Astronomy & Celestial Mechanics
the sky, computed
Concept

Angular momentum — where it appears

The conserved quantity of motion under a central force, equal to the rate at which the radius vector sweeps area, times twice the mass. It is why every orbit lies in a plane, and its conservation is what makes an accretion disc's central problem the disposal of it rather than the loss of energy.

Named by 55 essays across 8 fields — each of them below, with the objects they name alongside it.

Equal areas in equal times, at eccentricity 0.65. Positions of an orbiting body at equal intervals of time, obtained by solving Kepler's equation. The two shaded sectors span the same interval and enclose the same area — a long thin one at the far end, a short fat one at the close approach.

Equal areas in equal times, which is angular momentum in disguise

Kepler's second law is a statement about the area a radius line sweeps. It looks like an odd thing to have noticed, and it turns out to be a conservation law arriving eighty years early.

orbits · Angular momentum Every orbit one force allows. Circle, ellipse, parabola and hyperbola, all sharing a focus and a closest approach. The eccentricity alone decides which one a body is on, and whether it returns.

Every orbit one force allows, and the number that picks between them

Circle, ellipse, parabola, hyperbola. A single inverse-square force permits exactly these four, and one number decides which — including whether the body ever comes back.

orbits · Conic sections The energy budget of an orbit at e = 0.7. Kinetic, potential and total energy per unit mass against distance from the primary, in units where GM = 1. The total is a horizontal line — it depends only on the semi-major axis — and the vis-viva relation is that statement solved for the speed.

One equation for the speed anywhere, and the eccentricity is not in it

The vis-viva relation gives the speed at any point of any orbit from two numbers. What it leaves out is the surprise — the shape of the orbit does not appear at all.

orbits · Vis-viva The effective potential, for three angular momenta. The radial motion of an orbiting body is one-dimensional motion in an effective potential: the attraction −GM/r plus the centrifugal term L²/2r² that the angular momentum contributes. The barrier at small radius is what stops a body with any angular momentum at all from reaching the centre, and the bottom of each well is the circular orbit.

The wall that angular momentum builds

A body falling toward a star almost never arrives. Sideways motion, which looks like a detail of the initial conditions, turns the attraction into a well with a wall around the middle of it — and the wall is why it costs more fuel to hit the Sun than to leave the solar system.

orbits · Effective potential Four averages of one distance, and the two of them that are the semi-major axis. The average distance of a body from its primary, against eccentricity and in units of the semi-major axis, computed four ways: averaged over time, over true anomaly, over eccentric anomaly, and as the harmonic mean in time. Every curve is a quadrature over the orbit — 2,048 panels uniform in eccentric anomaly, with Kepler's equation supplying the time weight — and not a closed form. Two of the four are exactly a at every eccentricity, which is why they are drawn as one line: the eccentric-anomaly average, because the mean of cos E over a turn is zero, and the harmonic mean in time, because the time weight cancels 1/r at every node before the sum begins. The other two are not: the time average is a(1 + e²/2), which rises to 1.4050 a at e = 0.9, and the true-anomaly average is a√(1−e²) — the semi-minor axis — which falls to 0.4359 a there. So a is the average distance in two senses out of four, and the ordering b ≤ a ≤ ⟨r⟩ₜ holds at every eccentricity with equality only on the circle. At Earth's e = 0.0167 the four agree to 0.014%, and at Mercury's e = 0.2056 the spread is 2.14%. The distinction is invisible for the planets and unavoidable for a comet, and it is the reason a quoted "mean distance" has to say which mean.

The average depends on what is being averaged

Four ways of averaging one orbit's distance from its primary give four different numbers, and only two of them are the semi-major axis. Which two is not a matter of convention, and the same arithmetic decides how much sunlight a planet receives in a year.

orbits · Orbital averages The tidal field is a difference. The pull of a distant body at each point of a sphere, minus its pull at the sphere's centre. What remains stretches along the line to the source and squeezes across it — two bulges, not one.

The tide is a difference, which is why there are two of them

The Moon pulls the ocean toward it. That explains one bulge. The second one, on the far side, is the whole of the physics — and it comes from subtracting.

gravitation · Tides The five Lagrange points at mass fraction 0.12. The five points at which a small body can keep station with two larger ones. The three on the line of centres are roots of a quintic and are unstable; the two forming equilateral triangles are stable for a sufficiently lopsided mass ratio.

Five places that keep station, in a problem with no solution

Three bodies under gravity cannot be solved. Restrict the problem slightly and five exact answers fall out anyway — three of them roots of a quintic, two of them perfect equilateral triangles.

gravitation · Lagrange points The two-body problem, and the one-body problem it is. Left: two bodies with mass ratio 0.4 on ellipses of eccentricity 0.5 about their common barycentre, the heavier one on the smaller orbit. Right: the same system as one body of the reduced mass on a single ellipse of the same eccentricity about a fixed centre, at the separation of the two. The right-hand curve is the point-by-point difference of the two left-hand curves, so the substitution is drawn, not merely stated.

Two bodies replaced by one that does not exist

The two-body problem is solved by turning it into a one-body problem about a fixed centre. The substitution is not an approximation — it is exact, and the body it invents has a mass no object in the system has.

gravitation · The two-body problem Three bodies, integrated. Three equal masses integrated forward under mutual gravity: the figure-eight choreography — all three bodies on one closed curve. Every point is a step of the equations of motion, and the total energy is conserved to 2.5e-11 across the run.

Three bodies, and what "no solution" actually means

The three-body problem is routinely called unsolvable. Trajectories are computed for it every day, exact periodic solutions are known, and both statements are true — the word is doing more work than it looks.

gravitation · The three-body problem Integrating the measured recession back: the Moon reaches the Earth 1.54 Gyr ago. The Earth–Moon separation and the length of the Earth's day, integrated backwards from the measured present recession rate of 3.83 cm per year. Constant-Q tidal friction makes a^(13/2) linear in time, so the history is a single line in a variable nobody plots, and it is calibrated to the laser-ranging measurement rather than to a modelled k₂/Q — the k₂/Q it implies is 0.0257, or Q = 11.6 for the Earth's k₂ of 0.299, which is a startlingly dissipative Earth. Run back at that rate the separation reaches zero 1.54 Gyr ago and crosses the Roche limit at 2.88 Earth radii only 4 years before it, so the drawing is cut off there rather than extrapolated. The Moon is 4.5 Gyr old, so this is a refutation and not a date: the present rate cannot have been the rate, and a mean Q of 34 — drawn dashed, reaching 4.51 Gyr — is the sort of value the age requires. Tidal rhythmites at 620 Myr put the day at 21.9 h and the Moon at 96.5 per cent of its present distance, and this history reads 20.1 h and 92.4 per cent — too fast and too close, which is the same failure the zero crossing is. Day length follows from total angular momentum, 23.93 h today, 9.84 h at half the present lunar distance and 4.97 h at the Roche limit, and depends on the separation alone: it is the same curve whatever Q is. The rate of lengthening the recession requires is 2.10 ms per century, against a tidal total of about 2.3 including the Sun's tide, which slows the Earth without moving the Moon, and an observed 1.75 from ancient eclipses and occultations — the shortfall being the Earth's moment of inertia falling as the mantle rebounds from the last glaciation.

A day five hours long

The tidal bulge leads, so the Earth's spin is being paid into the Moon's orbit. Run the measured payment backwards and two curves come out of one integration — a timeline that is refuted by the Moon's own age, and a day length that is refuted by nothing.

gravitation · Tides The path of the celestial pole over 25,772 years. The circle the Earth's rotation axis traces among the stars, at a radius equal to the obliquity, with the bright stars that fall near it and the years at which each is closest. Polaris is the pole star for a few centuries either side of now, and nothing else on the circle is nearly as close.

