numbers above it is not an optimisation at all but the only way the inequality is satisfied: a launcher throws away its lower stages because otherwise it does not arrive.

It works because the payload of one stage is the whole vehicle of the next, so the fractions multiply where the requirements add. Each of $n$ equal stages faces $\Delta v/n$ and delivers $\lambda_1$ of what it was handed, so the overall fraction is $\lambda = \lambda_1^{\,n}$. Splitting 9.4 km/s in two gives each stage 4.7, a mass ratio of 3.93 and a per-stage fraction of 0.189 — comfortably positive where the single stage was negative — and $0.189^2 = 3.59$ per cent of the lift-off mass reaches orbit.

The discarded structure is the point: it was accelerated to 4.7 km/s and no further, so the second stage’s mass ratio is computed against a burnout mass that does not contain the first stage’s tanks. Nothing about the engine or the exponential changed. What changed is which structure appears in which final mass.

Infinitely many stages is one stage with a worse engine #

Two stages do almost all the work available, and what follows has an exact form rather than merely a shape. As $n$ grows, $\lambda_1^{\,n}$ tends to a finite limit,

$$\lambda_\infty = \exp\left(-\frac{\Delta v}{v_e(1-\varepsilon)}\right),$$

which is 5.10 per cent at the numbers drawn. Reading that expression is the surprising part: it is the rocket equation with no structural term and the exhaust speed multiplied by $1-\varepsilon$, so a vehicle with infinitely many stages behaves exactly like one ideal stage made of nothing, flying an engine derated by 8 per cent. The structural coefficient, in the limit, is not a mass penalty but an efficiency penalty on the engine: at $\varepsilon = 0.08$ the tank costs precisely 28 seconds of specific impulse and nothing else.

Against that ceiling the increments are readable. Two stages reach 3.59 per cent; the step from two to three is 0.67 percentage points and from four to five 0.14. A fifth stage buys about a thirtieth of what the second bought while adding engines, a separation system and two more events that can fail, which is why no launcher has five. It is the shape a bi-elliptic transfer has as its intermediate apoapsis grows: a real saving converging on a limit, so the question is where splitting stops paying rather than whether to split.

What a real mass table weighs #

The family curves assume one engine and one $\varepsilon$ throughout, so the test of the argument is a vehicle whose stages were weighed separately.

Falcon 9 Block 5 lifts off at 549 tonnes and places 22.8 of them in low Earth orbit: a payload fraction of 4.15 per cent, on two stages. Its first stage is 25.6 tonnes dry against 395.7 of propellant, giving $\varepsilon = 0.061$; its second is 3.9 against 92.7, giving 0.040. Both beat the 0.08 the middle curve assumes, and the vehicle lands between the $\varepsilon = 0.06$ and $\varepsilon = 0.08$ curves at $n = 2$ — not a coincidence to be admired but the prediction met by the one number the drawing did not choose.

Those quantities are different kinds of measurement, and the differences matter. Dry masses are weighed on load cells before flight. Propellant loads are not weighed: they are computed from tank volumes and densities and checked against integrated flow, so the 0.061 carries the density’s uncertainty and the 25.6 does not. Specific impulse is thrust divided by mass flow on a test stand, corrected to vacuum. And the stage masses sum to 540.7 tonnes against a published 549 at lift-off, the 8.3 tonnes of difference being interstage, fairing and residuals — so the fraction is taken against the published mass with the discrepancy admitted rather than the table quietly summed. It also settles a question the stage count raises. A geostationary mission adds two burns of a transfer to the 9.4, and 13.3 km/s would sit well along the hero’s axis toward a third stage. None appears, because the second restarts — a stage that is not thrown away, which the equation counts as one with its propellant split between two burns.

Staging also buys a better engine #

The arithmetic above treats the exhaust velocity as a property of the propellant, the same for every stage. It is not, and the difference is a second argument for staging that the mass-fraction calculation misses entirely.

A rocket engine’s exhaust velocity depends on how far the nozzle expands the gas. Expanding further extracts more of the thermal energy as directed motion, so a longer, wider nozzle is a faster exhaust — up to the point where the gas leaves the nozzle at a pressure below the ambient, at which the flow separates from the walls and the engine loses thrust and shakes itself apart.

At sea level the ambient pressure is one atmosphere and the expansion ratio is limited to about 20 or 40. In vacuum there is no such limit, and ratios of 100 to 300 are routine, which for the same propellants buys 15 to 20 per cent more exhaust velocity — a kerosene–oxygen engine delivering an effective exhaust velocity around 2.9 km/s at the pad and 3.4 in vacuum, and a hydrogen–oxygen engine around 4.4 against nearly 4.5.

Since the velocity budget appears in an exponent, that is not a small effect. A stage that must operate from the pad is stuck with the small nozzle; a stage that only ever fires above the atmosphere is not, and the upper stages of every launch vehicle carry the enormous bell-shaped nozzles that would be unusable lower down.

