Astronomy & Celestial Mechanics
the sky, computed
The collection

Every essay — page 2

One idea per essay, ordered so that the earlier ones set up the later ones — but nothing here depends on being read in sequence. Essays 21–40 of 514.

Orbits

Kepler's three laws, and the family of curves a single force allows.

Flight time against semi-major axis, for a fixed 135° sweep. Lambert's theorem drawn: the time to fly between two points 1 and 1.524 AU out and 135° apart, against the semi-major axis of the orbit that does it. Nothing else about the orbit enters — not its eccentricity, not where its periapsis is, not how it is oriented — which is the content of the theorem and the reason a two-point transfer is a one-dimensional search rather than a six-dimensional one. Two branches: the lower one is the ellipse whose arc stays short of apoapsis, falling towards the parabolic floor at 103.2 days as a grows without limit; the upper one is the ellipse of the same size whose arc runs through apoapsis, rising without limit. They meet at a = s/2 = 1.2161 AU, 244.2 days, which is the minimum-energy transfer and the slowest ellipse available — every faster one is bigger. Each branch is monotone, checked point by point across the drawn range, so a horizontal line cuts each at most once: for a given pair of points and a given time there is exactly one ellipse, and at 260 days it is a = 1.2189 AU on the upper branch. The freedom a mission designer has is not in this picture: it is the choice of the two points, which is what a porkchop plot sweeps.

Two places and a clock decide the path

The time to fly between two points depends on the semi-major axis, the chord between them, and the sum of their distances — and on nothing else about the orbit. Not the eccentricity, not where periapsis is, not the orientation. Lambert's theorem is why an interplanetary launch date is the root of one equation.

8 figures · Lambert's problem The angle that has nowhere to be measured from. An eccentricity vector carried round a circle of radius 0.034 centred at 0.031 — which is what a secular perturbation does to one, a forced eccentricity with a free one turning about it. Below, the two components e cos ϖ and e sin ϖ, which are smooth, bounded and perfectly ordinary throughout. Above, the longitude of pericentre read off them, which is not: as the eccentricity passes its minimum of 0.0030 the pericentre sweeps through most of a circle, at up to 4080° per unit time against the 12° the free vector itself turns in the same interval. Nothing has happened to the orbit. The pericentre is a place on the orbit, and a nearly circular orbit does not have one — so ω, and Ω with it at zero inclination, are angles measured from a feature that is not there. The equinoctial elements are the pair drawn below, and a propagator written in them steps through this instant without noticing it.

The elements that stop existing

An orbit needs six numbers, and three of the usual six are angles measured from features a perfectly ordinary orbit may not have. At zero eccentricity there is no pericentre to measure from, and the arithmetic knows it.

8 figures · Orbital elements An eccentricity and an inclination trading, at 65° of mutual tilt. The secular equations integrated from a nearly circular orbit (e = 0.02) inclined at 65° to a distant perturber's plane, over three oscillations. Above, the eccentricity; below, the inclination, with the constant √(1−e²)cos i drawn as the flat line it is. The eccentricity climbs to 0.8380 and the inclination falls to 39.25° at the same instant, and neither is a coincidence: the product is fixed, so one can only rise as the other falls. That floor is the same for every starting tilt — at maximum eccentricity j = √(5/3)Θ, so cos i = √(3/5) and the inclination arrives at 39.23° whether the orbit began at 50° or at 89°. The closed form for a circular start is e_max = √(1 − (5/3)cos²i₀) = 0.8380, which contains nothing about the perturber — not its mass, not its distance. Those set the clock and not the amplitude, and the period here is 4.83 Kozai times. What the figure cannot show is what happens at the top of the cycle in a real system: at e = 0.838 the pericentre is 0.1620 of the semi-major axis, where tides, general relativity or a stellar surface all intervene, and the quadrupole picture ends.

An inclination that turns into an eccentricity

A distant companion cannot change an orbit's size or its energy. It can take a circular orbit tilted past 39.23 degrees and drive it to an eccentricity near one, and back, over and over — and the companion's mass and distance set only the clock.