The pole star has a shelf life, and the sky has a slow hand

The Earth's axis traces a circle among the stars once every 25,772 years. Polaris is at the pole now, was not four thousand years ago, and will not be in two thousand more.

sky · Precession Two responses to the same transfer, turning round at q = 0.79 and q = 1. What conservative mass transfer does to the orbit and to the lobe, plotted against the mass ratio of donor to accretor on a logarithmic axis. Both curves are logarithmic derivatives with respect to the donor's mass, so a positive value means the quantity shrinks as the donor loses mass and a negative one means it grows. The orbit's response is exactly twice the mass ratio less one, which follows from holding the total mass and the total angular momentum fixed and nothing else, and it crosses zero at equal masses: transfer from the heavier star draws the orbit in, transfer from the lighter one pushes it out. The lobe's response adds to that the change in the lobe's shape, and it crosses zero earlier, at a mass ratio of 0.788. Between those two crossings the orbit is still widening while the lobe is already closing. To the right of both, a donor that loses mass finds its lobe shrinking around it, which is the runaway the essay is about: the transfer narrows the valve it is flowing through.

The flow that narrows its own channel

Two stars close enough share a surface, and the point where that surface pinches is a valve. What comes through it changes the orbit, and the orbit changes the valve — with a sign that reverses at equal masses, which is why some binaries transfer quietly for a hundred million years and others tear themselves apart in a thousand.

stars · Mass transfer Accretion from a medium a body is moving through, against how fast it moves. The rate at which a gravitating body captures gas out of a medium it is ploughing through, divided by the rate it would capture at rest, against its Mach number, both axes logarithmic. The accretion radius is set by where the body's escape speed matches the speed of the gas relative to it, and that relative speed combines the sound speed with the motion in quadrature; the rate carries the square of that radius times the speed, which leaves one plus the square of the Mach number to the power minus three halves. Subsonic motion therefore costs almost nothing — the curve is flat below Mach one third — while supersonic motion costs the inverse cube of the speed, drawn here with a measured logarithmic slope of -2.99. The same focusing produces the wake and the drag: the body pulls a denser column behind it, that column pulls back, and the material closest to the axis is captured. Accretion and dynamical friction are not two processes but two accounts of one, which is why a body that grows by this mechanism is also being slowed by it.

The wake and the meal are one calculation

A body moving through gas gathers what passes inside the radius at which its escape speed matches the flow. That single radius sets both the drag it feels and the rate at which it grows, so accretion and friction are not two processes but two readings of one — and the rate falls as the inverse cube of the speed.

stars · Accretion An edge where two torques balance, 21 kilometres from the shepherd. Two torques on the edge of a ring, against distance from a shepherding moon, both axes logarithmic and both scaled by the same combination of surface density, radius and orbital rate so that only their shapes are being compared. The flat line is the viscous torque, which comes from collisions between ring particles and does not care how far away anything is; it is drawn for a kinematic viscosity of 12 square centimetres a second, within the range ring seismology gives. The falling line is the moon's, summed over the first-order resonances that crowd together as the gap narrows, which makes it an inverse cube. A flat curve and an inverse cube cross once, and the crossing is where an edge can sit: closer in the moon wins and pushes the material back, further out viscosity wins and the ring spreads. For Daphnis and the Keeler gap the balance lands 20.8 kilometres out against a measured half-width of 21, which is agreement to well inside the uncertainty on the viscosity — and it is the only handle anybody has on that viscosity, since the quantity being inferred is the collision rate among particles a metre across, a billion kilometres away.

An edge is a balance, not a boundary

A ring of particles spreads, because collisions move angular momentum outward. Something has to push back, and the something is a small moon whose torque falls as the inverse cube of the gap. A flat curve and an inverse cube cross once, and the crossing is the sharp edge — which is why an edge exists at all rather than a gradient.

gravitation · Planetary rings A line nothing spun up by accretion can lie above, and the millisecond pulsars beneath it. The period–period-derivative diagram with the spin-up line drawn on it. An accreting neutron star is torqued by the disc until its magnetosphere turns at the same rate as the material arriving there, which fixes an equilibrium period as a function of the magnetic field and the accretion rate. Eliminating the field between that relation and the dipole formula that every point in this diagram is already read through leaves a straight line of slope 1.33, drawn here for accretion at the Eddington rate — the fastest a star can be pushed. The 7 recycled pulsars all sit below it, which is what the figure is for: none of them was spun up faster than the limit allows, and their positions are a record of how much mass each one received rather than of how old it is. The young pulsars are in the opposite corner, above the line and to the right, spinning down from birth. The two populations are not two stages of one life. A star that reaches the bottom left has been fed by a companion for a hundred million years, which is why almost every millisecond pulsar has one and almost no young pulsar does.

A corner of the diagram that has to be earned

A pulsar spinning a thousand times a second cannot have been born that way and stayed that way, because its own radiation would have slowed it in a few million years. It got there by being fed, and the line it cannot lie above is where the accretion torque balances the magnetic one.

stars · Pulsars A gravity assist with a 68° turn. The velocity triangle of a flyby. In the planet's frame the spacecraft's speed is unchanged and only its direction turns; adding the planet's own velocity converts that turn into a gain in speed measured from the Sun.

The planet pays, and it shows

A gravity assist takes energy from a planet and gives it to a spacecraft, and the planet's loss is exactly the spacecraft's gain. For a two-tonne probe past Jupiter that loss is unmeasurable. Do it with a hundred Earth masses of icy debris and the same bookkeeping moves Neptune outward by several astronomical units.

spaceflight · Gravity assist An edge where two torques balance, 21 kilometres from the shepherd. Two torques on the edge of a ring, against distance from a shepherding moon, both axes logarithmic and both scaled by the same combination of surface density, radius and orbital rate so that only their shapes are being compared. The flat line is the viscous torque, which comes from collisions between ring particles and does not care how far away anything is; it is drawn for a kinematic viscosity of 12 square centimetres a second, within the range ring seismology gives. The falling line is the moon's, summed over the first-order resonances that crowd together as the gap narrows, which makes it an inverse cube. A flat curve and an inverse cube cross once, and the crossing is where an edge can sit: closer in the moon wins and pushes the material back, further out viscosity wins and the ring spreads. For Daphnis and the Keeler gap the balance lands 20.8 kilometres out against a measured half-width of 21, which is agreement to well inside the uncertainty on the viscosity — and it is the only handle anybody has on that viscosity, since the quantity being inferred is the collision rate among particles a metre across, a billion kilometres away.

A torque that nearly cancels

A planet embedded in a gas disc pulls on the material inside its orbit and outside it, and the two torques are almost equal and opposite. What survives the subtraction is a per cent of either, and it is still enough to carry a planet from where it formed to its star in less time than the disc lasts.

exoplanets · Planet migration The 2.3-hour spin barrier, and the small bodies that are allowed through it. Rotation period against diameter for a synthetic asteroid population, both axes logarithmic, with the horizontal lines marking where a body held together by nothing but its own gravity would fly apart. That limit is P = √(3π/Gρ) and it contains only the density: 3.30 hours at 1 gram per cubic centimetre, 2.33 hours at 2 gram per cubic centimetre, 1.91 hours at 3 gram per cubic centimetre. Size does not appear in it, which is what makes the figure's shape informative rather than obvious — a barrier that depended on size would be drawn as a slope, and a horizontal line crossing five decades of diameter is a much stronger statement. The observed population respects it. Above about two hundred metres nothing rotates faster than the 2.3-hour line, and the crowding just below that line is real: bodies pile up against a limit they cannot cross. Below two hundred metres the wall stops applying and the fast rotators appear, some of them turning in minutes. Nothing about gravity changes at that size. What changes is that a body small enough is a single coherent rock with tensile strength, while a body large enough is a pile of fragments with almost none, and the barrier is a measurement of which is which. The picture is a synthetic population drawn at random from a fixed seed rather than a catalogue, so the individual points are not asteroids; what is real is the barrier, its value, and the fact that only the smallest bodies are found beyond it.