So the upper stage is not merely a smaller version of the lower one, and separating them separates two design problems that pull in opposite directions. That is a reason for staging that survives even in the hypothetical case where the structural coefficient is zero, and it is why the second stage’s engine looks nothing like the first’s on every vehicle ever flown.

The compromise available to a vehicle that must fly through both regimes is an engine whose expansion adapts, and several such designs have been built and none flown operationally: an aerospike, whose exhaust is bounded on one side by the atmosphere itself, has the right behaviour at every altitude and has never survived the trade against two ordinary engines and a separation event.

The reason it loses is instructive: the adaptive nozzle saves a few per cent of exhaust velocity on one stage, and staging saves a factor in the mass ratio, so the thing being optimised is much the smaller of the two.

So the nozzle question is real and it is second order, which is the correct place for it in an essay whose subject is the exponent.

What the picture cannot show #

Equal stages, and no real vehicle has them. The drawn family divides $\Delta v$ equally among $n$ stages. Falcon 9’s differ in both exhaust speed and $\varepsilon$, so the optimal division is unequal and comes out of a Lagrange multiplier rather than out of division. The curves are also continuous in $n$, where a stage count is an integer; the dots are the only points a launcher can occupy.

ε is not a property of the material. A tank’s mass scales with its surface and its contents with its volume, so a small stage is structurally worse than a large one. Holding $\varepsilon$ fixed across the axis therefore flatters the right-hand end: a five-stage vehicle’s stages are each a fifth the size and each worse than the one drawn, so the real curve turns over where the figure still rises. The saturation is more severe than the drawing, not less.

Thrust is absent, and so is time. A stage must also lift itself, which needs a thrust-to-weight ratio above one and sets the gravity loss buried in the 9.4. A vehicle with a beautiful mass ratio and inadequate engines does not fly, and neither axis reports it.

Recovery is a structural coefficient. Legs, grid fins, thermal protection and reserved landing propellant are all dry mass in a stage that must be accelerated, so recovering a booster raises its $\varepsilon$ and moves the vehicle down the family. Whether it is worth paying depends on how often the hardware flies, which is on neither axis.

The same inequality on a heavier planet #

The wall compares two speeds and only one is about rockets, so changing the planet moves it.

Both planets are measured twice, by two methods, to get a density, and the same pair of numbers gives a launch requirement. On 55 Cnc e the circular speed alone is 16.3 km/s, twice the Earth’s, and the ceiling at $\varepsilon = 0.08$ and 350 seconds is 0.57 per cent however many stages are used: two of 8.15 km/s each barely clear the per-stage wall and deliver 0.020 per cent, three deliver 0.25, and a 549-tonne vehicle would place 1.4 tonnes in orbit rather than 22.8.

The general form is the memorable one. Because the ceiling is exponential in the requirement with a scale of $v_e(1-\varepsilon)$, every additional 3.16 km/s costs a factor of $e$ in payload, and no stage count buys it back. So there is no planet from which chemistry cannot escape, only planets from which it cannot escape usefully — which is what an exponential with a finite limit says, and less dramatic than a prohibition. The way out is to stop carrying the reaction mass, as a gravity assist does: it supplies velocity with no propellant and appears nowhere in the equation.

A programme that died of a structural coefficient #

Staging is older than the equation that explains it: Kazimierz Siemienowicz published a three-stage design in 1650, three centuries before anyone could say why it helped. Tsiolkovsky supplied the reason in 1929, in a paper on what he called rocket trains, thirty-two years after deriving the equation itself.

The other date belongs to the near-miss. Lockheed Martin’s X-33, demonstrator for a single-stage VentureStar, was cancelled in 2001 after its composite liquid-hydrogen tank delaminated on test — usually described as a failure of materials, which is exact in a way worth saying plainly: it was a failure of $\varepsilon$. The design closed only if that tank could be built of composite at a mass nobody achieved, and the aluminium substitute, a few tonnes on the wrong side of 1.5 percentage points, consumed the margin the vehicle existed to demonstrate. A 0.73 km/s shortfall is small enough to attract forty years of attempts and large enough that all of them ended this way.

Where the ladder goes next #

The rungs this anchor still owes begin with the optimal division of $\Delta v$ between stages of unequal $\varepsilon$ and $v_e$ — the calculus assumed away here by drawing equal stages — and continue into parallel staging and propellant crossfeed, where the stage boundary stops being a moment in time. Beyond those sit $\varepsilon$ as an engineering quantity, and the trade between recovery hardware and payload that decides whether a cheaper place to burn is worth the mass of reaching it. Orbital refuelling belongs on the same ladder and breaks its framing: a vehicle departing a filled depot resets the mass ratio, and this inequality never arises.