Theme

Everything is a conic

One force law, four curves, and the discovery that a comet and a planet differ only in how fast they were going.

An orbit at eccentricity 0.6. An orbit of eccentricity 0.6. The primary sits at a focus, offset from the centre by 0.6 of the semi-major axis, and the closest and furthest points differ by a factor of 4.00.

The orbit is an ellipse, and the Sun is not in the middle of it

Kepler's first law is usually drawn wrong. The interesting content is not the ellipse — it is the focus, and the fact that one of the two is empty.

$What an eccentricity does to the shape, and what it does to the offset. The fractional flattening 1 − b/a and the fractional focal offset c/a, against eccentricity. The offset is first order in e and the flattening is second, so at Earth's e = 0.0167 the outline is 0.014% from a circle while the Sun sits 1.67% of the semi-major axis off the centre — a factor of 120.$ Orbits

An orbit can look exactly like a circle and still not be one

Earth's orbit departs from a circle by fourteen parts in a hundred thousand. The Sun's offset from its centre is a hundred and twenty times larger, and everything interesting is in the offset.

A vector that does not change, drawn five times. The Laplace–Runge–Lenz vector A = v × L − GM r̂, constructed at five points of one orbit at e = 0.45 under a force ∝ r^−2, each drawn from the body rather than from the focus so that its length and direction can be compared point by point. Under the inverse square every one of them is the same vector: identical to machine precision in length and in direction. Its length is 0.4500, which is the eccentricity, and it points at pericentre — so the orbit's orientation is a conserved quantity and not a constant of integration, which is why the ellipse does not turn.

The third thing that is conserved

Energy fixes an orbit's size and angular momentum fixes its shape. Neither fixes which way it points — so a curve that closes needs a third conserved quantity, and only two force laws in the universe supply one.

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.

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

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

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

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.

The formula that exists, and is not used

Kepler's equation does have a closed-form solution — an infinite series in the eccentricity, written down by Lagrange. It converges up to e = 0.6627434 and not one part beyond, and that number has nothing to do with astronomy.

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

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

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

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.

The mass correction against the mass. Each planet's departure from the massless harmonic law, against its own mass in solar units, on logarithmic axes. The exact law puts every point on the diagonal. Jupiter and Saturn are the only planets whose mass correction is larger than the perturbations from everything else, and Saturn's measured departure has the opposite sign.

The third law is wrong by the mass of the planet

Kepler's harmonic law says the square of the period goes as the cube of the size. Newton's version has one more term in it, and the term is the orbiting body's own mass — negligible for a planet, decisive for a binary star, and the reason a period can be converted into a mass at all.

1I/ʻOumuamua: an orbit at e = 1.201, and the angle it turned through. The open branch of a conic at eccentricity 1.201 and periapsis 0.2559 AU, with the Sun at the occupied focus. Two numbers fix everything else: |a| = q/(e−1) = 1.273 AU, an impact parameter b = |a|√(e²−1) = 0.847 AU, an asymptote at ν∞ = arccos(−1/e) = 146.37° from periapsis, and a deflection of δ = 2ν∞ − 180° = 112.74° — which is the same statement as sin(δ/2) = 1/e. The asymptotes cross at |a|e = 1.529 AU from the Sun on the periapsis side, where a bound orbit's centre would be on the other. The speed left over at infinity is √(μ/|a|) = 26.40 km/s. Schematic in one respect: the drawing runs at about 119 px to the AU, so the Sun's own disc would be far smaller than its marker.

The orbit that has no period

Above an eccentricity of 1 the conic is open, the energy is positive and the semi-major axis is negative — and the vis-viva relation survives the sign change without a single alteration. What replaces the period is a speed, and that speed is what says where a visitor came from.

Aiming at a plane instead of at a planet

A spacecraft arriving at a planet is not aimed at a periapsis distance. It is aimed at a point in a plane perpendicular to its own incoming asymptote, because that is the one coordinate in which the miss distance responds linearly to a correction — and every navigation product ever published for a flyby is written in it.