A wall with no size in it

Spin a body held together by nothing but its own gravity, and past a certain rate it comes apart. The rate depends on density alone — 2.3 hours for the stuff asteroids are made of — and the observed population respects the limit exactly, for every body larger than a couple of hundred metres.

orbits · Rubble piles A lever 12 radii long, and the spin it removes. A magnetised stellar wind, drawn with the Alfvén surface at 12 stellar radii — the schematic distance at which the wind's inertia finally beats the field. Inside it the gas is forced to turn with the star, so every gram that leaves carries the specific angular momentum of the radius at which it broke free rather than of the surface it came from, and the lever squares: J̇ = (2/3) Ṁ Ω r_A². Beyond the surface the streamlines curve backwards, because angular momentum conservation makes the azimuthal speed fall as 1/r while the radial speed does not. With a moment of inertia coefficient of 0.073 and a mass loss of 2.3·10⁻¹⁴ solar masses a year, the star loses a fraction 2.3·10⁻¹⁴ of its mass and a fraction 3·10⁻¹¹ of its angular momentum in the same year — a ratio of 1,315, which is (2/3)(r_A/R)²/k² and nothing else. The e-folding time for the spin is 3.3·10¹⁰ years against 4.3·10¹³ years for the mass. Nothing here is to scale in one respect that matters: the wind's density falls by more than ten orders of magnitude across the drawn region, so the streamlines are drawn as though the flow were visible when almost none of it is.

A wind that takes no mass and all the spin

The Sun loses about a ten-thousandth of itself to its own wind over its whole life. It loses most of its rotation to the same wind, and the whole of that asymmetry is one geometric factor — the gas is forced to keep turning with the star until it is a dozen or more radii out from the surface it left.

stars · Magnetic braking An oscillation that does not matter, on a ramp that does. Stored angular momentum in a reaction wheel over 160 days at 550 kilometres, with a capacity of 25 newton metre seconds. The total environmental torque is 1.75e-4 newton metres, of which 35 per cent is taken to survive averaging over an orbit. The fast oscillation is the part that does not survive: it has the 95.6-minute orbital period, reaches 0.10 newton metre seconds, and returns to where it started every revolution, so it consumes capacity and nothing else. The ramp under it is the secular part, and its slope measured between two instants a whole number of orbits apart is 6.117e-5 newton metres, which is the secular torque and is how the figure checks itself. The wheel fills in 4.7 days and has to be emptied 33 times in the span drawn. Every attitude-controlled spacecraft in the collection lives on this sawtooth, and the vertical drops are the only part of it that costs anything: the store can be moved between wheels for nothing, and taken out of the vehicle only by pushing against something outside it.

The spin that has to be put somewhere

A spacecraft holding an attitude is not resisting a force. It is absorbing a slow, one-directional trickle of angular momentum from the gradient of gravity across its own body, from sunlight, from the last of the atmosphere — and every store it has for that trickle fills up.

spaceflight · Attitude control A halo is born with a spin of a few hundredths. The distribution of the dimensionless spin parameter λ = J|E|^½ ÷ (G M⁵ᐟ²) across dark matter haloes, drawn as a lognormal of median 0.035 and logarithmic width 0.5, with the disc scale length each λ implies printed along the lower axis. The distribution is required to integrate to one and to peak at 0.0272, which is the median times e raised to minus sigma squared, and is the signature of a lognormal rather than of a bell curve drawn to look like one. λ is small because a halo is supported by random motion rather than by rotation: a value of 0.035 means the halo turns at about three and a half per cent of the rate it would need to hold itself up centrifugally. It is also nearly independent of halo mass, which is what points at a common origin. Mapping it to a disc through R_d = λ R₂₀₀/√2 with all of the specific angular momentum retained, a 10¹²-solar-mass halo of radius 206 kiloparsecs gives 5.1 kiloparsecs at the median, against the 2.6 kiloparsecs the Milky Way's disc actually has. The gap is not a failure of the estimate; it is the measurement that the baryons arrived with less spin per unit mass than the halo they arrived in.

A disc the size its halo was born with

A galaxy's mass says how much light it makes. It does not say how big it is. What sets a disc's size is a single dimensionless number describing how fast the dark halo around it happens to be turning — a number the disc had no part in choosing, distributed the same way for every halo mass in the universe.

galaxies · Galaxy spin From six gravitational radii to one. The radius of the innermost stable circular orbit against the dimensionless spin a = Jc/GM², in units of GM/c², for orbits prograde and retrograde with the hole's rotation. Both curves are the Bardeen–Press–Teukolsky expression and are checked at the three places it has exact values: 6 at zero spin, and 1 and 9 at the extremal limit. The separation is the observable consequence of frame dragging — space near the hole is itself circulating, so an orbit going the same way can stay closer before it becomes unstable, and one going the other way cannot come as close as a non-rotating hole allows. The prograde branch is required to fall and the retrograde branch to rise at every step drawn, which is a claim about the direction of the effect rather than about its size. The marked spin of 0.998 is not the extremal value but the equilibrium a hole fed by a thin disc actually reaches, because photons emitted by the disc are preferentially captured on retrograde orbits and spin the hole down again. Nothing here depends on what the hole is made of: two numbers fix the whole geometry, and this figure is the first of them holding still while the second moves.

The second number a black hole has

A black hole in equilibrium is described by its mass and its spin, and nothing else. The mass decides how strongly it pulls. The spin decides how much light a kilogram of infalling matter can emit before it disappears — and between the two extremes that figure changes by a factor of seven.

gravitation · Black hole spin A torque that stops when the patch lets go. Left, the mechanism: a protogalactic patch drawn as an ellipsoid of axis ratios 1:0.72:0.5, with the principal axes of the surrounding tidal field drawn across it at 30 degrees to its own. The torque is proportional to the difference of the patch's principal moments times the sine of twice that angle, so it is exactly zero when the two sets of axes agree — checked at both alignments — and largest at forty-five degrees. A spherical patch takes no torque whatever the field around it does, which is why the spin of every galaxy begins as a statement about its shape. Right, the angular momentum against time in units of the turnaround time: in linear theory the torque acts on a patch still expanding with the universe and the angular momentum grows as the first power of time, measured off the drawn curve as t^1.000. At turnaround the patch detaches from the expansion, its quadrupole shrinks, and the torque switches off — so a galaxy's spin is fixed before it has collapsed at all, by neighbours it will never interact with again. What the figure cannot show is the sign: the same mechanism gives no preferred direction, and the observed near-absence of alignment between neighbouring galaxies' spins is the check on that.

Spin acquired before there was anything to spin

Every galaxy turns, and nothing in a smooth expanding universe turns. The rotation was applied while the material was still a mildly overdense patch spread across megaparsecs — torqued by the tidal field of its neighbours, growing steadily with time, and switching off the moment the patch stopped expanding.

cosmology · Tidal torque theory A period below which every orbit is round. Orbital eccentricity against period for binaries in four clusters of 0.125, 0.625, 6, 4 billion years, with the eccentricities drawn from one seeded distribution and then damped by exp(−age/τ), where τ rises as the 5.333 power of the period. Each cluster shows the same thing: below a boundary period nothing survives eccentric, above it the original distribution is untouched, and there is almost nothing in between because the timescale is so steep. The boundary is a clock. It moves as the three-sixteenths power of the age, which the figure checks against the drawn curves, and the calibration puts it at 6.5 days at 125 million years, 8.8 at 625 million and 12.5 at four billion — against measured cut-offs near 7.2, 8.5 and 12.5 days in the Pleiades, the Hyades and M67. The boundaries are also read back off the plotted points rather than trusted, and required to move outward with age. This is the cleanest measurement of tidal dissipation in ordinary stars that exists, and its cleanliness comes from the ages: a cluster's age is read off its main-sequence turn-off and owes nothing whatever to the tide being measured.

