Astronomy & Celestial Mechanics
the sky, computed
Concept

Specific impulse — where it appears

An engine's exhaust speed divided by standard gravity, quoted in seconds by a convention that makes a velocity look like a time. A chemical engine reaches about 450 seconds and an ion engine about 3,000, which is the whole reason an electric transfer is cheaper in propellant while being dearer in velocity change.

Named by 13 essays across one field — each of them below, with the objects they name alongside it.

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.

spaceflight · Rocket equation 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.

spaceflight · Rocket equation 20% more Δv, spread over 26 revolutions. A continuous tangential thrust from a circular orbit of radius 1 to one of radius 6.611, integrated from dr/dt = 2r·a_T/v with the primary's GM set to 1. The spiral costs |v₁ − v₂| = 0.6111 in units of the inner circular speed, against 0.5076 for the two-impulse Hohmann drawn on the same pair of circles — 20.4% dearer, and the excess is exactly the Oberth advantage the impulsive transfer collects by burning where the vehicle is moving fastest and the spiral throws away by burning everywhere. The revolution count follows from the thrust level and nothing else, and the count drawn here is not a real one: 26 revolutions at an acceleration of 0.0015 of the inner orbit's own gravity, chosen so that the spiral can be seen at all. A real electric transfer runs nearer 3·10⁻⁵, which is 1,296 revolutions of a curve no page could resolve. Halving the thrust doubles both the turns and the time and leaves the Δv exactly where it is. That separation is the whole reason low thrust is flown at all — the Δv is worse and the propellant is not, because the exponential in the rocket equation is over Δv/(g₀Isp) and an electric engine's Isp is the larger number by more than this 20%.

The transfer that costs more the gentler it is

A continuous spiral from low orbit to geostationary needs a fifth more velocity change than the two-burn transfer, because the thrust is never at the one place it is worth most. It is flown anyway, and by a wide margin — because the exponential in the rocket equation changed sides.

spaceflight · Low-thrust transfer An oscillation that does not matter, on a ramp that does. Stored angular momentum in a reaction wheel over 160 days at 550 kilometres, with a capacity of 25 newton metre seconds. The total environmental torque is 1.75e-4 newton metres, of which 35 per cent is taken to survive averaging over an orbit. The fast oscillation is the part that does not survive: it has the 95.6-minute orbital period, reaches 0.10 newton metre seconds, and returns to where it started every revolution, so it consumes capacity and nothing else. The ramp under it is the secular part, and its slope measured between two instants a whole number of orbits apart is 6.117e-5 newton metres, which is the secular torque and is how the figure checks itself. The wheel fills in 4.7 days and has to be emptied 33 times in the span drawn. Every attitude-controlled spacecraft in the collection lives on this sawtooth, and the vertical drops are the only part of it that costs anything: the store can be moved between wheels for nothing, and taken out of the vehicle only by pushing against something outside it.

The spin that has to be put somewhere

A spacecraft holding an attitude is not resisting a force. It is absorbing a slow, one-directional trickle of angular momentum from the gradient of gravity across its own body, from sunlight, from the last of the atmosphere — and every store it has for that trickle fills up.

spaceflight · Attitude control 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.

spaceflight · Rocket equation 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.

spaceflight · Rocket equation 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.

spaceflight · Rocket equation One square root that raises the orbit and turns it. The velocity budget from a 300 km circular orbit to geostationary, against the plane change carried out along the way. Edelbaum's closed form — √(v₁² + v₂² − 2v₁v₂cos(½πΔi)) with Δi in radians — puts the whole continuous manoeuvre in one square root, and at Δi = 0 it collapses to |v₁ − v₂| = 4.651 km/s, which is the spiral's cost with no plane change in it. The two-impulse curve puts its rotation into the circularisation burn at apoapsis, where the vehicle is moving at 1.608 km/s and a rotation is cheap. At 28.5° the continuous transfer costs 5.951 km/s against the impulsive 4.256 — the plane change adds 1.300 to one and 0.363 to the other. That is the opposite of the usual claim that low thrust turns for free. It turns continuously, which is not the same thing: the gain is that the propellant is not the budget, and the Δv is worse here as it is everywhere else.

One square root that raises the orbit and turns it

Edelbaum put the plane change inside the same radical as the raise, and the half-pi in its cosine is the whole result — a continuous turn costs π/2 times an impulsive one below 140° and less above it.

spaceflight · Low-thrust transfer There is a best exhaust speed, and the calendar picks it. Payload fraction against exhaust speed for a 11 km/s mission, at three thrusting durations, with a power plant of 0.025 kilograms per watt of jet power. The vehicle is payload plus power plant plus propellant and the arithmetic is one line: λ = e^(−Δv/c) − (αc²/2t)(1 − e^(−Δv/c)), the first term the rocket equation and the second the mass of the machine that makes the jet. They pull opposite ways. A slow exhaust burns propellant; a fast one needs power, and the power per newton rises in proportion to c, so the plant's mass rises as c². Neither end is where anybody builds, and the optimum is 3,211 s over 200 days, 6,523 s over 700 days, 11,424 s over 2000 days — the same mission, the same Δv, and the best engine for it changes by a factor of 3.6 depending only on how long there is to do it. A gridded ion engine at 3,100 seconds sits at 30.4 km/s, which suits the shortest of these and is slow for the longest. What the curve cannot show is that α is not a constant either: a solar array's mass per watt falls as the fourth power of the distance from the Sun, so the same vehicle is a different point on this plot at Mars and at Jupiter.