Ceres from five directions and no distance. Ceres seen five times over 41 days, from an Earth on a circular orbit, reduced in the plane. Each sighting gives a direction and no range, so the object is somewhere on its sight line; the five lines here span 1.37° of geocentric arc altogether, and Earth's own motion supplies the only baseline there is — 0.403 AU of its 0.691 AU of travel lies across the sight lines. A planar orbit is four numbers, so five angles over-determine it and one orbit comes out: a = 2.7658 AU, e = 0.0785. That is the answer and not an input — the sightings were generated at a = 2.7658 AU and e = 0.0785, and the solve, which sees only the directions and the dates, returns them to 3e-12. The two shaded sectors are what closes the determination: between the first and middle sightings the radius vector sweeps 0.2993 AU² in 21.0 days and between the middle and last 0.2853 AU² in 20.0 days, a ratio of 1.04918 against a time ratio of 1.04918. Slide all three crossings out along their sight lines together and that equality fails at once, so it fixes the distance by itself, with no propagation anywhere in the argument — and it gives a = 2.7658 AU over again. The two dashed curves are candidates that thread the same three sight lines at 85% and 108% of the recovered distance: a = 1.86 AU at e = 0.35, sweeping its areas in a ratio 4.5% wrong; and a = 5.61 AU at e = 0.49, sweeping its areas in a ratio 2.9% wrong. A few per cent in the distance is an orbit of another kind, which is the same fact the conditioning panel measures: one arcsecond of angle error moves a by 0.60% on this arc.

Five directions and no distance among them

An image of a moving point records an angle and throws the range away, so an orbit has to be assembled out of angles alone. How many angles are needed is not a detail of the method — it is the whole of what a determination is.

Four conics through one periapsis, drawn by one solve. Distance from the Sun against time for four orbits sharing a periapsis of 0.5 AU, at e = 0.6, e = 1, e = 1.0001, e = 1.4, over 900 days. Every point on every curve came from the same universal Kepler solve — no branch on the conic class anywhere in it — and each curve was then checked against the classical solution of its own kind at the midpoint: Kepler's equation at e = 0.6 agrees to machine precision; Barker's cubic at e = 1 agrees to machine precision; e sinh H − H at e = 1.0001 agrees to machine precision; e sinh H − H at e = 1.4 agrees to machine precision. The curve to read twice is e = 1.0001: over this arc it lies within 0.02% of the parabola and is indistinguishable from it, and it is the only one of the four whose fate the drawing cannot show. What the classical parameterisation costs there is arithmetic rather than impossibility: at the midpoint of this arc, e sinh H − H throws away 3.3 of its sixteen digits to cancellation, against 0.2 at e = 1.4 — enough to matter to an ephemeris and not enough to stop a plot.

The one solve that does not ask which conic it is

Kepler's equation is for ellipses, Barker's cubic for parabolas, and a hyperbolic sine for the rest — three parameterisations of one motion, each worst exactly where its neighbour takes over. The universal variable removes the question, and the removal is not a convenience.

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.

One squaring, and the singularity is gone. Left, a harmonic oscillator: an ellipse centred on the origin, marked at 24 equal steps of its own phase. Right, the same points after squaring as complex numbers, u ↦ u². The image is an ellipse with the origin at a focus — checked here by the focal property, r₁ + r₂ = 2a to nine figures at four points — with semi-axes (A²+B²)/2 = 2.5000 and AB = 1.5000 and the focus at (A²−B²)/2 = 2.0000, which is ae exactly. Three things follow at once. The angle doubles, so one turn of the oscillator is one whole orbit traversed twice as fast in phase; the equal phase steps on the left arrive as the eccentric anomaly on the right, which is why that anomaly and not the true one is what the equations want; and the collision at r = 0, where the inverse square is infinite, is the point u = 0, where the oscillator has a perfectly ordinary velocity. The singularity was a property of the coordinates.

The singularity that is a change of variable

The Kepler problem blows up at zero separation, and a fixed-step integrator falls apart long before it gets there. Divide time by the radius and the equations become a harmonic oscillator — exactly, for every conic at once.

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.

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.

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.

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.

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.

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.

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.