A cut-off period that is an age

Plot eccentricity against orbital period for the binary stars of one cluster and the picture has a wall in it. Below a certain period every orbit is circular; above it, the original spread survives untouched. The wall moves outward as the cluster ages, and where it stands is a measurement of how stars dissipate a tide.

orbits · Circularisation Who holds the mass, and who holds the spin. The solar system's two ledgers on one logarithmic axis, each row a body or a group of them, with the pale bar its share of the mass and the dark bar its share of the angular momentum. Both columns are computed rather than quoted: the Sun's spin from 0.07 M R² Ω at a 25.38-day rotation, each planet's orbit from M √(GM☉ a (1 − e²)) with its own semi-major axis and eccentricity, and both sums are required to close to one part in a billion. The Sun holds 99.866 per cent of the mass and 0.61 per cent of the angular momentum. Jupiter holds 0.095 per cent of the mass and 61.1 per cent of the angular momentum, so a body a thousandth of the system by weight carries most of its rotation. The four inner planets together account for 0.0016 of it. A cloud collapsing to make this system had to move nearly all of its spin outward onto a small fraction of its mass, and the ledger is what that operation looks like when it is finished. What the figure cannot show is where the transfer happened, because everything that carried it away has either fallen in or left.

Ninety-nine per cent of the mass and none of the spin

The Sun holds 99.87 per cent of the solar system's mass and 0.6 per cent of its angular momentum. Jupiter holds a thousandth of the mass and three-fifths of the spin. That is not a curiosity of accounting — it is the record of the single operation that had to succeed before a star could form at all.

orbits · Angular momentum Every pair arrives circular. Eccentricity against gravitational-wave frequency for four binaries of 30 and 30 solar masses, each starting at 0.2 astronomical units with an eccentricity of 0.3, 0.7, 0.9, 0.99. Both axes are logarithmic and the tracks run left to right as the orbit shrinks. The curves are Peters' closed solution a(e), and they are checked against Peters' differential equation at three eccentricities on each track rather than against the integral they came from. The ordering is preserved — a pair that starts rounder stays rounder — but by the time the orbit is radiating at 10 hertz, where a ground-based detector begins to hear it, the four eccentricities are 8.8·10⁻⁸, 5.6·10⁻⁷, 3.6·10⁻⁶, 1.5·10⁻⁴. All four are far below anything a detector could measure. That is the figure's whole content and it is a strong statement: radiation reaction removes angular momentum faster, relative to energy, than a circular orbit would need, so eccentricity is destroyed on the way in. A binary observed to be eccentric in band therefore cannot have spent long shrinking quietly, and must have been put on that orbit recently — by a third body, or in the crowded centre of a cluster. The figure assumes the two bodies are points and nothing else acts on them, which is exactly the assumption an eccentric detection would refute.

Every pair arrives circular

Gravitational radiation drains a binary's energy and its angular momentum at rates that do not keep step, and the mismatch destroys eccentricity far faster than it shrinks the orbit. A pair that starts at 0.99 and spirals in from a fifth of an astronomical unit is round to better than a part in a hundred million by the time a detector can hear it.

gravitation · Gravitational waves Two measured numbers, and everything else on the page derived from them. 25 pulsars in the plane of period against period derivative, at their catalogued values. Only the two axes are measurements; the three families of contour are models. Constant surface field runs at slope −1 because B ∝ √(PṖ), constant characteristic age at slope +1 because τ = P/2Ṗ, and the two families cross the population at right angles — which is why a single dot fixes both. The Crab sits at 3.8·10¹² G and 1257 years, and its true age is 972; the millisecond pulsars at the lower left have fields ten thousand times weaker and characteristic ages of billions of years, because they were spun back up by a companion long after they died. The line at the lower right is the death line, B/P² below which the model says no pair production and therefore no radio emission — and J2144−3933 is drawn below it, an 8.5-second pulsar that is radiating anyway.

A neutron star born turning too slowly

Collapse an iron core a few thousand kilometres across down to twelve, and conservation of angular momentum multiplies its rotation rate by about eighteen thousand. A model with no transport in it delivers a newborn pulsar at the break-up limit; the ones that are observed turn twenty times slower, which is a measurement of the core before it fell.

stars · Supernovae An arm that moves angular momentum outwards, and is spent doing it. Where a 2-armed spiral pattern of 25 km/s per kiloparsec takes angular momentum from the disc and where it gives it back, against galactocentric radius, for the Milky Way's rotation curve. The resonances are found on that curve rather than placed: the inner Lindblad resonance at 1.91 kiloparsecs, corotation at 8.67, the outer Lindblad resonance at 13.80. The lower curve is the angular momentum removed per unit radius and the upper one what the wave delivers; between them runs the flux the wave carries, which is flat across the whole region where nothing is resonant, because a wave that is not interacting with anything simply travels. The two deposits are required to cancel to a part in a million — the pattern is a conveyor and keeps nothing — so the disc's total angular momentum is unchanged while its distribution is not. That is the sense in which a spiral is a machine: it moves mass inwards by moving angular momentum outwards, which is the same operation an accretion disc performs and the same one a protostellar disc must perform to make a star. The cost is paid in random motion. Every exchange heats the stellar disc, a hotter disc supports a weaker wave, and the pattern that did the work is the thing the work destroys. What the figure cannot show is the pattern's own lifetime, which is the unsettled part of the subject.

An arm that is undone by the work it does

A spiral pattern takes angular momentum from the inner disc and delivers it to the outer, which lets mass move inwards without violating anything. It is paid for in random motion, and a disc with too much random motion cannot carry a wave — so the pattern destroys the conditions it needs to exist.

galaxies · Spiral structure The last parsec, priced in stars. The two timescales that shrink a pair of 10⁸-solar-mass black holes at the centre of a merged galaxy, against their separation in parsecs, both axes logarithmic, for a stellar velocity dispersion of 200 km/s and a central density of 500 solar masses per cubic parsec. Ejecting stars hardens the binary at a rate proportional to the separation, so that timescale grows as the orbit shrinks; gravitational radiation goes as the fourth power of the separation, so its timescale collapses. Neither alone finishes the job and the crossing of the two is where the answer is set. With the loss cone kept full, the crossing is at 2.9e-2 parsecs and the total is 4.2·10⁸ years, inside a Hubble time. With the supply of low-angular-momentum orbits emptied by a factor of 100 — which is what happens in a smooth spherical nucleus, because the stars that could interact have already been thrown out and two-body relaxation refills the orbits far too slowly — the crossing moves outward to 7.3e-2 parsecs and the total becomes 1.7·10¹⁰ years. That is the final-parsec problem, and it is not a problem about gravity: it is a problem about supply. Real nuclei are not spherical, and the figure cannot show what a triaxial potential does, which is to keep feeding the binary orbits it has not already used.

The last parsec, and the stars that are not there

Two galaxies merge and their central black holes sink towards each other, and then stop. From about a parsec apart, friction no longer works and gravitational radiation is not yet strong enough — and the only mechanism in between throws away the stars it depends on faster than they can be replaced.

galaxies · Galactic nuclei A wobble that should have stopped seventy years ago. Left, the path of the Earth's rotation pole across its own crust over 13 years, as the sum of two circular motions: the 433-day Chandler wobble at 150 milliarcseconds and the annual wobble at 90. The spiral is a beat, and its period measured off the drawn path is 6.39 years against the 6.39 the two frequencies require. Right, the same path's radius against time. Two numbers in this figure are the argument. The first is the Chandler period itself: a rigid Earth of dynamical ellipticity 0.0032737 would wobble freely at 305 days, and the observed 433 is 42 per cent longer because the Earth deforms under its own wobble and the oceans move with it — the period is a measurement of the planet's elasticity, made by watching a free motion rather than by forcing anything. The second is the damping: at a quality factor of about 100 the wobble should decay in 38 years, and it has been running for as long as anyone has watched. Something is exciting it continuously, and the excitation is fluctuating pressure at the bottom of the ocean and in the atmosphere. What the figure cannot show is the excitation itself, which is not periodic and is only visible statistically.