8 figures · Kozai–Lidov The same thrust is worth 2.08 times more at perigee, and out of plane it is worth nothing at all. The rate of change of the semi-major axis under a unit acceleration in each of the three directions, against position around an orbit of eccentricity 0.35, from Gauss's variational equations. The along-track curve carries the factor p/r = 1 + e cos f and therefore peaks at perigee, where the same impulse is worth 2.08 times what it is worth at apogee — the whole of the Oberth effect, arriving as a term in a differential equation rather than as an argument about kinetic energy. The radial curve is antisymmetric about apoapsis and integrates to exactly zero over a revolution: pushing outwards for half an orbit and being pushed back for the other half changes the energy by nothing, which is checked here by quadrature and comes out at -9.0e-18. And the out-of-plane response is identically zero at every point of the orbit, because W is perpendicular to the velocity and does no work. An orbital plane can be rotated without touching the energy, and that is why a plane change is so expensive: none of what is spent goes anywhere useful.

Which direction moves which element

Resolve a small force into three components and Gauss's equations say exactly what each one does. The out-of-plane component can rotate an orbit and can never change its energy; the along-track component owns the semi-major axis outright and is worth more at perigee than at apogee by a factor that is pure geometry.

8 figures · Perturbations Earth's eccentricity is a sum of 8 sinusoids. The eccentricity of Earth over 800 thousand years, from the Laplace–Lagrange solution for all eight planets — the secular matrix built from the JPL masses and semi-major axes, symmetrised, and diagonalised by Jacobi rotations. It runs between 0.0035 and 0.0436, and it has no period, because it is a sum of 8 incommensurable frequencies. The two largest contributions to this planet are the modes at 3.73 and 7.33 arcseconds per year, drawn as the flat lines: those are constants, and everything moving in the figure is their beat. The check is the quadratic form ½ΣΛe², which a symmetric secular matrix conserves exactly and which drifts by 6.7e-16 across the whole interval — computed from the same curves the figure draws and from nothing the eigenvalues were fitted to. The true angular momentum deficit, Σ Λ(1 − √(1−e²)), drifts by 7.1e-4, and that difference is not an error either: the two agree only to fourth order in e, and Mercury at 0.206 supplies almost all of the gap.

No planet has an eccentricity of its own

Strip the short-period terms out of the planetary equations and what is left is a linear system. Its eigenvectors are modes of the whole solar system, and the number a catalogue quotes for a planet's eccentricity turns out to be a reading of a clock rather than a property of the planet.

8 figures · Secular theory One admissible root, 0.01% from the truth. Gauss's reduction of three directions to a distance, drawn as the two relations whose intersection it is. Three observations of Ceres on days 0, 20, 40 of an arc, generated from its elements and used only as sight directions — no range, no radial velocity. The rising curve is geometry: the heliocentric distance a candidate at geocentric distance ρ₂ would have, r₂² = ρ₂² + 2ρ₂(R₂·L̂₂) + R₂², which contains no dynamics at all. The falling curve is dynamics: ρ₂ = A + µB/r₂³, with A and B built from the three sight vectors, the three observer positions and the three times, and containing no orbit. Eliminating ρ₂ between them gives r₂⁸ + a r₂⁶ + b r₂³ + c = 0 — an eighth-degree equation, from a problem with exactly as many equations as unknowns. Here they cross once at a positive ρ₂, at r₂ = 2.5893 AU against the true 2.5890. The other 2 real roots are rejected not by fitting but by sign: the ρ₂ each implies is negative, and an object behind the observer was not the thing observed.

Three observations and no orbit at all

Three directions in space give six numbers for the six elements of an orbit, which sounds like a solved problem. The algebra that solves it is of the eighth degree, and for a near-Earth asteroid three perfect observations can be consistent with three different orbits.