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

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.

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.

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.

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.

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.

$The Kirkwood gaps. Asteroid numbers against semi-major axis, with the resonant radii marked. Each gap sits where the orbital period is a simple fraction of Jupiter's, and each of those radii is computed from the harmonic law rather than placed by eye.$ Gravitation

Resonance clears a gap in one place and locks a moon in another

When two orbital periods are in a simple ratio, small tugs stop averaging away and start accumulating. Sometimes that empties a region entirely. Sometimes it holds three moons together for the age of the solar system.

$Zero-velocity curves, mass fraction 0.15. Level sets of the Jacobi constant in the frame that rotates with the two bodies. A particle with a given value of C is confined to the side of its own curve where the kinetic energy would be positive; as C falls the curves open, first at the inner point between the bodies, then behind the smaller one, and finally around the whole system.$ Gravitation

The curve that says where a body cannot go

The restricted three-body problem has no solution and one conserved quantity. That quantity is enough to draw a boundary the body can never cross — without integrating anything, without knowing where it started, and for all time.

G: 14 determinations in two families. Published determinations of G, each with its quoted one-sigma interval, sorted into two families — torsion balance, in one form or another, against beam balance, pendulum, atom interferometry. The shaded band behind each family is that family's inverse-variance weighted mean: 6.67435 ± 0.00004 across 11 of them, against 6.67343 ± 0.00009 across 3. The difference is 0.00092 ± 0.00010 10⁻¹¹ m³ kg⁻¹ s⁻², which is 9.3 standard deviations, computed here from the quoted errors alone. The arithmetic is the same one the Hubble figure uses and here it should be distrusted, because the scatter inside each family already exceeds what the intervals allow: eleven torsion-balance determinations spread over 500 parts per million with quoted intervals of 12 to 130 cannot all be right, whatever the difference between the families comes to. That is why the recommended value's uncertainty is expanded far beyond any single experiment's rather than being the weighted combination drawn here — the disagreement is between laboratories using the same method, not between methods.

Nothing in the sky is weighed in kilograms

The Sun's gravitational parameter is known to eleven significant figures. The Sun's mass is known to five. The two statements are about the same object and the difference between them is a constant measured in basements, which is the worst-determined fundamental constant in physics.

Energy error over 240 revolutions, at one step size. The relative error in total energy against revolution number, for three integrators run on the same Kepler orbit at e = 0.5 with the same step of 200 per revolution. The exact energy is a constant, so every curve here is the method rather than the problem. Euler climbs steadily: its energy at the end is 106.5% wrong, and the orbit it draws has spiralled outwards. Runge–Kutta 4 begins 4.7e+2 times more accurate than leapfrog and ends at 2.22e-4, having grown by a factor of 10 across the run: the error is SECULAR. Leapfrog oscillates inside a band and stays there — worst error 2.62e-3, and the second half of the run is no worse than the first, which is measured here rather than claimed. That is the property that decides whether a five-billion-year integration means anything, and it is not accuracy: a symplectic method is the exact solution of a Hamiltonian a step-size away from the intended one, so its energy cannot wander, while a more accurate non-symplectic method has no such constraint and eventually wanders further.

Wrong about where, and right about how much

Runge–Kutta is the more accurate method and loses energy steadily; leapfrog is cruder and its energy error never leaves a band. Over five billion years only one of those properties survives — and neither method knows where the planet is.

The contours a planetary encounter cannot cross. Contours of the Tisserand parameter with respect to Jupiter in the plane of semi-major axis and eccentricity, drawn for a coplanar orbit. An encounter with Jupiter moves a comet along one of these curves and never across one, because T is what the encounter conserves. The heavy contours are at T = 3 and T = 2, and they are the boundaries the comet families are defined by: T > 3 means no encounter is possible at all, since v∞²/v_J² = 3 − T and a negative squared speed is not a trajectory; 2 < T < 3 is the Jupiter family; below 2 the approach speed exceeds Jupiter's own orbital speed and the orbits are the nearly isotropic ones. The shaded boundary is the crossing condition — an orbit whose pericentre is outside Jupiter's, or whose apocentre is inside it, never meets the planet whatever its T. The five comets are placed at their JPL elements and labelled with the T the literature quotes; all five with inclination sit off the coplanar contours by exactly the cos i in the definition, which is why 1P/Halley's is negative — a retrograde orbit meets Jupiter at nearly twice Jupiter's speed.