A wobble that should have stopped

The Earth's rotation pole wanders across its own crust in a circle a few metres wide. A rigid Earth would do it in 305 days; it takes 433, and the difference is a measurement of the planet's elasticity. At the observed damping it should have died out within a human lifetime, and it has not.

sky · Precession A hundredfold expansion is a four-order slowdown. Equatorial rotation speed against radius for three stars leaving the main sequence at 2, 10, 100 km/s, on the single assumption that nothing exerts a torque. Both axes are logarithmic. The specific angular momentum is held fixed along each track and checked at four radii rather than assumed, so the speed falls as the reciprocal of the radius and then faster, as the moment of inertia coefficient slides from 0.073 on the main sequence to 0.02 in a centrally condensed giant envelope. A star at 100 km/s crosses the 8 km/s line — the usual boundary for calling a giant a rapid rotator — at 45.6 solar radii, and everything larger is slower. That is why the observed giants are almost all under two kilometres a second, and why the one or two per cent that are not cannot be explained by anything the star did on its own: the angular momentum has to have been delivered, by a swallowed companion or a merger. The figure assumes no mass loss, which for the largest radii drawn is the weakest of its assumptions.

A surface that slowed because the star grew

A star leaving the main sequence expands by a factor of a hundred. Conserve its angular momentum and its equator slows by four orders of magnitude, which is why every red giant is a slow rotator. One or two per cent are not, and a giant turning at fifty kilometres a second has to have been given the angular momentum by something else.

stars · Stellar evolution A vanishing field that changes the answer completely. The growth rate of the magnetorotational instability against wavenumber, both in units the orbital frequency sets: the vertical axis is the growth rate divided by Ω, the horizontal is the wavenumber multiplied by the Alfvén speed and divided by Ω. The maximum is exactly three quarters of the orbital frequency and it does not depend on the field strength at all — it occurs at kv_A = √15Ω/4, so a weaker field simply moves the fastest-growing wavelength to a longer one. The curve vanishes above kv_A = √3Ω, which is the only thing the field decides: modes shorter than that are stabilised by magnetic tension. An unmagnetised Keplerian disc is Rayleigh-stable and would never accrete; a disc with a field a millionth of the strength needed to matter dynamically grows a mode that doubles in about a fifth of an orbit. The limit is discontinuous, which is the reason this was found late and by algebra rather than early and by observation.

The weakest field changes the answer

A Keplerian disc is stable by every hydrodynamic test there is. Add a magnetic field of any strength whatever — a millionth of what would matter dynamically — and it becomes violently unstable, at a growth rate of three quarters of an orbit per radian that does not depend on the field at all.

gravitation · Magnetorotational instability Where the disc stops, and the spin that follows from it. Two radii against accretion rate, for a neutron star with a 10⁸-gauss field. The falling curve is the magnetospheric radius, where the field's stress on the disc matches the rate at which the flow carries angular momentum inward; it goes as the accretion rate to the minus two sevenths, which is a weak enough dependence that the factor of 1000 in supply drawn here moves the boundary by a factor of 7.2. The horizontal lines are corotation radii for three spin periods — the radius at which the disc orbits as fast as the star turns. Above corotation the field is spinning the gas faster than it wants to go and flings it out; below, the gas is faster and spins the star up. So the crossing is an attractor, and a star accreting steadily walks to the period where the two coincide. That period goes as the field to the six sevenths and the rate to the minus three sevenths — which is why a neutron star that has swallowed a tenth of a solar mass from a companion comes out at a few milliseconds, and why the millisecond pulsars have fields ten thousand times weaker than the young ones.

The period a star is pulled towards

A magnetised neutron star does not let a disc reach it. The disc stops where the field's stress wins, and if that radius lies inside the corotation radius the star is spun up while outside it the star is spun down — so there is one period at which nothing changes, and an accreting star walks to it.

gravitation · Accretion The path of a spin across the sphere of fixed momentum. A body with principal moments 1, 8, 8.6, started about its axis of least inertia with a 3° wobble and an internal energy sink strong enough that the whole motion fits in 80 turns and each circuit of the path can be seen. The disc is the near hemisphere of the sphere of fixed angular momentum, seen from a direction between all three body axes, with the near end of each axis marked. Each thin curve is a contour of kinetic energy on that sphere — a polhode, one of the paths the angular momentum can follow in the body with no dissipation — and the thick curve is the separatrix through the intermediate axis, which divides motions that circle the axis of least inertia from motions that circle the axis of greatest. The coloured path is what the integration did: solid on the near hemisphere, dashed where it passes behind. It starts at the open dot and ends at the filled one.

A spin that left the axis it was given

The first American satellite was spun about its long axis, like a rifle bullet, and soon after launch it was tumbling end over end. Nothing outside it had pushed. A body that cannot change its angular momentum but can lose energy has exactly one place to end up, and a long body spun about its length is as far from that place as a spin can be.

spaceflight · Attitude control A spin about the middle axis of a 1:2:3 body, turning over every 2.9 turns. A body with principal moments of inertia 1, 2, 3, spun about its intermediate axis with a hundredth of its angular momentum knocked onto the axis of least inertia, integrated with no dissipation and no external torque. The curves are the components of the angular momentum along the three body axes, as fractions of its fixed size. The intermediate component stays near one for 1.5 spin periods, then swings through zero to minus one — the body turns over, end for end — and keeps doing so every 2.9 periods, 14 times in the span drawn. Energy and angular momentum are both conserved throughout, the energy to better than one part in a billion; nothing is being lost and nothing drives the flips. A spin about the intermediate axis is an equilibrium like a pencil balanced on its point, and the smallest disturbance grows exponentially, here by a factor of e every 0.28 spin periods, until it carries the body to the opposite equilibrium and back.

A wingnut that turns over on its own

Spin a rigid body about the axis whose moment of inertia is neither the largest nor the smallest and it turns end over end, again and again, with nothing pushing it and nothing lost. The flip was noticed aboard a space station in 1985 and was already implicit in equations written in 1765. How long it waits is a logarithm, and no care in setting up the spin can make the logarithm infinite.

spaceflight · Attitude control A tumble removed with a coil and a compass, at 500 km. The rotation rate of a small spacecraft — principal moments 0.0067, 0.041, 0.043 kg m², the proportions of a three-unit cubesat — tumbling at 8.8° a second after release, against orbits at 500 km, with nothing to control it but magnetic coils driven by the B-dot law: a dipole opposite to the rate of change of the field measured aboard, capped at 0.2 A m². The field is a dipole tilted 9.2° from the Earth's axis and turning with the Earth. In a polar, 97.4°, orbit the rate settles at 0.12° a second over the last orbit drawn, passing 1° a second after 0.58 orbits; in a 51.6° orbit the rate settles at 0.12° a second over the last orbit drawn, passing 1° a second after 0.32 orbits; in an equatorial orbit the rate settles at 2.44° a second over the last orbit drawn. The law needs no knowledge of the spacecraft's attitude: a tumbling body sees the Earth's field swing round in its own frame, and a dipole opposing that swing produces a torque that removes the part of the spin perpendicular to the field. The polar orbit does not reach zero. It settles at 0.94 of twice the orbital rate, 0.13° a second, and twice the orbital rate is how fast the field direction itself turns round a polar orbit: a body turning with the field sees little change to oppose. An equatorial orbit keeps the field pointing nearly the same way all the way round, so the spin about it is reached only through the dipole's tilt and the Earth's turning, and 28 per cent of the starting rate is still there at the end.

