Generator

The plane-change generator

What 28.5° of plane change costs, at three orbital speeds

The cost of rotating an orbital plane, 2v sin(Δi/2), against the angle turned, for the three speeds a transfer to geostationary orbit passes through: 7.669 km/s in a 400 km circular orbit, 3.075 km/s once circular at 42,164 km, and 1.618 km/s at the apogee of the ellipse between them. All three are √(μ/r) or vis-viva at μ⊕ = 398,600 km³/s². A 28.5° turn therefore costs 3.775, 1.514 or 0.797 km/s depending only on where it is done — the apogee figure is 21% of the low one, and that fraction is the speed ratio, so it is the same at every angle. Past 23.9° a turn in the low orbit costs more than the 3.176 km/s that leaves Earth from it, and a full reversal costs 2v = 15.34 km/s, 4.8 times the escape burn.

9 essays call plane-change. 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.

$What 28.5° of plane change costs, at three orbital speeds. The cost of rotating an orbital plane, 2v sin(Δi/2), against the angle turned, for the three speeds a transfer to geostationary orbit passes through: 7.669 km/s in a 400 km circular orbit, 3.075 km/s once circular at 42,164 km, and 1.618 km/s at the apogee of the ellipse between them. All three are √(μ/r) or vis-viva at μ⊕ = 398,600 km³/s². A 28.5° turn therefore costs 3.775, 1.514 or 0.797 km/s depending only on where it is done — the apogee figure is 21% of the low one, and that fraction is the speed ratio, so it is the same at every angle. Past 23.9° a turn in the low orbit costs more than the 3.176 km/s that leaves Earth from it, and a full reversal costs 2v = 15.34 km/s, 4.8 times the escape burn.$ Spaceflight

The cheapest place to turn

Rotating an orbital plane costs 2v sin(Δi/2), and the only quantity in that expression that a mission controls is the speed. Where the turn is made therefore matters more than how far it turns.

Where the fuel goes, and it is not where a satellite points. Left, the orbit pole of a geostationary satellite, in degrees from the Earth's. The Sun and the Moon between them carry it round a circle of radius 7.4° in 53 years, and a satellite launched into the equatorial plane starts on the rim of that circle rather than at its centre — so its inclination climbs from zero at 0.88° a year, reaches 14.8° after 27 years, and comes back. Right, what holding it costs. A plane change of 0.88° at 3.07 km/s is 47.1 m/s a year; holding the longitude against the equatorial bulge, computed from the same resonant term that makes the longitude a pendulum, is 1.8 m/s a year. North–south is 96% of the budget, and a satellite that gives up on it does not fail — it starts tracing a figure of eight on the sky 1.8° tall in the first year, which a fixed dish cannot follow and a steerable one can. Retiring at the end of the propellant is therefore a choice about which service ends first.

Spaceflight

The orbit that has to be paid for every year

A geostationary satellite is not in equilibrium in any direction. The Sun and Moon tilt its plane by 0.85 degrees a year, the Earth's equatorial ellipticity makes two longitudes stable and two unstable, and the end of a satellite's life is the end of its propellant.

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

Rotating an orbit by flying away from it first. Total Δv against the angle turned, in units of the circular speed of the starting orbit, for two ways of rotating an orbital plane at a fixed radius. The single combined burn does everything at once and costs √(v₁² + v₂² − 2v₁v₂cos Δi); the three-burn route raises the apoapsis to 200 starting radii, turns there where the speed is only 0.7 per cent of what it was, and comes back down. Below 48.9° the single burn is cheaper and the two extra burns are not worth paying for. Above it the three-burn route wins, and it wins by more the larger the angle: at 90° it costs 0.8314 against 1.4142, a saving of 41 per cent. The mechanism is the one thing worth carrying away. A plane change costs 2v sin(Δi/2) and is therefore proportional to the speed at which it is done, so the cheapest place to turn is the slowest place available — and a vehicle can make a slow place by climbing, at a cost that does not grow with the angle while the rotation's cost does. That is why the crossover is an angle rather than a distance, and why it exists at all. What the figure does not price is time: the round trip to 200 starting radii takes 2015 times the period of the starting orbit, which for a low Earth orbit is months.

Spaceflight

Flying further away in order to turn

A plane change costs the speed at which it is done, so the cheapest place to turn is the slowest place available — and a vehicle can make a slow place by climbing. Above about thirty-nine degrees the round trip pays for itself, and the crossover is an angle rather than a distance.

Which inclinations a launch site can reach, and which it cannot. Orbital inclination against launch azimuth for three sites, from cos i = sin A cos φ. Due east is the only azimuth that gives the minimum, and that minimum is the latitude itself: Kourou 5.2°, Kennedy 28.5°, Baikonur 45.6°. Everything below the shaded line is unreachable from the highest-latitude site by any azimuth at all, and getting there costs a plane change afterwards — 5.94 km/s from a 400 km orbit to reach the equator from 45.6°. Baikonur in fact flies no lower than 51.6° rather than its 45.6°, and the extra 6.0° is overflight constraint rather than mechanics: the azimuth that would give 45.6° sends the spent stages over places they may not fall on.

Spaceflight

A plane change paid at the worst speed there is

A launch reaches an inclination fixed by its latitude and its azimuth, and no azimuth reaches an inclination below the latitude. Getting there afterwards means turning at orbital speed, which is the most expensive place available — so a site's latitude is a floor no trajectory removes.

Spaceflight

A plane change borrowed from the Moon

Turning an orbit's plane by the engine costs a fixed fraction of the spacecraft's speed, which is why it is done as far from the Earth as possible. Farther still, at the Moon's distance, a spacecraft is moving so slowly that the Moon can turn its velocity through tens of degrees for nothing — but a flyby close enough to turn it hard also flings it outward, and the useful flyby is a gentle one, in a window of heights the arithmetic picks out.

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.

Spaceflight

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

A turn spread where the climb leaves room

An electric spacecraft spiralling from low orbit to geostationary has to remove its launch inclination on the way, and the classic steering law tilts its thrust out of the plane by the same angle all the way round each orbit. But out-of-plane thrust turns the plane only near the nodes, while in-plane thrust raises the orbit equally well everywhere. Letting the tilt follow the nodes — leaning hard where turning works and lying flat where it does not — saves 150 metres a second on a transfer from Cape Canaveral, with no coasting, because the climb fills exactly the parts of each orbit where the turn would be wasted.

Where on the orbit the turning is done, and where the shadow falls. How effectively out-of-plane thrust turns a circular orbit's plane at each point round the orbit, measured from the ascending node: it goes as the absolute value of the cosine of that angle, greatest at the two nodes and zero halfway between them. The shaded bands are the Earth's shadow on an orbit 6,378 km up with the Sun in the orbit's plane, 60 degrees of the circle, placed once over a node and once between the nodes. Over the node the shadow switches the engine off where it would have done the most turning, and the orbit loses 25% of its turning; between the nodes it switches the engine off where it was doing little, and the loss is 7%.

Spaceflight

A turn that costs more when the shadow sits on the node

An electric spacecraft turning its orbit's plane does most of the turning near the two nodes, where a push out of the plane tilts it most. Its engine runs on sunlight and stops in the Earth's shadow, and the shadow covers a sixth of the orbit wherever it happens to fall. When it falls on a node it removes a quarter of each orbit's turning, and a law that concentrates its thrust at the nodes loses half. The node drifts one way and the Sun the other, so the shadow slides across the nodes over weeks — and a transfer's length depends on the day it starts.

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