8 figures · Orbit determination 5 transfers through the same two points in the same 1400 days. Time of flight against semi-major axis for every transfer through two points 135° apart at 1 and 1.524 AU, with the revolution count running from 0 to 2. Each count contributes two branches, and for N ≥ 1 the pair folds: the time has a minimum at a = 1.2426 AU for one revolution — only 2.2 per cent above the minimum-energy value of 1.2161, which is why the horizontal axis is the excess over that value and logarithmic — so a flight time above it is met twice and below it not at all. Reading the crossings of the 1400-day line off the drawn curves gives 5 of them — 0 revs high, 1 rev low, 1 rev high, 2 revs low, 2 revs high — which is 2N + 1 with N = 2, and the count is a property of the time rather than of the geometry. That is the practical content: a root-finder started from a single guess returns one of these 5 and gives no sign that the other 4 exist, and the cheapest of them is often not the one nearest the guess.

One time of flight and five ways round

Lambert's theorem says two positions and an interval fix the transfer. Allow the transfer to complete whole revolutions and that stops being true: the flight time folds, one number admits five arcs, and the cheapest of them is usually not the one a solver started nearest to.

8 figures · Lambert's problem Mars to five metres and Neptune to five thousand kilometres, in the same file. Present-day heliocentric position uncertainty for each planet, in kilometres, with the range component marked separately below it. The two differ because a transponder measures a distance along the line of sight and says nothing about the two directions across it, so a planet with an orbiter is known radially some 17 times better than it is known altogether. Neptune is 10⁶ times less well determined than Mars and only 20 times further away, which is the whole point: the accuracy is a property of the observations, not of the geometry. Mars has carried a transponder almost continuously since 1976; Neptune has been visited once, in 1989, and everything else known about it is meridian-circle astrometry covering 1.07 of one orbit. The two ice giants are the only entries here whose ephemerides are still limited by nineteenth-century technology, and the only cure is a spacecraft.

The table that is a fit

A planetary ephemeris is not evaluated from Kepler's laws and is not evaluated from a theory. It is a numerical integration whose starting conditions were least-squares fitted to a century and a half of observations, and its accuracy is a property of those observations rather than of the mathematics.

10 figures · Ephemerides After one revolution the error is 3π times longer than it is wide, and after 300 it is 2827. The two semi-axes of a fitted orbit's position uncertainty, against elapsed revolutions, for a solution whose semi-major axis is uncertain by 12 kilometres. The radial extent does not grow at all: a body on a slightly larger orbit is slightly further out and stays so. The along-track extent grows linearly, because δn/n = −(3/2)δa/a makes a semi-major-axis error into a mean-motion error and a mean-motion error into a phase that runs away — a·δM = 3πN·δa after N revolutions. The ratio is 3π ≈ 9.42 after a single revolution and 2827 after 300, which is why an asteroid recovered after one apparition is found within a few arcseconds of its predicted place along its own track and could be a long way from it in time. Every consequence of this in practice — that an impact probability is a one-dimensional integral rather than a volume, that a keyhole is an interval, that the next observation worth taking is the one across the track rather than the one that fits best — is a restatement of these two lines diverging.

An error that is nearly all in one direction

A fitted orbit's uncertainty is not a ball. Within a few revolutions it has collapsed onto a line along the track, because an error in the size of an orbit is an error in its period and an error in period is a phase that runs away — which is why an impact probability is an integral along a curve rather than over a volume.

8 figures · Orbit determination Every model curve has slope −1, and four measurements agree on κ to 1.5×. Semi-major-axis drift against body diameter, for a thermal recoil in which a fraction κ = 0.085 of the absorbed sunlight comes back out along-track. The three curves are the same expression at 1, 1.6, 2.5 astronomical units, and each has a slope of exactly −1: the acceleration is the absorbed power divided by the mass, which is a cross-section over a volume, so it falls as one over the size and nothing else on this axis changes it. A kilometre-wide body drifts a few metres a year; a ten-metre one drifts hundreds. The four filled marks are the bodies whose drift has actually been measured as a fitted parameter in an orbit solution, and they do not lie on any single curve because each carries its own density, distance and obliquity. What they agree about is the number beside each: solve every measured drift for the efficiency that would produce it and the four answers are 0.084, 0.085, 0.089, 0.129 — a factor of 1.5 apart, for a quantity that could in principle have been anything from zero to a fifth. That agreement is the evidence that the mechanism is understood, and it is the only evidence there is, because the thermal conductivity that sets κ has never been measured for any of them.

