Astronomy & Celestial Mechanics
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Series

Rocket equation — the series

8 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. How much of a rocket has to be propellant. Propellant fraction against the velocity change bought, for three real propellant combinations. The curves approach 1 and cannot reach it, so every extra kilometre per second costs a larger share of what remains — which is why staging exists.

    The exponential that decides what can be flown

    A rocket carries its own reaction mass, so every kilogram of propellant has to be accelerated by the propellant beneath it. The result is exponential, and it is the reason spaceflight is hard.

    part 1 · spaceflight
  2. Payload fraction against the number of stages, for 9.4 km/s at three structural fractions. Overall payload fraction λ = λ₁ⁿ against the number of equal stages sharing 9.4 km/s, for structural coefficients 0.06, 0.08, 0.12 and an exhaust speed of 3.432 km/s (I_sp 350 s). At ε = 0.08 a single stage has a negative payload fraction, −1.67%: the mass ratio 15.5 it needs leaves less than the tankage weighs. That is the wall, and it is the structural fraction rather than the exponential — one stage tops out at vₑ ln(1/ε) = 8.67 km/s, 0.73 km/s short, so each curve begins where λ₁ crosses zero. At ε = 0.06 the same engine does clear 9.4 km/s in one stage, with 0.50% of the vehicle as payload — the wall moves with the tank, not with the engine. Two stages then buy 3.59% and five 4.66%, against a ceiling of 5.10% at infinitely many: the step from two stages to three is 0.67 points of payload, and the step from four stages to five is 0.14 points of payload — which is why a launcher is built with two or three stages and not with five. Falcon 9 Block 5 is plotted at its own mass table: 22.8 t of 549 t is 4.15%, and its per-stage structural coefficients are 0.061 and 0.040, both better than the 0.08 the family curve assumes.

    The stage that has to be thrown away

    The rocket equation does not forbid a single stage from reaching orbit. The tank does — by 0.73 km/s out of 9.4, which is close enough that the question stayed open for forty years and expensive enough that it was never once answered in flight.

    part 2 · spaceflight
  3. Where the 9.4 km/s should be divided between two unequal stages. Overall payload fraction against the share of the 9.4 km/s given to the first stage, for a vehicle whose two stages are not alike: kerosene and oxygen, first stage at ε = 0.06 and I_sp 300 s, so vₑ = 2.942 km/s; hydrogen and oxygen, second stage at ε = 0.09 and I_sp 450 s, so vₑ = 4.413 km/s. The curve falls to zero at both ends, because a stage asked for too much Δv has a negative payload fraction and the vehicle does not close. The maximum is at 26.8 per cent — 2.52 km/s in the first stage and 6.88 in the second — and it delivers 5.130 per cent of the lift-off mass as payload against 4.240 per cent for an equal division. The gain from optimising is 21.0 per cent of the payload — worth having on a vehicle whose payload is five per cent of its mass, and small next to the difference the second stage's exhaust speed makes. What the shape says is more useful than where the peak is: moving ten points either side of the optimum still delivers 4.905 per cent, so the penalty for getting the split wrong is 4.4 per cent of the payload. A penalty that small is why launch vehicles are staged on structural and operational grounds — where the tank domes go, which engines exist, what can be transported by road — and the calculus is run afterwards to check that nothing has been left on the table. Note also which way the optimum leans: the better stage is the second, and it is given more of the work, because a share of Δv bought at a higher exhaust speed costs less mass.

    The split that is not an equal split

    Two stages sharing a velocity budget do not share it evenly, and the calculus that divides it hands more of the work to whichever stage has the better exhaust speed. The optimum is interior, it is worth about a fifth of the payload, and at this mission it is flat enough that nobody designs to it.

    part 3 · spaceflight
  4. Exhaust speed against the molecular weight of the exhaust. The ideal exhaust speed of a converging–diverging nozzle, √(2γ/(γ−1) · R_uT_c/M · [1 − (p_e/p_c)^((γ−1)/γ)]), against the mean molar mass of the exhaust, at three chamber temperatures — 2200, 3000, 3600 K — with γ = 1.2 and an expansion to 1 per cent of chamber pressure. Every choice a propellant makes enters through two symbols, and the speed goes as the square root of their ratio: four times the chamber temperature doubles it, and a quarter of the molar mass doubles it too. That symmetry is the point. A cooler flame with a lighter exhaust beats a hotter one with a heavier, and the marked combinations show it — hydrogen + oxygen at M = 10, T_c = 3500 K, 441 s; methane + oxygen at M = 20.5, T_c = 3550 K, 310 s; kerosene + oxygen at M = 23, T_c = 3670 K, 298 s. The hottest flame drawn belongs to kerosene + oxygen and the fastest exhaust to hydrogen + oxygen, which are not the same entry. Hydrogen's advantage is not that it burns hot; it burns slightly cooler than kerosene. Its advantage is that the mixture is run fuel-rich on purpose, so the exhaust carries unburnt hydrogen and its mean molar mass falls to about 10 rather than water's 18 — buying more in the denominator than it loses in the numerator.