The number that survives the encounter

A comet that passes Jupiter comes away with every orbital element changed. One combination of them is not changed, and it is enough to recognise the comet afterwards, to sort the comet families, and to say where a spacecraft can and cannot go.

The wake, computed from the streamlines that make it. Left: 26 streamlines past a point mass, in the mass's own frame, each integrated from far upstream with the same speed and a different impact parameter, and mirrored about the axis. Nothing is drawn to converge — every track is the hyperbola its own impact parameter gives it, and they cross downstream because an attraction focuses. Right: the density that focusing produces at 3.2 focusing radii behind the mass, as (b/y)(db/dy) — the Jacobian of the map from starting radius to arrival radius — which peaks at 15.53 times the background at 0.03 radii off the axis. The overdensity is behind the mass, and that is the entire mechanism: the wake pulls backwards on the body that made it. What the figure cannot show is the steady state, because it has no time in it: a real wake is continuously replenished, and the drag is the sum over an infinite train of these encounters, which is where Chandrasekhar's logarithm comes from.

A drag with nothing to drag against

A massive body moving through a sea of light ones raises an overdensity behind itself and is pulled back by it. The force does not depend on the masses of the background bodies at all — and it is strongest at one particular speed.

A feeding zone, and the spacing it forces

The radius that decides what a planet may keep also decides what it could reach while it was growing — and measuring the gaps between planets in that unit turns a distribution spanning three orders of magnitude in astronomical units into a band a factor of ten wide, with a floor that is partly a theorem and partly a fit.

$GW150914: 33 Hz to 250 Hz in 0.22 seconds. The strain of GW150914 — two black holes — through the last 0.22 seconds before merger, computed from the quadrupole sweep at the chirp mass its fit returned, 28.716 solar masses, and drawn at the luminosity distance it returned, 440 megaparsecs. Two things rise together and neither is free to rise on its own: the frequency goes from 33 Hz to 250 Hz, and the envelope — the outer curve — grows by a factor of 3.9, because the amplitude goes as f^2/3 and nothing else in it changes over so short a span. The vertical axis is in units of 10⁻²¹, so the peak here is a fractional length change of about 2.9·10⁻²¹: over the four kilometres of an interferometer arm that is 1.2·10⁻¹⁷ metres, a thousandth of the width of a proton. The chirp mass is not fitted to the amplitude at all — it comes from the spacing of these zero crossings, which is why it is the best-determined number in the whole event and why the distance, which does come from the amplitude, is the worst.$ Gravitation

A distance with no ladder under it

The frequency sweep of an inspiral fixes the chirp mass with no distance in it, and the amplitude then gives the luminosity distance directly, because one expression fixes both. That is a distance measured with nothing calibrated beneath it — and its error budget is one angle.

A detector the size of the galaxy

At a nanohertz no instrument can be built, so the clocks already in the sky are used instead. The signal is in no single pulsar's data — it is in the correlation between pairs as a function of the angle between them, and that curve has no free parameters at all.

A step that must not be adapted

A symplectic integrator's bounded energy error is a property of a fixed step. It is conserving a Hamiltonian a step-size away from the intended one, and changing the step changes which Hamiltonian — so refining the step at a close encounter, which is the first thing anybody does, destroys the only property the method was chosen for.

The observed sky

The Sun is a bad clock, by up to sixteen minutes

Solar noon and twelve o'clock are not the same instant, and the discrepancy runs through a fixed annual cycle. It has two causes, one from the shape of the orbit and one from the tilt of the axis.

The observed sky

The loop a planet does not make

Mars stops in the sky, backs up for ten weeks, and goes on. Neither orbit reverses anywhere — the loop belongs to the difference of two position vectors seen from one of them, and its width and duration are fixed by the ratio of the two radii and by nothing else.