An orbit moved by heat

A rotating body re-radiates absorbed sunlight from the hemisphere that has had time to warm, so the recoil is not aimed at the Sun. The resulting force is a few parts in ten billion of gravity, it is the only orbital force whose sign depends on which way the body spins, and it has been measured to four figures.

8 figures · Non-gravitational forces The divisor is 0.129″/yr and the theory's own error is 0.24. Six frequencies of the secular solar system on one logarithmic axis, in arcseconds per year. The top two are the pair whose near-equality is the whole story: the perihelia of Mercury and Jupiter separate at 1.333″ a year, the nodes of Mercury and Venus at 1.462, and the difference of those two differences is 0.129 — a resonant argument that turns once every 10.0 million years. A term with that argument in the disturbing function acts in one direction for five million years at a stretch, which is what pumps Mercury's eccentricity, and it is the reason the inner solar system's Lyapunov time is what it is. The bottom three bars are why this figure exists. The divisor is smaller than the corrections the theory that computes it leaves out. Relativity contributes 0.4298″ a year to g₁ alone — the same 43 arcseconds a century that broke Newtonian gravity — which is 3.3 times the divisor; the fourth-order terms in the eccentricity that Laplace–Lagrange truncates come to about 0.24″, which is 1.8 times it; and the second-order solution computed on this page gets 0.35″, missing the published value by more than the value itself. A theory cannot bound what it cannot resolve. Laplace's proof that the eccentricities stay bounded is a proof about a system whose frequencies are constants, and the frequency that decides the question is not one.

The bound that holds only in the linear theory

Laplace proved the planetary eccentricities bounded, and the proof is a proof about a linearised system with constant frequencies. One combination of those frequencies is nearly zero — and it is smaller than the terms the linearisation threw away, which is why the stability of the solar system is a probability rather than a theorem.

9 figures · Secular theory A 128.8-million-year-old collision, dated from the shape of a scatter plot. The Erigone family: 165 members drawn at their diameters and their proper semi-major axes, with inverse diameter up the page. The cloud is a V, and the V is a clock. Each member has been drifting in semi-major axis ever since the collision at a rate that goes as one over its diameter, with a sign set by which way it spins — prograde outward, retrograde inward — so after 130 million years the small members have moved far and the large ones have barely moved at all. Plotted against 1/D that envelope is a straight line through the family's centre, and its slope is the drift rate for a one-kilometre body multiplied by the elapsed time. Fitting the two edges of the points actually drawn here returns 128.85 million years against the 130 the members were generated from. The rounding at the bottom is not an artefact: it is the ejection velocity, some 15 metres per second, which every member got at the moment of the collision and which is the same for all sizes. The picture cannot show the interlopers — background asteroids that happen to lie inside the V and have nothing to do with the family — and it cannot show the members that have drifted into a resonance and left the belt entirely, which is the reason the oldest families have the softest edges.

A collision dated by a scatter plot

Nothing in the solar system carries a date. A collisional family does — because a force that depends on a body's size has been pushing its fragments apart ever since, so the cloud is a V whose slope is an elapsed time, and one of those dates is confirmed by fossil meteorites in Swedish limestone.

8 figures · Asteroid families A clock that is a straight line for three billion years and then is not. The lunar chronology function: craters of a kilometre or more per square kilometre against the age of the surface, with the count logarithmic and time not. The dashed line is the present impact rate extrapolated backwards, and it accounts for the whole curve up to 3.1 billion years — over that entire range, dating a surface is dividing a crater count by a constant. The rate itself, read off the slope of the drawn curve at the present day, is 8.4·10⁻⁴ craters per square kilometre per billion years, which over the whole Moon is about 32 new craters of a kilometre or more per million years. Past three and a half billion years the exponential term takes over and the curve turns almost vertical: ground that is 4.1 billion years old carries 26 times the crater density of ground 3.5 billion years old, for a difference in age of six hundred million years. Most of the craters on the Moon were made in a small fraction of its life, and nothing that happened after them is recorded anything like as densely. The six marked ages are laboratory measurements on returned rock, and they are what makes the curve a chronology rather than a shape — the Moon is the only body whose crater counts and whose radiometric ages have ever been measured on the same square kilometre.

