The push that needs nothing to push against

In January 1920 the New York Times published an editorial about Robert Goddard’s proposal to send a rocket to the Moon. Its central objection was that a rocket in vacuum would have nothing to push against, and that Goddard appeared to lack “the knowledge ladled out daily in high schools”.

The objection is worth taking seriously rather than laughing at, because it is a definite physical claim and it makes a measurable prediction: thrust should fall as the air thins.

The whole of it, as one conservation law #

A rocket and its exhaust are one isolated system. Nothing enters or leaves; the propellant that was inside is now outside, and both are still part of the system. So the total momentum does not change, and if the rocket started at rest the total momentum is zero for ever.

That is the entire physics. The vehicle goes forward because the exhaust goes backward, and no medium appears anywhere in the arithmetic.

The frame in which that is easiest to see is the one riding with the centre of mass of vehicle-plus-exhaust, which never moves at all. What accelerates is one part of the system; the other part accelerates the other way; and the mass-weighted average of the positions goes on exactly as it was. Nothing outside the system has been touched, which is the content of “nothing to push against” stated as an arithmetic fact rather than as a slogan.

Where the extra thrust comes from #

If the atmosphere is not being pushed against, why does it appear in the thrust at all?

Because it presses on the nozzle. The gas inside the bell pushes outward on the walls; the atmosphere outside pushes inward. Integrating both over the whole engine leaves a residue proportional to the exit area — the one place where there is ambient pressure outside and exhaust pressure inside, across a plane of area $A_e$:

$$F = \dot{m}v_e + (p_e - p_a)A_e.$$

The first term is the momentum flux of the exhaust and does not know about the air at all. The second is a pressure difference across a plane, and it grows as $p_a$ falls, reaching its maximum in vacuum.

Pressure is a rate of arrival of momentum, which is what makes the two terms in that expression the same kind of quantity rather than an addition of unlike things. The first counts momentum carried out by the exhaust. The second counts momentum delivered back to the exit plane by whatever molecules happen to be outside it. Remove the outside molecules and the second term goes up, because there is nothing pushing back — which is why the same engine is more efficient in vacuum than at sea level, and why a nozzle is designed for one altitude.

An engine designed for vacuum has a much larger exit area than one designed for sea level, and it is not usable at sea level at all: the exhaust over-expands, the atmosphere pushes the flow off the nozzle wall, and the resulting flow separation shakes the engine apart. That is a real design constraint and it is the exact opposite of needing air to work against.

The demonstration that settles it in a lecture theatre #

The argument above is complete and it convinces nobody who has not already been convinced, so it is worth recording the experiment.

Goddard did it in 1920, immediately after the editorial. He fired a small powder rocket inside a bell jar evacuated to a few per cent of an atmosphere and measured the impulse from the recoil of the mounting. The thrust was higher in the partial vacuum than in air, by about the amount the exit-area term predicts, and he published the result the same year in a Smithsonian paper whose relevant section is titled simply as a test of the reaction in vacuum.

The demonstration that reaches further is cheaper still. Sit on a wheeled chair on a smooth floor and throw a heavy object. The chair moves backwards. Nothing about the air was involved; the same throw made in a vacuum chamber gives the same recoil, and the thrown object is the exhaust. A rocket is that, repeated a hundred thousand times a second, with the thrown object being a few grams of hot gas.

What makes the wrong account persistent is not that the evidence is hard to obtain. It is that “pushing” is a word for a contact interaction, and a body accelerating with nothing in contact with it looks as though something has been left out. The thing left out is the exhaust, which is easy to overlook because it is going away.

Why the nozzle widens #

The thrust expression has an exhaust speed in it and says nothing about how a gas is persuaded to leave that fast. The answer is the one piece of the machine that looks wrong from the outside, and it is worth having, because the shape of a rocket engine is the shape of one inequality.