A tumble stopped by the field it tumbles through

A small satellite leaves its deployer tumbling, and the first thing most of them do is stop, using nothing but a magnetometer and three coils. The law they run needs no idea where the satellite is pointing. What it cannot do, at any instant, is touch the spin about the local field line — so how much tumble survives is decided by how much the field's direction changes along the orbit, and the stillness it reaches is defined by the field rather than by the stars.

spaceflight · Attitude control The same well after the primary has lost 45 per cent of its mass. Two effective potentials for one body: the solid curve before the primary loses mass and the faint one after, both at the same angular momentum, because a central force of any strength exerts no torque. The well shallows and its floor moves out from r = 1.00 to 1.82. The body's own level moves with it — from E = -0.420 to -0.127 — and the two horizontal lines are drawn where the radial action is conserved, which puts the turning points at 0.71–1.67 before and 1.30–3.03 after. The ratio between them is 2.3333 in both, so the orbit is the same shape at a larger size: everything about the body's path has scaled and its eccentricity of 0.4 has not moved. That is what a slow change leaves behind, and it is not what a sudden one leaves.

The well moves, and the body keeps its share of it

When the Sun becomes a white dwarf it will throw away half its mass, and every planet's orbit will swell by the same factor. Their eccentricities will not change at all — provided the loss is slow, and the only meaning "slow" has here is slow compared with one orbital period.

orbits · Effective potential A rubble pile that splits below a mass ratio of 0.204 can lose its piece; above it, the piece stays. The total energy of two spherical components of equal density in contact, spinning together at the rate at which their mutual gravity just holds them against the spin, against the mass ratio of the smaller to the larger, in units of G m₁²/R₁. The energy is the kinetic energy of the rotating pair minus their mutual gravitational binding. When the smaller piece is a small fraction of the whole, the spin carries more energy than the binding and the total is positive: a body spun to breakup that sheds a fragment of that size has enough energy for the fragment to escape entirely, becoming a separate asteroid on a nearly identical orbit. The total changes sign at q = 0.204. Above that ratio the pair cannot separate without an energy source; it stays as a binary, orbiting and eventually synchronising, or re-accretes. As q goes to zero the energy tends to 0.2 G m₁²/R₁, the rotational energy of the primary alone at its breakup rate. Nothing in the threshold depends on the size or the density of the body — it is a pure number from the geometry of two touching spheres — and asteroid pairs sharing an orbit have been found overwhelmingly with estimated mass ratios below it.

A split that decides whether the piece can leave

A rubble pile spun past its limit splits in two, and whether the smaller piece escapes or stays in orbit is not a matter of luck. Two touching spheres spinning at their shared limit have positive total energy only when the smaller is less than 0.204 of the larger's mass — a number with no size and no density in it. Below it the pieces can become a pair of asteroids on nearly identical orbits; above it, a binary. And the larger the piece that leaves, the slower the body left behind.

orbits · Rubble piles A ceiling the outer orbit imposes. The greatest eccentricity a cycle can drive an inner orbit to, against the mutual inclination it started at, for four ratios of the inner orbit's angular momentum to the outer one's. The bottom curve is the test-particle case and it is the familiar closed form, √(1 − (5/3)cos²i₀), reaching one at ninety degrees — recovered here by a solver that knows nothing about it, which is the check. Every other curve is lower. The reason is conservation: the cycle works by moving angular momentum out of the inner orbit, and it has nowhere to put it but the outer one. A test particle can give up all of its angular momentum because the outer orbit is infinite by comparison; an inner binary carrying 1 times as much as its companion cannot, because tilting the outer orbit by the required amount changes the mutual inclination and shuts the cycle down. At ninety degrees the ceiling falls from 1.000 to 0.866. The pericentre a mechanism can deliver is bounded by what the perturber can absorb, and the test-particle formula is an upper bound rather than a prediction.

A ceiling the outer orbit imposes

The closed form for how far a cycle can drive an eccentricity assumes the inner orbit is a test particle, so the angular momentum it gives up goes nowhere. A real inner binary of comparable mass has to be paid for: the outer orbit absorbs what the inner one loses, tilts, and changes the very inclination the cycle depends on.

orbits · Kozai–Lidov A spin that prefers the axis in the middle. Where the spin of a tidally torqued patch points, measured against the three principal axes of the tidal field acting on it: the distribution of the absolute cosine between the spin and each axis, from 20,000 patches with random shapes in one field of principal values −0.75, 0.09, 0.66. Principal values are those of the second derivative of the potential, so a positive one compresses. An isotropic spin would give three flat lines at one. The spin avoids the most compressive and the most stretching axes and prefers the intermediate one, because the torque about each axis is proportional to the difference of the field's principal values about the other two, and the intermediate axis is the one whose two neighbours differ most: squared, those weights are 0.11, 0.66, 0.24 of the total. Fitted as Lee and Pen's alignment parameter, the directions give a = 0.617. The preference is real and it is weak — the densest bin is 3.37 times the isotropic value — which is the whole difficulty of using spins to map a field.

A spin that remembers the field and not its sign

Tidal torque theory says where a galaxy's spin should point, not only how large it should be. It points along the middle axis of the tidal field, it is exactly perpendicular to an ideal filament, and it is an even function of the field — so a map of spins can find the directions of the cosmic web and can never say which of them is collapsing.

cosmology · Tidal torque theory Where a disc pushes a planet outward. The sign of the total torque on a low-mass planet, as a function of the power-law indices of the disc's surface density (Σ ∝ r^−α, horizontal) and temperature (T ∝ r^−β, vertical), from the fitting formulae of Paardekooper and collaborators for an adiabatic disc with γ = 1.4 and the horseshoe torque unsaturated. Above the line the total is positive and the planet migrates outward; below it the planet migrates inward. The line is straight because every part of the torque is linear in the two indices, and it is steep in α because a shallow density profile strengthens the vortensity part and weakens nothing. The minimum-mass nebula (α = 1.50, β = 0.50) sits at a total of −3.76 in units of Γ₀/γ; The viscously heated inner disc (α = 0.60, β = 0.90) sits at a total of +0.74 in units of Γ₀/γ; The irradiated outer disc (α = 1.07, β = 0.43) sits at a total of −2.65 in units of Γ₀/γ. The minimum-mass nebula every textbook starts from is on the inward side by a wide margin, and a disc heated by its own accretion is on the outward side: the same planet can migrate either way depending on how the gas is heated, which the Lindblad torque alone could never say.

The gas that turns round beside the planet

The two spiral wakes a planet raises nearly cancel, and what decides the direction it drifts is a third, smaller torque from the gas sharing its orbit. That gas makes a U-turn in front of the planet and behind it, and whether the turn pushes the planet outward depends on how the disc was heated and on a viscosity nobody has measured.

exoplanets · Planet migration Circularised at twice the closest approach. Eccentricity against semi-major axis, measured in units of the planet's Roche distance 2.16 Rₚ(M⋆/Mₚ)^⅓, for planets delivered on orbits of eccentricity 0.97 with periastra of 1.2, 1.6, 2.2, 3 Roche distances. Tides raised on the planet dissipate orbital energy at periastron and exert almost no torque over the rest of the orbit, so the orbit's angular momentum is nearly conserved while its energy is removed: a(1 − e²) stays fixed and each track runs down and to the left along it. Each ends on a circle of radius q(1 + e₀), which for e₀ near one is twice the periastron — 1.2 → 2.36, 1.6 → 3.15, 2.2 → 4.33, 3 → 5.91. The shaded region above the curve is where the periastron is inside the Roche distance and the planet is stripped on its first close passage. So the channel cannot deliver a circular orbit inside twice the Roche distance: the inner edge of the hot Jupiters is a prediction, stated in units that depend on each planet's own radius and mass rather than on its period.

