Astronomy & Celestial Mechanics
the sky, computed
Concept

Plane change — where it appears

Rotating an orbit's plane without changing its size or shape. It costs 2v sin(Δi/2), so the cost is proportional to the speed at which it is done and nothing else about the vehicle — which makes where the burn happens matter more than how far it turns.

Named by 5 essays across one field — each of them below, with the objects they name alongside it.

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.

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.

spaceflight · Plane change 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.

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 · Plane change 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.

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.

spaceflight · Plane change 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.

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 · Plane change 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

Named alongside it

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

ΔvCircular velocityHohmann transferOrbital inclinationVis-vivaApoapsisGeostationary orbitLaunch azimuthOberth effectBi-elliptic transferElectric propulsionEscape velocity

All concepts