    Choosing a propellant is choosing a molecular weight

    An exhaust speed is the square root of a chamber temperature divided by a molecular weight, so a cooler flame with a lighter exhaust beats a hotter one with a heavier. Hydrogen wins the rocket equation and loses the tank, and the two cannot be optimised separately.

    part 4 · spaceflight
  5. Where the missing kilometres a second go. The two losses along a gravity-turn ascent, against the vehicle's thrust-to-weight ratio at lift-off, from an integration of the trajectory rather than from a table. Every run spends the same 9400 m/s of ideal Δv and every one is flown as the same manoeuvre: one pitch kick, solved by bisection so that the vehicle is horizontal at burnout, and thereafter zero angle of attack so that gravity alone turns it — which makes the steering loss identically zero and the comparison a fair one. What differs is how much of the Δv survives as speed. The gravity loss is ∫g sin γ dt, the part of the thrust spent holding the vehicle up rather than accelerating it, and it falls as the thrust rises because a vehicle that leaves quickly spends less time doing it: 1263 m/s at T/W = 1.15 against 381 at 2.2. The drag loss is ∫(D/m) dt and it rises, because the same haste means reaching high speed lower down where the air is: 119 m/s against 1874. The sum is least at T/W ≈ 1.5, at 1198 m/s, and the minimum is shallow — which is why real vehicles cluster between 1.2 and 1.5 and none of them is there because of this curve. At the marked 1.5 the 9400 m/s of ideal Δv leaves the vehicle at 8202 m/s and 42 km, against a circular speed of 7884 m/s there, so the 1198 m/s of loss is the whole of the answer to why orbit costs about 9.4 km/s when orbital speed is under 7.9.

    Orbit costs 7.8 and a launch buys 9.4

    The gap between orbital speed and the velocity change a launcher spends is not overhead. It is three integrals along the ascent, only one of which can be reduced by flying better, and the two that can be traded move in opposite directions.

    part 5 · spaceflight
  6. The mass ratio an interstellar probe needs, and the exhaust that decides it. Mass ratio against final speed, for four exhaust speeds given as fractions of c, on a logarithmic vertical axis. The relativistic rocket equation replaces the velocity change with the rapidity artanh(β), which is the quantity that adds when velocities are combined, so the mass ratio is exp(c·artanh β / vₑ) and the dashed curves are the Newtonian exp(βc/vₑ) for comparison. The two agree wherever the speed is small and part company above about a third of c, with the relativistic answer always the dearer of the two. What the figure is really about is which curve a mission sits on: reaching 0.95c needs a mass ratio of 3.29e+26 at vₑ = 0.03c, 9.02e+7 at vₑ = 0.1c, 2.57e+2 at vₑ = 0.33c, 6.24e+0 at vₑ = 1c. A fusion drive with a realistic exhaust speed sits at the left of that list and the numbers are not engineering numbers. Even the photon rocket — vₑ = c, the fastest exhaust physics permits, and requiring the propellant to be converted entirely to directed radiation — needs a mass ratio of 1.73 to reach half of c and 4.4 to reach nine tenths. The equation never forbids a speed. It prices one, and the price is exponential in a quantity that itself runs to infinity.

    An equation that does not break at the speed of light

    Relativity replaces the velocity change in the rocket equation with the rapidity, which adds where velocities do not. The equation therefore never forbids a speed — it prices one, and the price is exponential in a quantity that itself runs to infinity.

    part 6 · spaceflight
  7. Where to stage, if the first stage is coming home. The payload a two-stage rocket puts into low orbit, as a share of its lift-off mass, against how much of the 9.4 km/s it needs is supplied by the first stage, with exhaust speeds of 3 and 3.4 km/s and structural fractions of 0.05 and 0.06. Thrown away, the first stage is best asked for 4.1 km/s and the rocket carries 3.5%. Landed downrange — 0.02 more of its mass in legs and fins, and propellant kept for an entry and a landing burn — the best split moves down to 3.1 km/s and the payload falls to 3.0%. Flown back to the launch site, which first cancels the stage's downrange speed, it moves to 2.5 km/s and 2.7%. A recoverable first stage is asked to do less, and the upper stage does the rest.

    A booster that flies home pays with the stage above it

    A first stage that lands carries legs, fins and the propellant to bring itself back, all of which the rocket equation charges at the exponential rate it charges everything else. Recovered from a rocket designed to be thrown away, it costs a fifth of the payload. Recovered from a rocket designed around it, the toll falls by a third — because the best place to separate a stage that is coming home is earlier and slower, and the stage above it does the rest.

    part 7 · spaceflight
  8. How much thicker the cable must be at the top. The ratio of an Earth elevator cable's cross-section at synchronous orbit to its cross-section at the ground, as a power of ten, against the specific strength of its material, for safety factors of 1 and 2. The taper is exp(ΔΦ/(σ/ρ)) with ΔΦ = 48.4 MJ/kg, the same exponential as the rocket equation. With a safety factor of 2: high-strength steel (0.25 MJ/kg) beyond 10³⁰; Kevlar (2.5 MJ/kg) about 6.7×10¹⁶; carbon fibre (3.6 MJ/kg) about 4.8×10¹¹; Zylon (3.8 MJ/kg) about 1.2×10¹¹; nanotube yarn, today (4 MJ/kg) about 3.3×10¹⁰; ideal nanotube (50 MJ/kg) about 7. Only a material with tens of megajoules per kilogram — near the ideal strength of a carbon–carbon bond, ten times anything yet made in bulk — gives a cable that is not absurd.

    A cable with a rocket equation of its own

    A cable hanging from synchronous orbit to the equator would let a climber reach orbit on electricity, with no rocket at all. But the cable has to hold its own weight, and a cable that holds its own weight at constant stress must thicken exponentially with the potential it spans. The exponent is the same 48 megajoules per kilogram a rocket must supply, divided by the material's strength per unit mass. The rocket equation has not been escaped; it has moved from the propellant to the rope.

    part 8 · spaceflight

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