The solar system to three figures, and not one distance in it. Left, why an inferior planet's wandering is a measurement. At greatest elongation the sight line from the Earth is tangent to the planet's orbit, so the angle at the planet is a right angle and a/a⊕ = sin ε — no distance anywhere in the argument, only the angle between two directions. Venus reaches 45.4°–47.1°, giving 0.7224 AU against the modern 0.72333. Right, every planet Copernicus could see, derived this way and by the synodic route for the outer ones — 1/P = 1/E − 1/S for the year, then the harmonic law for the distance — plotted against the catalogue. Mercury is the interesting failure: its elongation runs from 17.9° to 27.8° rather than sitting still, so the method returns a range, 0.307 to 0.466 AU, and the true 0.3871 lies inside it. That spread is not an error in the method; it is Mercury's eccentricity being measured by a technique that assumed a circle.

The observed sky

The solar system measured from inside one orbit

Venus never appears more than 47 degrees from the Sun. That single angle gives its orbital radius as a fraction of the Earth's, with no distance measured anywhere — and every other planet gives one up as easily.

$20 closure phases, unmoved by an atmosphere that ruins every baseline. Closure phase measured against closure phase true, for all 20 triangles of a 6-antenna array observing a binary 3.2 mas apart with a flux ratio of 0.35. Each antenna has been given an independent atmospheric phase of 65° rms, which corrupts the individual baseline phases by 102° rms — several times the 31.2° the source itself produces, so no single visibility phase in this simulation carries usable information. Every point here nonetheless sits exactly on the diagonal: the per-antenna terms cancel identically round any triangle, and the largest departure over all 20 is 2.5e-14 degrees, which is round-off. The price is in the counting. 6 antennas give 15 baseline phases of which 5 are consumed by the unknowns, so of the 20 triangles only 10 closures are independent — a fraction (N−2)/N = 0.667 of the phase information. For two antennas that fraction is zero and there is no closure at all; the Event Horizon Telescope's image rests on quantities of this kind and on no absolute phase whatever.$ Starlight

A phase that survives what corrupts it

An atmosphere over each antenna adds an unknown to the phase of every baseline that antenna takes part in. Sum the phases round a triangle and every one of those unknowns cancels identically — which is the reason an image can be made across ten thousand kilometres, and the reason it has no position on the sky.

A distance of 52.0 parsecs with nothing underneath it. Two ways to a distance for the same pair. The orbital parallax needs no iteration and no assumption: a double-lined spectroscopic orbit gives the relative orbit's linear size as (K₁+K₂)P√(1−e²)/2π sin i = 0.2268 AU, an astrometric orbit gives its angular size as 4.36 milliarcseconds, and the ratio is 52.0 parsecs — a length divided by an angle, with no step of the distance scale below it and no property of the stars assumed. The curves show the dynamical parallax, the version available when only one spectrum can be measured: guess the mass sum, take the linear size from the harmonic law, divide by the angular size, convert the apparent magnitude to an absolute one and read a new mass sum off a mass–luminosity relation. Three starting guesses spanning a factor of 10 in mass converge to the same distance in 8 passes and agree to 0.001 per cent. It converges because the distance depends on the assumed mass only as its cube root — the measured exponent here is 0.3333 — so a factor of two in the mass is 26 per cent in the distance, and one pass removes most of that. What it converges to is not the orbital parallax: the iteration settles at 54.2 pc against 52.0, 4.2 per cent away, because the fixed point is set by the mass–luminosity relation and the apparent magnitude rather than by anything measured about this orbit. The same insensitivity that makes it converge is why it is never better than the relation it leans on.

Two orbits of one pair, and a distance falls out

Measure the same binary spectroscopically and astrometrically and the orbit comes back twice — once as a length in kilometres and once as an angle on the sky. The ratio is a distance that owes nothing to parallax, nothing to a standard candle, and nothing to any assumption about the stars.

A Hohmann transfer, 2.6 to 1 in radius. Two circular orbits and the ellipse that touches both. The first burn raises the far point to the outer orbit; the second, half an orbit later, circularises. The costs are computed from the vis-viva relation.