A planet circularised at twice the distance that would destroy it

A giant planet thrown onto an orbit of eccentricity 0.97 by a distant companion loses its orbital energy to tides at every close passage and keeps almost all of its angular momentum. The orbit therefore shrinks to a circle at exactly twice its closest approach — and since a closest approach inside the Roche distance destroys the planet, the hot Jupiters delivered this way should begin at twice the Roche distance, measured in units of each planet's own size.

exoplanets · Planet migration The separation of the Milky Way and Andromeda, run forward. The distance between the Milky Way and Andromeda against time from now, in billions of years, integrated from today's separation of 770 kpc and radial approach of 110 km/s, for four values of Andromeda's sideways (tangential) velocity: 0, 17, 57, 100 km/s. The haloes are extended Hernquist spheres of 1.15 and 1.5 trillion solar masses, and dynamical friction drains the orbit at each passage. At 0 km/s the first approach is so nearly head on that the centres pass within 10 kpc and the pair merges at 4.0 Gyr. At 17 km/s the first approach is so nearly head on that the centres pass within 10 kpc and the pair merges at 4.8 Gyr. At 57 km/s the first pericentre is 117 kpc at 4.5 Gyr, and the pair merges at 9.2 Gyr. At 100 km/s the first pericentre is 321 kpc at 4.8 Gyr, and the pair has not merged by 14 Gyr. The measured sideways speed runs from about 17 km/s (Hubble) to about 57 km/s (Gaia), and the outcomes across that range differ by billions of years.

A collision decided by a speed across the sky

Andromeda is coming toward the Milky Way at a hundred and ten kilometres a second, and that part is measured to a few per cent. Whether the two galaxies merge in the next ten billion years depends on how fast Andromeda is also moving sideways — a drift of fifteen millionths of an arcsecond a year, measured twice, with answers of seventeen and fifty-seven kilometres a second. One is a near head-on collision in under five billion years, the other a coin toss.

galaxies · Galaxy interactions What it costs to follow a turning grapple point. The velocity a chaser must spend each minute to hold station on a grapple point carried round by a flat spin, ω²r × 60 s, against the spin rate in degrees per second, for points 1, 3, 10 m from the spin axis, both axes logarithmic. The dashed line is what holding station 10 m from a target in a 400 km orbit costs in the same units, from the orbital dynamics alone: 2.3 mm/s per minute. At 1° per second, a point 3 m out costs 5 cm/s per minute — 24 times the orbital term; at 5° per second, 1.4 m/s per minute, 595 times. The cost rises as the square of the rate, so a target spinning twice as fast is four times as expensive to follow, and above a few tenths of a degree per second the orbit — the whole subject of the approach up to this point — has become the small correction.

A grapple point that moves faster than the orbit

Every rendezvous problem up to the last few metres is orbital mechanics: two bodies in nearly the same orbit, drifting apart and together under gravity. A derelict satellite tumbling at a few degrees a second changes the problem entirely. Holding station on a point of its surface costs more each minute than a week of orbital station-keeping, the moment to grab it comes and goes in seconds, and a free tumble never repeats — so the orbit becomes a small correction to a problem in rotation.

spaceflight · Rendezvous A fall into the Sun from a standing start. The distance of a body released at rest 1 AU from the Sun, as a fraction of its starting distance, against time as a fraction of the fall's total duration, 64.6 days: the exact radial orbit (solid) and a fall at the constant acceleration felt at the start (dashed). The exact fall begins exactly like the constant one and finishes far faster, because the pull grows as the inverse square of the shrinking distance: it reaches the Sun at 79 per cent of the time the constant pull would take. The curve is a cycloid, and the total time is a fifth of the circular orbital period at the starting distance — exactly 1/(4√2) of it — because a straight fall is half of an ellipse whose semi-major axis is half the starting distance.

A fall is half an orbit with no width

Stop a planet dead and it falls into the Sun, in a time that is not a quarter of its year or a half but exactly one over four times the square root of two. The fall is an ellipse squeezed to a line, it spends four-fifths of its duration in the outer half of its path, and in a uniform cloud every shell of it arrives at the centre at the same moment — which is where the free-fall time of star formation comes from.

orbits · Conic sections The plane a moon's orbit turns about, between equator and ecliptic. The inclination of the Laplace plane — the plane about which a satellite's orbit precesses — to the Earth's equator, against distance from the Earth in Earth radii, for an obliquity of 23.44°. Close in, the Earth's equatorial bulge dominates and the plane is the equator; far out, the Sun dominates and it is the ecliptic, 23.44° away. The dashed curve uses today's bulge, which makes the transition at 8.4 Earth radii. The solid curve lets the bulge grow with the faster spin the Earth had when the Moon was closer — a 5.3-hour day at 5 Earth radii, from the system's angular momentum alone — which moves the transition out to about 14. The Moon today, at 60.3 Earth radii, has a Laplace plane within 0.0022° of the ecliptic, so its 5.15° inclination is entirely free: nothing forces it.

A tilt the Moon should not have

A moon's orbit precesses about a plane that sits on its planet's equator close in and on the planet's own orbit far out. A Moon born in the Earth's equator and carried outward by the tides should have followed that plane all the way to the ecliptic, because the angle it keeps to it is an adiabatic invariant. It is inclined by five degrees instead, and slow recession cannot have put them there.

orbits · Angular momentum Four numbers that carry a position and a velocity round an orbit. The Lagrange coefficients f, g, ḟ and ġ — g and ḟ scaled by the mean motion so that all four are pure numbers — against time as a fraction of the period, for an ellipse of eccentricity 0.6 and periapsis 0.5 AU, each computed from one universal-variable solve starting at periapsis. The new position is f times the old position plus g times the old velocity, and the new velocity ḟ times the old position plus ġ times the old velocity: the state at any time is a linear combination of the state at the start, with no semi-major axis, eccentricity or anomaly formed. f and ġ return to one and g and ḟ to zero after a full period, which is the orbit closing. At every one of the 401 instants the combination f ġ − ḟ g equals one to 6·10⁻¹⁵, which is the conservation of angular momentum written as the area of a parallelogram.

Four numbers that carry a state without an orbit

Once the universal variable is solved, a body's position and velocity at any later time are a linear combination of its position and velocity now, with four coefficients and no orbital elements formed at all. The four obey one identity exactly — the conservation of angular momentum, written as an area — and its failure measures the error a calculation has made without knowing the answer.

orbits · Universal variables Five clocks around the black hole at the centre of the Galaxy. Timescales against distance from a black hole of 4.3·10⁶ solar masses at the centre of a cluster whose stellar mass inside r is 10⁶ (r/1 pc)^1.5 solar masses, made of stars of 1 solar mass. Ordinary two-body relaxation (orange) is slow everywhere, 8.4·10¹⁰ years at 0.01 pc. Orbits that close are nearly Keplerian and keep their orientation until they precess — through the stars' own mass farther out, through relativity close in — and over that coherence time the torques between them add rather than random-walk. That makes scalar resonant relaxation (blue) faster than two-body relaxation inside about 0.57 pc, and vector resonant relaxation (green), which turns orbital planes and is not stopped by precession within a plane, faster still: 3.5·10⁷ years at 0.1 pc, against the 6·10⁶-year age of the young stars there. Coefficients of order one are set to one; the slopes are the content.

Torques that add because the orbits hold still

Near a massive black hole every star's orbit is almost a fixed ellipse, repeated thousands of times before it precesses. The torques the other stars exert on it therefore point the same way orbit after orbit, and they add in proportion to time instead of cancelling in a random walk — so angular momentum near the hole relaxes hundreds of times faster than ordinary encounters allow, and orbital planes faster still.

gravitation · Two-body relaxation The pitch wheel, dumped into the field at four inclinations. Momentum stored in the pitch wheel of an Earth-pointing satellite at 500 km, over 3 days, under a steady pitch torque of 0.2 μN m, with coils of at most 0.2 A m² dumping it into the Earth's field under a law of gain 0.0005 per second. Without coils the store grows in a straight line to 51.8 mN m s. In orbits inclined at 0°, 28.5°, 51.6°, 97.4° it ends at 7.51, 1.18, 0.54, 0.40 mN m s. In the equatorial orbit the field lies close to the pitch axis, so only a small share of the stored momentum is across the field where the coils can remove it, and the store climbs until that share, times the gain, matches the torque.

