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 13 essays across 2 fields — 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 The same porkchop with Mars's orbit tilted, and the ridge that appears. Contours of departure energy C₃ for transfers from the Earth to Mars, over departure date and flight time, with Mars's orbit inclined by its real 1.85° to the ecliptic and every point a solved three-dimensional Lambert problem. The dashed line is where the transfer angle is exactly 180°. Along it the contours pile up into a ridge reaching C₃ of 346 km²/s² and more, against 7.9 at the cheapest point: a transfer half-way round the Sun has to lie in a plane through the Sun and two nearly opposite points, and reaching a target a fraction of a degree out of the ecliptic tilts that plane towards the pole. The ridge splits the plot into two basins — Type I transfers, less than half-way round, below and to the left, and Type II, more than half-way, above — and the cheapest transfers sit in one basin or the other, never on the line between them.

A wall across the porkchop

Draw the cost of going to Mars with Mars's orbit in the Earth's plane and the porkchop plot is a smooth pair of basins. Tilt Mars's orbit by its real 1.85 degrees and a wall appears between them, hundreds of times higher than the valley floor, along the line where the transfer goes exactly half-way round the Sun. The wall is a plane change forced by geometry, its height changes from one launch window to the next, and a single burn half-way to Mars removes it.

spaceflight · Launch windows Heliocentric latitude bought three ways, for the same launch. The highest inclination to the ecliptic a spacecraft can reach, against the hyperbolic excess speed it leaves the Earth with. Launched directly into an inclined orbit at 1 AU (dashed), it pays 2v⊕ sin(i/2) — 15.4° for 8 km/s and 23.2° for 12. Launched on a transfer to Mars whose plane is tilted — the wall across the porkchop plot, where transfers near half a turn are forced steeply out of the plane — it reaches latitude i half-way, for 13.7° at 8 km/s and 21.5° at 12: the same list price, starting from the cost of getting to Mars at all. Launched to Jupiter, it can turn there: a flyby rotates its Jupiter-relative velocity, and the steepest heliocentric orbit a flyby at 6 Jupiter radii can leave it on climbs to 90° for a 12 km/s launch, reaching the 79° of Ulysses's orbit (dotted) from about 10.6 km/s — close to the launch speed that mission actually used. Inclination bought near the Earth costs the Earth's orbital speed; bought at Jupiter it costs Jupiter's, less than half as much, and the flyby pays.

The wall sells inclination at the dearest price

Transfers to Mars that go almost exactly half-way round the Sun are forced into planes tilted steeply out of the ecliptic, which is why they form a wall of expensive dates across every launch chart. A mission that wanted to leave the ecliptic could use the wall on purpose. It would be buying inclination at the Earth's orbital speed — the dearest place to buy it — and the routes that reach the Sun's poles have always gone somewhere slower to make the turn.

spaceflight · Launch windows The plane change wants a gentle flyby. The lowest inclination to the Moon's orbit that one flyby can leave a spacecraft in, against the flyby's height above the Moon, for a spacecraft arriving from an orbit of perigee 6678 km inclined 51.6°: with no constraint on where the new orbit goes (dashed), and requiring the new perigee to lie between the atmosphere and 20,000 km, so that the spacecraft comes back to the Earth (solid). Unconstrained, any flyby closer than tens of thousands of kilometres can bring the inclination to nearly zero. But a close flyby turns the velocity so far that it also flings the spacecraft outward, and no aim point brings it back near the Earth; only from about 9,500 km up does a returning orbit become possible. Between that height and about 18,000 km the flyby can both bring the spacecraft back and leave its orbit nearly in the Moon's plane; further out still, the turn is too small to remove the inclination. The useful flyby is a gentle one, in a window.

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.

spaceflight · Plane change Three sails against the two routes that have flown. Inclination reached against years since launch. The Venus staircase and the Jupiter step are the arithmetic of the launch-window figures — repeated Venus flybys at a v∞ of 15 km/s reaching 23°, and a single Jupiter flyby after 1.2 years reaching 90° on a heavy launch. The sails, of lightness number 0.020, 0.040, 0.080, spiral in to 0.48 AU — flat lines, 4.6, 2.3, 1.1 years — and then climb at a constant rate until they are polar. They pass the Venus route after 9.0, 4.5, 2.2 years; only the lightest catches Jupiter inside the decade, and none ever stops climbing.

A climb out of the ecliptic paid for in orbits

A solar sail can tilt its orbit without propellant by leaning against sunlight and changing which way it leans twice an orbit. Each orbit then adds the same few degrees whatever its size, so the climb runs faster the closer to the Sun the sail goes. Nothing in the orbit fixes how close that can be. The sail film's temperature does.

spaceflight · Launch windows 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 An iteration that slows toward the straight-through transfer. The contraction ratio of the corrected-target iteration — how much each pass shrinks the miss — against the transfer angle, for a transfer between orbits of 51.6° inclination at 400 and 420 km, with the Earth's bulge at its real strength and at 10 and 30 times it, on a logarithmic scale. The iteration converges where the ratio is below 1 (dashed). At the real strength it shrinks the miss 1857-fold per pass at 150°, 83-fold at 178° and 8-fold at 181°: always convergent, ever slower. At 10 times the strength it fails between 181° and 184°; at 30 times it fails between 182.5° and 192°. The slowing is geometric: near 180° the transfer's plane is set by a sideways shift of the aim point, so correcting the bulge's sideways miss tilts the whole arc.

A correction that tilts the arc it meant to aim

A transfer designed with Lambert's two-body solution and flown through the Earth's flattened field misses its target, and the standard repair is to aim at the target minus the miss and solve again. That converges fast — until the transfer angle approaches 180°. There the two end points and the Earth's centre are nearly in line, the arc's plane is decided by a sideways shift of the aim, and correcting the bulge's sideways miss rotates the whole arc. The iteration slows by a factor of a hundred, and with a larger perturbation it runs away. Correcting only within the plane fixes it, at the cost of a plane change paid at arrival.

orbits · Lambert's problem A thrust that leans out of the plane at the nodes. The yaw of the thrust — its angle out of the orbit's plane — against the argument of latitude round one revolution, at the start, half-way through and near the end of a low-thrust transfer from a 7.726 km/s orbit to geostationary speed (3.075 km/s) that also removes 28.5° of inclination. Edelbaum's law (dashed) holds the yaw at one size through each revolution, switching sign at the antinodes. The optimal law (solid) has tan β proportional to |cos u|: the thrust lies in the plane at the antinodes, where out-of-plane thrust would mostly swing the node, and leans out furthest at the nodes, where it turns the plane best — 44° half-way through the transfer, against Edelbaum's 34° all round. Raising the orbit is equally efficient anywhere, so the thrust never needs to coast: the in-plane share fills exactly the parts of the orbit where turning is poor.

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.

spaceflight · Plane change 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%.

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.

spaceflight · Plane change

Named alongside it

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

ΔvInclinationGeostationary orbitOrbital inclinationCircular velocityElectric propulsionGravity assistHohmann transferLow-thrust transferVis-vivaApoapsisArgument of latitude

All concepts