The cheapest way between two orbits, and why it is so slow

Two burns and a long coast is the least fuel that will move a spacecraft between two circular orbits. It is also, for anything beyond the Moon, an unreasonably long wait.

Circular and escape speed. Orbital and escape speed against distance, in units of the circular speed at the surface. The escape curve is the circular one multiplied by the square root of two, at every distance without exception.

The speed that does not come back, and the √2 that separates it

Escape speed is exactly the square root of two times circular speed, at every distance from every body. Being in orbit is already 71% of the way to leaving.

$How much of a rocket has to be propellant. Propellant fraction against the velocity change bought, for three real propellant combinations. The curves approach 1 and cannot reach it, so every extra kilometre per second costs a larger share of what remains — which is why staging exists.$ Spaceflight

The exponential that decides what can be flown

A rocket carries its own reaction mass, so every kilogram of propellant has to be accelerated by the propellant beneath it. The result is exponential, and it is the reason spaceflight is hard.

Total Δv against the radius ratio. Total transfer Δv, in units of the starting circular speed, against the ratio of the two circular radii. The Hohmann transfer is cheapest at small ratios; the bi-elliptic transfers overtake it, and the limiting one — an intermediate apoapsis taken to infinity — crosses at a ratio of 11.94. Above about 15.6 every bi-elliptic transfer beats the Hohmann.

Going too far in order to arrive cheaply

The Hohmann transfer is the cheapest two-burn route between circular orbits. Past a radius ratio of 11.94 the cheapest route is three burns, and it goes far beyond the destination first.

Along a contour is free; across one costs, and 6 resonances sit on this one. Perihelion against aphelion, in units of Jupiter's orbit, with contours of constant v∞ — which is the Tisserand parameter through v∞² = (3 − T)v_p². Every contour crosses the line r_p = r_a = 1, because a spacecraft has to be at the planet's orbit to have an encounter there at all, and every point on one contour is reachable from every other point on it with no propellant: a flyby rotates the v∞ vector without changing its length, which moves the pump angle and slides the spacecraft along the curve. A burn is the only thing that moves it between curves, and that is the whole economy of a gravity-assist tour. The diagonals are resonant orbits — 1:1, 3:2, 2:1, 5:2, 3:1, 4:1 with the planet — and each is a straight line of slope −1 because a resonance fixes the semi-major axis and r_p + r_a = 2a. They matter because a spacecraft on one comes back to the same place at the same time as the planet, which is what makes a second encounter possible without waiting for a chance alignment. The marked points are where the v∞ = 0.3 contour meets each: a tour is a walk along the highlighted curve from one to the next, and the arithmetic that decides whether it can be walked is the turn one flyby delivers. At 1.35 Jupiter radii and 3.9 km s⁻¹ that turn is 163°, so the longest step drawn here needs 0.1 encounters — which is why a real tour has dozens of flybys and why the ones with the largest steps are the ones that need a deep-space manoeuvre in between.

The same planet, three times

A flyby cannot change the encounter speed, only its direction, so a tour has to be designed in the space of what is conserved. One pass moves a spacecraft along a single curve and no further than the planet can bend it — and reaching a distant target means walking that curve, returning to the same planet again and again.

What a 1 km/s burn is worth, against where it is spent. A vehicle arriving at Jupiter with an excess speed of 5.6 km/s, burning 1 km/s along its velocity at one point of the hyperbola. The vertical axis is the excess speed it leaves with. Spent at the surface the burn is worth 12.33 km/s of departure speed; spent far away it is worth 7.20. The energy bought is v·Δv, so the same propellant is worth 5.9 times as much at the bottom of the well — and nothing about the rocket has changed.

The same burn is worth more when moving fast

A rocket firing for ten seconds delivers the same change of speed wherever it is. It does not deliver the same change of energy, because energy is quadratic in speed — so the identical burn buys six times as much at the bottom of a gravity well as at the top, and every escape manoeuvre ever flown is arranged around that fact.

