Astronomy & Celestial Mechanics
the sky, computed
Generator

The hohmann generator

A Hohmann transfer, 2.6 to 1 in radius
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.

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.

16 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.

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. 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.

 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. 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.

 What it costs to reach Mars faster than the Hohmann transfer. The total velocity change to go from a 400 km orbit round the Earth to a 400 km circular orbit round Mars (solid), and the heliocentric part alone (dashed), against flight time, for transfers that leave the Earth's orbit tangentially and cross Mars's before reaching their far point; both planets are on circular, coplanar orbits. The right-hand end is the Hohmann transfer: 259 days and 5.65 km/s. Shortening it to 200 days costs 0.82 km/s more; to 150 days, 3.4 km/s more; below a hundred days the cost climbs past twenty. The curve leaves the Hohmann point with zero slope — the cheapest transfer is a minimum, so the first days saved cost almost nothing — and turns steeply upward only once the transfer ellipse has become long and thin. For the fastest transfers the orbit-to-orbit cost is below the heliocentric one, because a burn made deep in a planet's gravity well adds more to the speed at which the craft leaves it than the burn itself. Spaceflight

The first weeks off a Hohmann transfer are nearly free

The Hohmann transfer to Mars takes 259 days and is the cheapest route there is. Because it is a minimum, the cost of going faster starts flat — ten days saved costs twenty metres a second — and only climbs steeply once weeks have become months. For a crew, the time is itself a cost, paid in radiation at about two millisieverts a day.

 Two ways out of a transfer orbit. The eccentricity against the semi-major axis of a spacecraft spiralling out from a geostationary transfer orbit — perigee 300 km up, apogee at geostationary height — under a continuous thrust of 0.2 millimetres a second squared, integrated through hundreds of revolutions, for two steering laws. Thrust always along the velocity (red) raises the orbit efficiently but leaves its shape almost unchanged, reaching the geostationary semi-major axis after 64 days still with an eccentricity of 0.59 — an orbit swinging between 17,000 and 67,000 km. A law that points the thrust to reduce a weighted distance from the target in both elements (blue) circularises as it raises, and reaches a circular geostationary orbit in 129 days. Spaceflight

The thrust that rounds an orbit turns once a revolution

An electric engine pushes gently for months, and the obvious way to point it — along the velocity — raises an orbit as fast as possible and leaves its shape almost untouched. Starting from an elongated transfer orbit, that is the wrong job. Rounding the orbit off needs the thrust forward at apogee, backward at perigee and sideways in between, a full turn of the thrust vector every revolution, and the law that does it can be written down from the equations for how each element changes.

 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. Spaceflight

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.

 The propellant a long departure burn wastes, under three ways of pointing it. The gravity loss of a departure from a 300-km orbit to a hyperbolic excess speed of 3 km/s with an exhaust speed of 4.41 km/s, against the stage's initial thrust-to-weight ratio, both logarithmic, for thrust held along the velocity (orange), held fixed in space at the best direction (gold) and turned at the best constant rate (blue). At a thrust-to-weight of 0.05: 1264 m/s, 3597 m/s and 1521 m/s. At 0.2: 166 m/s, 328 m/s and 154 m/s. At 0.8: 12 m/s, 20 m/s and 10 m/s. No one of the three is best everywhere. For burns of a few minutes to half an hour the constant-rate turn beats following the velocity by up to a tenth; for the longest burns, which sweep most of an orbit, following the velocity wins, because the direction that does most good turns faster as the vehicle gains speed and no constant rate keeps up. A fixed attitude, the simplest to fly, costs 88 per cent more than the best turn even for the strongest stage drawn, and several times more for weak ones. The true optimum, which no simple family contains, lies below all three and is computed numerically for each mission. Spaceflight

No single way to point a long burn

A departure burn that lasts tens of minutes wastes propellant because it cannot all happen at the lowest, fastest point of the orbit, and how much it wastes depends on where the engine points while it runs. Three simple rules — follow the velocity, hold one direction in space, turn at a steady rate — are each best somewhere and none everywhere. A steady turn saves up to a tenth for burns of minutes; following the velocity wins for burns that last most of an orbit; holding still costs twice as much as either.

 Every orbit turns by about the same angle, whatever its size. How far the Earth's equatorial bulge turns an ellipse with a 300-km perigee in one revolution, in degrees, against its semi-major axis, logarithmic: the line of apsides (solid) and the node (dashed), for inclinations of 28.5, 51.6, 98°. The turn per orbit depends on the semi-latus rectum, which for an elliptical orbit lies between one and two perigee radii, so it changes by less than a factor of four from a near-circular orbit to one reaching the Moon's distance: at 28.5°, 0.56° of apsides per orbit at 8,000 km and 0.20° at 300,000 km. Per day the large orbit turns far less, because it completes so few revolutions; per revolution, almost the same. Spaceflight

An orbit lost to a burn that did not fire

A weak stage leaving the Earth splits its departure into several burns at perigee, coasting round a staircase of ever larger ellipses between them. If one burn does not fire, the stage goes round once more. That costs a revolution of time — an hour and a half early in the staircase, more than a day at the top — and a revolution of turning by the Earth's bulge, which comes out at nearly the same angle for every step. The angle is cheap to undo at the slow apogee of a large orbit. The time is not, when the target is the Moon.

 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. Spaceflight

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.

The whole library · All essays