The one axis the field will not take

A satellite that keeps one face towards the Earth can empty its reaction wheels into the Earth's magnetic field with coils, using no propellant. Its own rotation, once an orbit, cancels every steady torque except those about one axis, the pitch axis normal to the orbit. In an orbit near the equator the field lies along that same axis, and a coil cannot push about the direction of the field.

spaceflight · Attitude control In a mixed field the torque follows the dipole and ignores the rest. The torque of a field that is part dipole and part quadrupole, with the total surface strength fixed, as a fraction of a pure dipole's, against the fraction of the field strength in the dipole, for wind magnetisations of 300, 3,000, 30,000, using the broken power law fitted to simulated mixed-field winds. Above a dipole fraction of 0.39, 0.24, 0.15 respectively, the torque is exactly what the dipole component alone would give: the quadrupole has fallen below the dipole long before the wind reaches its Alfvén radius, and contributes nothing to the lever. Below it, the quadrupole sets a floor. A star whose surface is mostly quadrupole by strength can therefore be braked almost entirely by its small dipole — and a measurement of the total field says little about the torque unless it says how much of the field is dipolar.

A lever set by the dipole and not the field

A star's wind carries away its spin because the magnetic field holds the gas in step with the surface out to the Alfvén radius, and the torque goes as the square of that radius. How far out the field can hold depends less on how strong it is than on how it is arranged: a quadrupole of the same surface strength holds the Sun's wind to half the distance and brakes it at a fifth of the rate. In a mixed field only the dipole reaches far enough to matter — so the field that brakes a star is the part of it that is hardest to measure.

stars · Magnetic braking Errors the identity sees and errors it cannot. Propagations of an ellipse of eccentricity 0.6 through 0.37 of a period, spoiled four ways at sizes from 10⁻¹³ to a hundredth, plotted by their true relative position error across and by how far their coefficients miss f ġ − ḟ g = 1 up (both logarithmic; the bottom row is a miss below the arithmetic's resolution). A solve stopped early and coefficients jostled at random break the identity in proportion to the damage. A wrong time step and a wrong gravitational parameter do not break it at all, at any size — errors up to 0.009 of the position sit on the floor — because each produces an exact Kepler map, for a different question. One equation on four numbers leaves three directions unchecked, and two of them are the two inputs every propagation depends on.

A check that cannot see the clock

The four Lagrange coefficients that carry an orbit's state forward satisfy one identity exactly, and a propagation that breaks it is wrong by about as much as it breaks it. But one equation on four numbers leaves three directions unchecked, and two of them are the inputs every propagation depends on: the time and the strength of gravity. A state carried forward for the wrong time, or under the wrong gravitational parameter, satisfies the identity perfectly — and a solver can be written so that its own errors take exactly that form.

orbits · Universal variables The rhythms in the rock, set by the Moon's distance. The periods of the climatic precession cycles — the precession constant plus the perihelion frequency of each of the planets that dominate the Earth's eccentricity — and of the obliquity cycle, the precession constant plus the node frequency s₃, against the Earth–Moon distance. Today they are 23.8, 22.4, 19.0 thousand years for precession and 41 for obliquity. At 341,000 km they were 14.7, 14.2, 12.8 and 20.0. The planetary frequencies are held at today's values, which the chaos of the inner solar system makes uncertain beyond a few tens of millions of years; the precession constant is the term that moves.

A rhythm in the rock that measures the Moon

The Earth's axis precesses because the Sun and Moon pull on its equatorial bulge, and two-thirds of the pull is the Moon's. When the Moon was nearer and the day shorter, the axis turned faster, and the climate cycles that precession drives were shorter. Sediments laid down 1.4 billion years ago record those cycles — and with nothing in the rock but layer thicknesses, the ratio of two rhythms gives the Moon's distance at the time.

sky · Precession Precession cycles grow as the inverse of the lowest frequency heard. The number of times the orbital plane precesses between a detector's lowest frequency and the last stable orbit, against that lowest frequency, both on logarithmic scales, for pairs of 36 + 29 solar masses, 10 + 8 solar masses, 8 + 1.4 solar masses with spins tilted into the orbital plane. The count grows as the inverse of the lowest frequency — the slopes are −1.07, −1.02, −1.01. Above 20 Hz, the limit of the ground-based detectors that have heard every merger so far, the heaviest pair precesses 0.9 times, the middle 4.4 and the lightest 6.2. Above 5 Hz the counts are 4.9, 18.7, 25.5; above 1 Hz, 26, 95, 129.

Precessions bought one hertz at a time

A merging pair of black holes whose spins are tilted makes its orbital plane wobble, and the wobble is the only part of the signal that separates one hole's spin from the other's. How many wobbles a detector records is a simple formula: two plus three halves of the mass ratio, divided by the total mass and by the lowest frequency the detector can hear. The detectors that have heard every merger so far stop at twenty hertz, where a typical heavy pair wobbles less than once.

gravitation · Black hole spin A tilt that wanders up, one passing body at a time. The inclination of the Moon's orbit to the plane it started in, against the number of lunar-mass bodies that have passed the Earth–Moon pair at 5 km/s within 300 Earth radii, with the Moon at 30 Earth radii, for four different sequences of passages. Each passage tilts the orbit by a fraction of a degree in a random direction, so the inclination performs a random walk and grows as the square root of the number of passages: after 1,000 the four runs stand at 3.8, 1.8, 3.9, 0.2 degrees. The dashed line is the Moon's present inclination, 5.145 degrees. Whether a given history reaches it is a matter of chance; how far a typical one gets is set by the passages' number, mass and speed.

A tilt built from a thousand near misses

The Moon's orbit is tilted five degrees to the ecliptic, and slow tidal recession cannot have done it. One candidate is the leftover debris of planet formation: bodies the size of the Moon, passing the young Earth–Moon pair, each giving its orbit a kick of a fifth of a degree in a random direction. A random walk of a few thousand such kicks reaches five degrees — and the bodies that would supply them are the same population whose impacts left the Earth's mantle its gold.

orbits · Angular momentum Two old points, and the history they draw back to the Moon's birth. The Earth–Moon distance against time before the present, with the estimates from tidal rhythmites (620 million years), cyclic sediments (1.4 billion) and banded iron formation (2.46 billion, 321.8 thousand km, ±6.2). Today's tidal efficiency, scaled as the inverse sixth power of the distance, brings the Moon to the Earth 1.54 billion years ago. The efficiency fitted between the two oldest points — 21% of today's — held further back brings it out from the Earth 4.78 billion years ago, close to the Moon's age of about 4.5 billion. The constant law is an assumption; what the points measure is the average between them.

A Moon measured in banded iron

Layers of iron-rich sediment laid down 2.46 billion years ago in what is now Western Australia keep time with the Earth's precession, and the precession's period says how far away the Moon was: about 321,800 kilometres, with a day of seventeen hours. Together with an older-is-closer point from 1.4 billion years ago, it measures how hard the tides worked between them — about a fifth as hard as today. Held further back, that efficiency brings the Moon out from the Earth a few hundred million years too early, which says the young tides worked harder than the middle-aged ones.

sky · Precession

Named alongside it

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

EccentricityMoment of inertiaAccretionAngular momentum transportAttitude controlPlanet migrationRoche limitSpecific angular momentumTorqueAccretion discDynamical frictionEffective potential

All concepts