The cost of going to Mars, against when to leave and how long to take. Contours of departure energy C₃ over a grid of 64 × 56 solved Lambert problems: each point is a departure date, a flight time, and the unique single-revolution conic that connects the two planets between them. The cheapest transfer on this grid costs 5.1 km²/s², leaving in Mar 2001 with a flight time of 220 days, against 8.7 for the idealised Hohmann transfer between circular orbits of the same radii. A launch window is a region on this plane, not a moment, and its shape is what a launch period is negotiated against.

Two dates decide a mission

Given where a spacecraft leaves from, where it is going, and how long it may take, there is exactly one orbit joining the two. Solving that problem over every pair of departure and arrival dates produces a contour map, and the shape of the contours is what a launch window actually is.

$What 28.5° of plane change costs, at three orbital speeds. The cost of rotating an orbital plane, 2v sin(Δi/2), against the angle turned, for the three speeds a transfer to geostationary orbit passes through: 7.669 km/s in a 400 km circular orbit, 3.075 km/s once circular at 42,164 km, and 1.618 km/s at the apogee of the ellipse between them. All three are √(μ/r) or vis-viva at μ⊕ = 398,600 km³/s². A 28.5° turn therefore costs 3.775, 1.514 or 0.797 km/s depending only on where it is done — the apogee figure is 21% of the low one, and that fraction is the speed ratio, so it is the same at every angle. Past 23.9° a turn in the low orbit costs more than the 3.176 km/s that leaves Earth from it, and a full reversal costs 2v = 15.34 km/s, 4.8 times the escape burn.$ Spaceflight

The cheapest place to turn

Rotating an orbital plane costs 2v sin(Δi/2), and the only quantity in that expression that a mission controls is the speed. Where the turn is made therefore matters more than how far it turns.

$Payload fraction against the number of stages, for 9.4 km/s at three structural fractions. Overall payload fraction λ = λ₁ⁿ against the number of equal stages sharing 9.4 km/s, for structural coefficients 0.06, 0.08, 0.12 and an exhaust speed of 3.432 km/s (I_sp 350 s). At ε = 0.08 a single stage has a negative payload fraction, −1.67%: the mass ratio 15.5 it needs leaves less than the tankage weighs. That is the wall, and it is the structural fraction rather than the exponential — one stage tops out at vₑ ln(1/ε) = 8.67 km/s, 0.73 km/s short, so each curve begins where λ₁ crosses zero. At ε = 0.06 the same engine does clear 9.4 km/s in one stage, with 0.50% of the vehicle as payload — the wall moves with the tank, not with the engine. Two stages then buy 3.59% and five 4.66%, against a ceiling of 5.10% at infinitely many: the step from two stages to three is 0.67 points of payload, and the step from four stages to five is 0.14 points of payload — which is why a launcher is built with two or three stages and not with five. Falcon 9 Block 5 is plotted at its own mass table: 22.8 t of 549 t is 4.15%, and its per-stage structural coefficients are 0.061 and 0.040, both better than the 0.08 the family curve assumes.$ Spaceflight

The stage that has to be thrown away

The rocket equation does not forbid a single stage from reaching orbit. The tank does — by 0.73 km/s out of 9.4, which is close enough that the question stayed open for forty years and expensive enough that it was never once answered in flight.

20% more Δv, spread over 26 revolutions. A continuous tangential thrust from a circular orbit of radius 1 to one of radius 6.611, integrated from dr/dt = 2r·a_T/v with the primary's GM set to 1. The spiral costs |v₁ − v₂| = 0.6111 in units of the inner circular speed, against 0.5076 for the two-impulse Hohmann drawn on the same pair of circles — 20.4% dearer, and the excess is exactly the Oberth advantage the impulsive transfer collects by burning where the vehicle is moving fastest and the spiral throws away by burning everywhere. The revolution count follows from the thrust level and nothing else, and the count drawn here is not a real one: 26 revolutions at an acceleration of 0.0015 of the inner orbit's own gravity, chosen so that the spiral can be seen at all. A real electric transfer runs nearer 3·10⁻⁵, which is 1,296 revolutions of a curve no page could resolve. Halving the thrust doubles both the turns and the time and leaves the Δv exactly where it is. That separation is the whole reason low thrust is flown at all — the Δv is worse and the propellant is not, because the exponential in the rocket equation is over Δv/(g₀Isp) and an electric engine's Isp is the larger number by more than this 20%.