For a compressible gas flowing steadily along a duct, conservation of mass and of momentum combine into a relation between the area and the speed:

$$\frac{\mathrm{d}A}{A} = (M^2 - 1)\frac{\mathrm{d}v}{v},$$

with $M$ the ratio of the local speed to the local speed of sound. The bracket changes sign at $M = 1$, and everything follows from that. Below the speed of sound, a narrowing duct speeds the flow up, which is what a garden hose does and what anybody would expect. Above it, a narrowing duct slows the flow down and a widening one speeds it up, because the gas is expanding and thinning faster than the duct is opening.

So a nozzle that is to deliver supersonic exhaust must have both: a contraction to bring the gas up to the speed of sound, a throat where it arrives at exactly Mach 1, and a bell beyond that in which it goes on accelerating as the area grows. The throat is not a restriction to be minimised. It is the one place in the engine where the flow is sonic, and its area is what sets the mass flow for a given chamber pressure — which is why an engine’s size is quoted by its throat and its expansion by the ratio of exit to throat.

That ratio is the design decision the ambient-pressure term above is about. A large expansion gives a low exit pressure and a high exhaust speed, which is what a vacuum engine wants; the same nozzle at sea level expands the flow below the ambient pressure, and the atmosphere pushes the flow off the wall. So the bell is chosen for the altitude the engine will spend its life at, and the fifteen per cent in the opening figure is the residue of that compromise.

What the logarithm does #

The conservation law gives Tsiolkovsky’s result immediately. Each element of propellant leaves at $v_e$ relative to the vehicle, so $m\,dv = -v_e\,dm$, and integrating gives

$$\Delta v = v_e \ln\frac{m_0}{m_f}.$$

The second property is the one that does the damage, and it is worth stating as arithmetic rather than as a shape. Going twice as fast on the same engine does not need twice the propellant; it needs the mass ratio squared, so a vehicle that was nine parts propellant to one part everything else becomes ninety-nine to one. Three times as fast makes it nine hundred and ninety-nine to one. No engineering removes a logarithm, and there is no size of vehicle at which the ratio improves — the equation contains no absolute mass anywhere.

A ratio of 21.8 means a vehicle that is 95.4 per cent propellant. Adding tanks to hold the propellant, and structure to hold the tanks, at the usual few per cent of the propellant’s mass, leaves nothing.

The gain from staging is not that smaller pieces are easier to build. It is that a stage which has finished burning is dead mass being carried by everything after it, and dropping it changes the mass ratio of what remains — the only quantity the logarithm contains.

Where the energy goes #

Momentum is conserved and energy is not, and the difference between them is what makes a rocket’s efficiency an odd quantity.

Kinetic energy divides itself between whatever is moving, and a rocket’s chemical energy divides between the vehicle and the exhaust it has just thrown away. Which of the two gets most depends on how fast the vehicle is already going, and the answer is not monotone: at low speed most of the energy ends up in the exhaust, and the share going to the vehicle peaks when the vehicle’s speed equals the exhaust speed.

At the start of a burn the vehicle is slow and the exhaust leaves at nearly the full $v_e$ in the ground frame, carrying almost all the kinetic energy away. Later the vehicle is fast and the exhaust, leaving at $v_e$ relative to the vehicle, may be nearly at rest in the ground frame — carrying almost no energy at all. The propulsive efficiency peaks when the vehicle’s speed equals the exhaust speed, and at that point every joule of chemical energy is going into the payload.

The arithmetic is worth doing once. Working in the ground frame, the exhaust element leaves with speed $v - v_e$, so its kinetic energy per unit mass is $\tfrac12(v - v_e)^2$ — which is zero when the vehicle is going at exactly the exhaust speed, and rises again as the vehicle goes faster still. Meanwhile the vehicle gains $v\,dv$ per unit mass of propellant burnt. The ratio of the two is the propulsive efficiency, it reaches one at $v = v_e$, and it is below a half for the first two thirds of a typical launch.

That is not a paradox about frames; it is the ordinary observation that the same push does different amounts of work depending on how fast the thing being pushed is already moving. It is also why a burn made at the bottom of a gravity well, where the vehicle is moving fastest, buys more energy than the same burn made high up.

