Astronomy & Celestial Mechanics
the sky, computed
Concept

Orbital inclination — where it appears

The angle between an orbit's plane and a chosen reference plane. Changing it is the most expensive manoeuvre in spaceflight, because the required velocity change is proportional to the orbital speed at the point of the burn.

Named by 10 essays across 4 fields — each of them below, with the objects they name alongside it.

An orbit at i = 42°, Ω = 35°, ω = 55°. The three orientation elements. The orbit is generated in its own plane and rotated by the standard sequence, so the inclination, the node line and the argument of periapsis are the angles that produced the drawn curve rather than labels applied to it.

Six numbers that fix an orbit for all time, and the sixth is the awkward one

Five of the orbital elements describe a curve that never changes. The sixth says where on it the body is, and it is the only one that has to keep being measured.

orbits · Orbital elements Three rotations, applied in order. An orbit of eccentricity 0.45 carried into space by the three orientation angles, one panel per rotation: the argument of periapsis ω = 40°, then the inclination i = 42°, then the longitude of the ascending node Ω = 55°. The last panel applies the same three angles in the reverse order and arrives somewhere else, because rotations about different axes do not commute.

Three rotations that put an orbit in space, and they do not commute

An orbit's orientation takes three angles. Give the same three angles in a different order and the orbit ends up somewhere else — which is why the convention is part of the data.

orbits · Orbital elements Phases are a viewing angle, not a shadow. A satellite at eight points of its orbit. Exactly half of it is lit at every one of them; what changes is how much of the lit half faces the centre. Nothing is in shadow except at an eclipse.

Phases are not shadows, and eclipses are

Half the Moon is lit at every instant of every month. The phase is which part of the lit half faces the Earth — and confusing that with a shadow is the commonest error in astronomy.

sky · Phases and eclipses Three periods that nearly share a multiple. How far the draconic and anomalistic months are from a whole number, after a whole number of synodic months, in hours. At 223 synodic months — 6585.321 days — both are within an hour of closing, which is what makes an eclipse repeat. The draconic residual is 0.87 hours and the anomalistic 5.19 hours.

The eclipse that repeats a third of a world away

Three lunar periods nearly share a multiple after 6,585 days. The word "nearly" is what makes eclipses predictable, and the leftover third of a day is what moves each repeat a third of the way round the Earth.

sky · Phases and eclipses 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 Two equilibria become four, and three worlds sit near the join. The Cassini equilibria of a spin axis, drawn against the ratio of its own precession rate to the rate at which its orbit plane turns, for an orbit inclination of 1.5 degrees. Each column of dots is the full set of obliquities at which the two precessions keep step at that ratio, found by root-finding rather than by tracing a remembered curve. Below α cos ε/|g| = 1.135 there are two such obliquities and above it there are four, and the figure checks both counts on either side of the join. The three marked bodies are placed by their own measured precession constants: the Earth with the Moon at 2.67, safely on the four-state side; the Earth without it at 0.86; and Mars at 1.06. Two of the three sit within a few tenths of the bifurcation, which is the whole reason their obliquities are not constants: near the join the equilibria are close together, the libration around them is wide, and a body pushed between neighbouring resonances wanders. The Moon's contribution to the Earth's precession constant is what moves the first mark away from that region, and the second mark is the same planet with that contribution removed. This is a two-frequency model of a many-frequency system, and the real chaos comes from the overlap of resonances it does not contain.

A tilt that is not a constant

The Earth's axis leans by 23.4 degrees, and that lean is what makes the seasons. It is also a dynamical variable with its own equilibria, its own resonances and its own chaos — and on Mars the same variable has swung between nearly zero and sixty degrees without anything having to happen.

sky · Obliquity 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 How many planets a star has is the hardest thing a catalogue measures. The multiplicity distribution a transit catalogue would contain, for systems that all truly hold 5 planets, at four mutual inclination dispersions. 40,000 systems are drawn per dispersion with an isotropic viewing direction and Rayleigh-distributed inclinations about a common plane, at semi-major axes of 12, 16, 21, 27, 34 stellar radii; the bars are conditioned on at least one planet transiting, which is what makes a system appear in a catalogue at all. At 0.5° of dispersion 33 per cent of the detected systems show all 5 planets and the mean apparent multiplicity is 3.13; at 10° it is 1.39, with 68 per cent of them showing exactly one. Every one of those systems has 5 planets. The entire difference between a catalogue of singles and a catalogue of compact multiples is one number that nothing in the light curve measures. And the two effects run in opposite directions: the fraction of stars showing any planet RISES with the dispersion — 8%, 9%, 12%, 19% across the four — because scattering the orbits gives more of them a chance to cross the line of sight, while the number seen per detected star falls by a factor of 2.2. A survey that scatters its systems finds more stars with planets and fewer planets per star, and neither number on its own says which has happened. What no figure here can show is the true dispersion, because the observable is the ratio of those two and a system with fewer planets and a tighter plane reproduces it exactly.

How many planets a star has is not a measurement

Draw five thousand identical five-planet systems, scatter their orbital planes by half a degree, and a third of the detections show all five. Scatter them by ten degrees and two thirds show exactly one. Every system has five.

exoplanets · Detection bias

Named alongside it

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

Circular velocityΔvPlane changeApoapsisAscending nodeEccentricityGeostationary orbitHohmann transferLaunch azimuthVis-vivaAnomalyArgument of periapsis

All concepts