A surface dated by counting holes in it

Every age quoted for a surface in the solar system outside the Earth — a Martian lava flow, a crater on Mercury, the ice of Europa — comes from counting craters and passing the count through one curve. That curve was calibrated on nine square kilometres of the Moon, and it is nearly a straight line for three billion years and then is not.

8 figures · Surface chronology An impulse delivered inside 0.5 AU, and a comet 2050 hours early. Above: Marsden's outgassing law, the factor g(r) that scales a comet's non-gravitational acceleration, against distance from the Sun over one orbit of a comet with perihelion at 0.336 AU and aphelion at 4.09. It is close to an inverse square inside the water snow line and then falls off a cliff, because water ice that is not being heated does not sublimate. Half the whole revolution's impulse is delivered inside 0.55 AU — a few weeks out of a 3.3-year orbit — so the force is effectively a kick at perihelion rather than a perturbation spread around the path. Below: what a kick of that kind does to the timekeeping. A transverse component changes the semi-major axis and so the period, by 2.5 hours per revolution here, and a constant change in the period accumulates as the square of the number of revolutions rather than in proportion to it. After 40 returns the comet arrives 2050 hours — more than 85.4 days — before an orbit fitted without the term predicts, and doubling the number of returns multiplies the discrepancy by 3.90. That is why the effect was found in the eighteen-twenties from nothing but arrival times, and a century and a half before anyone photographed a jet.

A comet that arrives a day early

Encke's comet returned two and a half hours ahead of prediction, every revolution, for decades before anybody could say what was pushing it. The force is a rocket — a few tonnes a second of vapour leaving the sunward side of a rotating nucleus, delivered almost entirely in the few weeks around perihelion, and it accumulates in the arrival time as the square of the number of returns.

8 figures · Non-gravitational forces

Gravitation

Two bodies pulling on each other, and everything that goes wrong at three.

Two bodies at a mass ratio of 3 to 1. Both bodies orbit their common centre of mass, on similar ellipses whose sizes are in inverse proportion to the masses — here 3 to 1, so the heavier body's path is 3 times smaller.

Neither body is still, and the wobble is how planets are found

A planet does not orbit its star. Both orbit a point between them, and the star's share of that motion is small, measurable, and the reason thousands of planets are known.

9 figures · The two-body problem Why a shell pulls a point inside it not at all. A double cone from a point inside a uniform shell. In the narrow-cone limit the far patch is 2.80 times further away and 2.80 times wider — the same number, checked to a part in a thousand before this figure is drawn. Its mass is greater by the square of that ratio and its pull weaker by the same square, so the two cancel in every direction.

A sphere pulls exactly like a point, and the proof is a pair of cones

Every orbit ever computed treats the Sun as a dot. That is not an approximation — for a spherical body it is exact, and the reason is a cancellation between two patches of a shell.

9 figures · Shell theorem 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.

8 figures · 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.

9 figures · 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 rather than asserted.

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.

8 figures · The two-body problem The tide across a moon, against the moon's own gravity. The tidal acceleration across a satellite and the satellite's own surface gravity, both in units of that surface gravity, against distance from the primary in planet radii. The tide falls as the inverse cube and the self-gravity does not fall at all, so they cross once — at 2.23 radii for the density ratio drawn. Inside the crossing the tide wins and a body held together only by its own weight comes apart.

The distance at which a moon stops holding together

The tide across a body falls as the inverse cube; the body's own gravity does not fall at all. There is therefore exactly one crossing, and Saturn's rings end within a few per cent of it.

8 figures · Tides

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