An orbit’s shape is set by its energy and angular momentum, so a change of speed at one point changes the whole orbit — and for a given $\Delta v$ the change is largest where the vehicle is already fastest, because kinetic energy goes as the square of the speed and the cross term $mv\,\Delta v$ is what the burn buys. That is the Oberth effect, and it is the reason a probe to the outer planets fires its engine at closest approach rather than anywhere else.

The alternative that has no propellant to carry #

Since what matters is momentum thrown away, and a photon carries momentum $E/c$, light will do.

Light has a pressure, so a beam of it is a propellant with the highest exhaust speed there is and by far the worst thrust per watt: 3.3 newtons per gigawatt. A photon rocket has a mass-ratio problem no chemical rocket approaches, and it has none of the propellant problem at all. That is the trade the whole subject is about, stated at its extreme.

Between the extremes sit ion engines: exhaust speeds of thirty kilometres per second, thrusts of tens of millinewtons, and a mass ratio of 1.4 to reach orbital velocity against a chemical rocket’s 21.8. They cannot lift their own weight and they are the only practical way to move a useful payload across the solar system.

And the far end of the logarithm behaves as the near end does. The energy of a body rises without bound as its speed approaches $c$, so the relativistic rocket equation is written in rapidity rather than in speed — a quantity that adds, where speeds do not — and reaching a given rapidity still costs a mass ratio of $e$ raised to it. Nothing in the argument changes; only the variable that adds.

What it costs #

The exhaust speed has been a constant. It is set by the temperature and molecular weight in the chamber and by the nozzle’s expansion ratio, and it varies through a burn as the chamber conditions and the ambient pressure change. Every quoted specific impulse is one of a family.

The vehicle has been a point. A real rocket has propellant sloshing in tanks, a centre of mass that moves as the tanks empty, and a thrust vector that has to be steered to keep the two aligned. None of it changes the momentum accounting and all of it decides whether the vehicle stays pointed anywhere.

Gravity and drag have been left out entirely. The Δv to reach low orbit is about 7.8 km/s of orbital speed plus roughly 1.5 to 2 of losses to gravity and drag during the climb, which is where the 9.4 in the figures comes from. Those losses depend on the trajectory and are the reason a launch vehicle pitches over rather than going straight up.

The angle that throws furthest with no drag is forty-five degrees, and no launch vehicle uses anything like it. A rocket’s trajectory is chosen to spend as little time as possible fighting gravity while gaining horizontal speed, which is a different optimisation entirely — the gravity loss is an integral of $g\sin\theta$ over the burn, and it is minimised by pitching over early. The difference between the two answers is the clearest single instance in this essay of how much a “negligible” term can change a result.

The one number engineering can move #

Everything above turns on the exhaust speed, and it is worth saying what it is made of, because the answer is not what most people expect.

For an ideal expansion the exhaust speed goes as

$$v_e \propto \sqrt{\frac{T_c}{\mathcal{M}}},$$

the chamber temperature over the mean molecular weight of the exhaust. Both matter and only one of them can be pushed far: chamber temperatures are already at the limit of what a cooled wall will survive, and doubling the temperature buys only a factor of $\sqrt2$. The molecular weight is the free variable, and it is the reason hydrogen is the fuel of choice.

Hydrogen burnt with oxygen releases less energy per kilogram of mixture than several alternatives. What it produces is water, molecular weight eighteen, against the twenty-two or so a kerosene engine manages — and the square root of that ratio is most of the difference between an exhaust speed of three kilometres per second and one of four and a half. Engines built on it go further: they burn deliberately fuel-rich, at something like six parts oxygen to one of hydrogen rather than the eight that would consume both completely, precisely because the unburnt hydrogen left in the exhaust drags the mean molecular weight down faster than the wasted fuel drags the temperature down.