The engine is chosen by the calendar

A chemical stage's exhaust speed is fixed by chemistry. An electric one's is a dial, and turning it up costs power — so there is a best setting, and it is decided by how long the mission has rather than by how far it is going.

spaceflight · Low-thrust transfer A sail has to be tilted, and tilting it throws most of it away. The thrust on an ideal flat sail, resolved into the orbit frame, against the angle between the sail's normal and the sunline. The force is along the normal and goes as cos²α — one cosine for the area the sail presents to the light, one for the momentum the reflection returns along the normal — so the radial component goes as cos³α and the transverse one as cos²α sin α. A sun-facing sail has no transverse push at all. Its thrust is purely outward and falls as 1/r² exactly as solar gravity does, so it merely replaces μ with μ(1 − β): the orbit stays the same conic with a smaller central mass, and the vehicle raises nothing. Every manoeuvre a sail makes it makes by tilting, and the transverse push peaks at 35.26° — arctan(1/√2), differentiated rather than tabulated — where it is 0.385 of the face-on force, or 2/(3√3). Two thirds of the thrust is the price of pointing any of it somewhere useful. The lightness number β is the sail's whole specification, radiation pressure and gravity both falling as 1/r² so their ratio is a constant: IKAROS, 2010 at 1607 g/m² gives β = 9.5e-4; LightSail 2, 2019 at 156 g/m² gives β = 9.8e-3; a 5 µm film with no structure at 7 g/m² gives β = 0.219, against the 1.53 g/m² at which the Sun would push as hard as it pulls. What no figure here can show is the thing a sail actually has instead of a rocket equation, which is nothing: the exponential that limits every other vehicle is absent, and what limits this one is a structure that has to hold a square kilometre of film flat.

A drive with no rocket equation

Radiation pressure and solar gravity both fall as the inverse square, so their ratio is a constant of the vehicle. A sun-facing sail therefore only rescales the central mass — it has to be tilted to do anything, and the best tilt throws away sixty-two per cent of the thrust.

spaceflight · Low-thrust transfer What a departure loses to taking time. The gravity loss of a departure burn — the velocity a rocket spends beyond what an instantaneous impulse would need — against its initial thrust-to-weight ratio, on logarithmic axes, for departures from a 300-km orbit to hyperbolic excess speeds of 0.5, 3, 6 km/s, an exhaust speed of 4.41 km/s and thrust held along the velocity. The impulsive costs are 3.21, 3.60, 4.74 km/s. At a thrust-to-weight of 1.2 the losses are 4, 5, 11 m/s; at 0.1, 298, 396, 708; at 0.03, 1485, 1785, 2546. For a stage of high thrust the loss falls as about the inverse square of the thrust (slope -1.98 at the high end), because the burn arc shrinks in proportion and the loss grows as the square of the arc.

A burn that cannot happen at one point

Every transfer cost in orbital mechanics is computed as though the engine fired for an instant. A real departure burn runs for minutes or hours, and the vehicle travels a long arc while it does — climbing, slowing, and delivering thrust ever further from the point where it would have done the most good. The extra velocity that costs is the gravity loss, and it grows as the square of the arc: negligible for a strong stage, ruinous for a weak one, and recoverable by splitting the burn at the price of time.

spaceflight · Orbital transfer The cost of turning an orbit, against how much of each revolution the engine runs. The velocity spent per unit of inclination change, relative to an instantaneous turn at a node, when a small out-of-plane thrust is applied only in arcs centred on the two nodes, against the half-width of each arc (blue). Thrusting all the time costs π/2, 1.571; arcs of ±60° cost 1.209; ±30°, 1.047; ±10°, 1.0051. The price is time: the engine is off for the rest of each orbit, and the turn takes longer by the inverse of the duty times the efficiency (orange, right-hand scale) — relative to a push that was always fully effective, 1.57 times as long for continuous thrust, 3.14 for ±30°, 9.0 for ±10°. Most of Edelbaum's penalty is spent in the quarter-orbits near the points where a push turns the node instead of the plane.

A plane turned only where turning is cheap

A low-thrust engine that turns an orbit's plane continuously pays π/2 times what an impulse at the node would cost, because for half of every revolution it is pushing where a push mostly swings the node instead of tilting the plane. Thrusting only in arcs about the nodes recovers almost all of that: arcs of ±30 degrees cost five per cent more than an impulse, at the price of taking twice as long. It is how geostationary satellites hold their planes, and it stops working for large turns, where a chord is shorter than any arc.

spaceflight · Plane change How hard a hovering spacecraft pulls an asteroid. The velocity change a spacecraft hovering beside a 200-m asteroid gives it in a year by gravity alone, in millimetres per second, against the hovering distance from the asteroid's centre in asteroid radii, for spacecraft of 5 t, 20 t, 50 t. The pull depends on the spacecraft's mass and the distance, not on the asteroid's mass. A 20-tonne tractor at 1.5 radii gives 1.9 mm/s a year; at 3 radii, a quarter of that. Closer is better, and the limit on how close is set by the exhaust: inside the shaded region the engines would have to point so steeply outward to keep their 20° plumes off the surface that holding station costs more than the pull is worth.

A tow with no rope

A spacecraft parked beside an asteroid pulls it by gravity alone, with a force set by the spacecraft's mass and its distance and nothing else. Twenty tonnes hovering a hundred and fifty metres from the centre of a two-hundred-metre asteroid moves it by two millimetres a second in a year. That is the gentlest engine there is, and its limit is not the pull but the exhaust: the spacecraft must thrust away from the rock to stay put without blasting it, and the angle that takes decides how close it can hover.

spaceflight · Low-thrust transfer

Named alongside it

The objects these essays reach for when they reach for this one.

ΔvLow-thrust transferMass ratioElectric propulsionExhaust velocityOberth effectOrbital transferPropellantRocket equationStagingHohmann transferPayload fraction

All concepts