Generator

The hohmann generator

A Hohmann transfer, 2.6 to 1 in radius

Two circular orbits and the ellipse that touches both. The first burn raises the far point to the outer orbit; the second, half an orbit later, circularises. The costs are computed from the vis-viva relation.

10 essays call hohmann. The drawing above is what it returns with no arguments at all; every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about.

Where it is called

Every figure listed here is the same construction drawn at different numbers, so a correction to one is a correction to all of them.

Spaceflight

The cheapest way between two orbits, and why it is so slow

Two burns and a long coast is the least fuel that will move a spacecraft between two circular orbits. It is also, for anything beyond the Moon, an unreasonably long wait.

$How much of a rocket has to be propellant. Propellant fraction against the velocity change bought, for three real propellant combinations. The curves approach 1 and cannot reach it, so every extra kilometre per second costs a larger share of what remains — which is why staging exists.$ Spaceflight

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.

Catching a target 40° ahead. A phasing manoeuvre. Dropping into an orbit 6% lower shortens the period to 0.9553 of the target's, so the chaser gains 16.1° each lap and closes 40° in 3 revolutions. Speeding up would have lost ground instead.

Spaceflight

Catching up by slowing down, which cost Gemini 4 its fuel

To reach something ahead in the same orbit, a spacecraft must fire backwards. Pointing at the target and thrusting makes the gap grow, and a crew found that out in orbit before anyone had flown the correct manoeuvre.

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%.

Spaceflight

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.

Down by 280 km, and faster by 164 m/s. A circular orbit at 400 km with a ballistic coefficient of 100 kg/m², integrated down to 120 km through a tabulated atmosphere at solar minimum and solar maximum. At solar min it takes 1.2 years; at solar max it takes 147 days — a factor of 2.9 for the same satellite in the same orbit, decided by an eleven-year cycle nobody controls. The rising curves are the orbital speed on the right-hand scale, and they are the point: the drag force is opposite the motion and takes energy out, and the body goes faster, from 7673 to 7836 m/s. There is no contradiction in it. The specific energy is −μ/2a, so removing energy shrinks a, and the circular speed √(μ/a) rises when a falls; the kinetic energy gained is exactly half the potential energy lost, and the other half is what the air took. Every point on every curve was integrated from da/dt = −(ρ/β)√(μa), and the speed at each point is √(μ/a) at that point rather than a separate model.

Spaceflight

An orbit that speeds up as it is slowed down

Drag takes energy out of a satellite and the satellite goes faster. There is no paradox in it, only a sign — and the same sign makes a re-entry date a space-weather forecast rather than an orbital computation, which is why Skylab was predicted for 1983 and came down in 1979.

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

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.

$The best split of a plane change between perigee and apogee, for three turns. How much is saved by moving part of the plane change into the perigee burn, against how much is moved, for turns of 15°, 28.5°, 51.6°. Doing the whole rotation at apogee is the standard answer and it is not the cheapest one: at perigee the turn is bought as a small correction to a burn that is happening anyway, so the first fraction of a degree is nearly free while the apogee saving is linear. Each curve therefore rises to an interior maximum — 10 m/s at 1.35° for a 15° turn, 25 m/s at 2.23° for a 28.5° turn, 40 m/s at 2.88° for a 51.6° turn — and falls back through zero at about twice that split. The saving is small against a 4.78 km/s budget, and it is free.$ Spaceflight

A rotation split between two burns

The standard answer is to do the whole plane change at apogee, where the vehicle is slowest. It is not the cheapest answer, and the reason is that a small part of the turn bought at perigee is a second-order correction to a burn that is happening anyway.

Spaceflight

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.

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

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.

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.

Spaceflight

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.

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