Deliberately failing to burn some of the fuel, in order to go faster, is a strange-looking decision and it is the correct one. It is also the clearest demonstration that a rocket is a momentum machine rather than an energy machine: what is wanted is mass leaving quickly, and light molecules leave quickly for a given amount of heat.

The convention the industry quotes this in is worth decoding once, since it appears on every datasheet. Specific impulse is the exhaust speed divided by the standard acceleration of gravity, so it comes out in seconds — 263 for the engine in the first figure at sea level and 304 in vacuum. There is no physics in the division; it is a historical artefact of quoting thrust per unit weight of propellant per second rather than per unit mass. Multiplying a specific impulse by 9.807 recovers the number that belongs in the rocket equation.

The correction, forty-nine years later #

The editorial that opened this essay was not retracted at the time. It was retracted on the day Apollo 11 was on its way to the Moon, in a three-sentence item noting that further investigation had confirmed the work of Newton in the seventeenth century, that it is now definitely established that a rocket can function in a vacuum, and that the Times regretted the error.

It is a good joke and it hides a better point. The original objection was not stupid; it was a definite, checkable, physical claim, and the reason it was wrong was available in 1920 to anybody willing to fire a rocket in a bell jar — which is what Goddard did that year. What kept the claim alive for half a century was not the difficulty of the experiment but that nobody who believed the claim thought an experiment was needed.

Where the model stops #

Nothing above says how fast the propellant can be made to leave. That is chemistry and nozzle design, and the exhaust speed is the one number in the rocket equation that engineering can move. It has moved by about a factor of two in a century and there is no prospect of another.

The exhaust has been treated as leaving in one direction. A real nozzle produces a cone, and the transverse components of the exhaust momentum cancel while their contribution to the energy does not. The resulting loss is quoted as a divergence efficiency, it is a per cent or two for a well-designed bell, and it is a straightforward instance of momentum being a vector and energy not being one.

And the staging arithmetic is more optimistic than any real vehicle. The model gives every new stage its engines, its tanks, its avionics and its separation hardware at the same structural fraction of a smaller propellant load, so payload fraction rises for ever with stage count. Put a fixed mass on each stage instead and the curve turns over — which is the real reason nothing with more than three stages has been worth its plumbing.

The picture of an energy budget as a landscape is the one every account of getting to orbit reaches for, and it is misleading in a specific way: what a launch has to supply is a speed, not a height. Reaching the altitude of the space station takes about two per cent of the energy. The other ninety-eight is the 7.8 kilometres per second needed to stay there, and a vehicle that went straight up to that altitude with no horizontal speed would fall straight back down.

What the pictures cannot show #

The ledger figure draws two curves whose sum is zero, and the flatness of the sum is the whole content. What it cannot show is why the sum is zero — that comes from Newton’s third law, which is a statement about pairs of forces and has no representation on a graph of momentum against burnt fraction.

Nor can the thrust figure show the flow. What produces the momentum flux is a supersonic expansion through a converging–diverging nozzle, with the gas accelerating as the area increases — which is the opposite of what an incompressible intuition expects and is the piece of the physics that makes the engine possible. The figure draws the resulting force against an ambient pressure and contains no gas at all.

Where this ladder goes next #

Three rungs stand on momentum. The first found that collisions are easier than forces, because the total is fixed whatever happens in the middle; the second found the point that keeps moving as if nothing had happened. This one applies the same accounting to a system that throws part of itself away, and finds that nothing outside the system is needed or involved.

The habit worth carrying away is about choosing the system. Almost every difficulty in this subject comes from drawing the boundary around something whose mass is changing. Include the exhaust and the mass is constant, momentum is conserved, and the answer falls out in three lines; exclude it and the second law has to be patched with a term whose derivation is exactly the calculation being avoided. The boundary is a choice, and it is nearly always the whole of the problem.

What is left on this ladder is what happens when the interaction cannot be localised in time at all. A continuous stream of momentum arriving at a surface is a pressure, and following the same accounting into a gas of $10^{23}$ particles turns a mechanical statement into a thermodynamic one — with the same third law doing the same work.