The transfer that costs more the gentler it is

A continuous spiral from low orbit to geostationary needs a fifth more velocity change than the two-burn transfer, because the thrust is never at the one place it is worth most. It is flown anyway, and by a wide margin — because the exponential in the rocket equation changed sides.

Down by 280 km, and faster by 164 m/s. A circular orbit at 400 km with a ballistic coefficient of 100 kg/m², integrated down to 120 km through a tabulated atmosphere at solar minimum and solar maximum. At solar min it takes 1.2 years; at solar max it takes 147 days — a factor of 2.9 for the same satellite in the same orbit, decided by an eleven-year cycle nobody controls. The rising curves are the orbital speed on the right-hand scale, and they are the point: the drag force is opposite the motion and takes energy out, and the body goes faster, from 7673 to 7836 m/s. There is no contradiction in it. The specific energy is −μ/2a, so removing energy shrinks a, and the circular speed √(μ/a) rises when a falls; the kinetic energy gained is exactly half the potential energy lost, and the other half is what the air took. Every point on every curve was integrated from da/dt = −(ρ/β)√(μa), and the speed at each point is √(μ/a) at that point rather than a separate model.

An orbit that speeds up as it is slowed down

Drag takes energy out of a satellite and the satellite goes faster. There is no paradox in it, only a sign — and the same sign makes a re-entry date a space-weather forecast rather than an orbital computation, which is why Skylab was predicted for 1983 and came down in 1979.

A corridor 0.78° wide. Peak deceleration against entry flight-path angle, from sixty-one integrated entries at 11 km/s and β = 250 kg/m². The steep edge is where the load reaches 12 g, at 6.05°. The shallow edge is skip-out: below 5.27° the vehicle passes through the upper atmosphere and leaves again at 8.55 km/s, having lost too little speed to be captured. The corridor between them is 0.78° wide, which at an approach speed of 11 km/s is a targeting problem measured in kilometres of periapsis, days out. The curve is steep everywhere, which is the other half of the difficulty: half a degree of aiming error is a factor of 1.46 in the load. Lift is what widens this, and no ballistic capsule has any.

A corridor a degree and a half wide

The peak deceleration of an entering vehicle contains no property of the vehicle at all. Only the speed and the angle of arrival decide how hard it is slowed — the ballistic coefficient decides where, and nothing decides whether.

three revolutions, on a turning Earth. The ground track of a circular orbit at 420 km and 51.64° inclination, over 3 revolutions, on an equirectangular graticule. The latitude is a sine wave bounded by ±51.64° exactly — sin φ = sin i sin u, so the inclination is the highest latitude the orbit ever passes over, and it is reached twice per revolution. Each successive pass is displaced west by (ω⊕ − Ω̇) × 92.90 min = 23.61°, of which 0.32° is the orbital plane's own regression and the rest is the planet turning underneath: the vehicle comes back to nearly the same place in inertial space and the place has moved. The period used is the nodal one, 92.899 min against the Keplerian 92.970: J₂ makes the two differ by 4.31 s, which is 0.018° of walk per revolution and 102° in a year — the difference between a repeat track and a track that used to repeat. The map is equirectangular and therefore wrong about area everywhere; what it is right about is longitude difference, which is the whole of what this figure measures.

The line under a satellite

A ground track is an orbit seen from a frame that is turning, so every pass lands west of the last one. The track closes only when two periods are commensurable — which turns "look at the same place every day" into a condition on the altitude.

Two outcomes, and a boundary with no width. 26 trajectories launched from one point beyond L₂, all at the one speed the Jacobi constant C = 3.5124 permits there, differing only in the direction they set off in. The heavy curve is the zero-velocity boundary at that constant — the region no trajectory of this energy may enter — and it is open at L₂ by the neck the trajectories are aimed at. 11 of the 26 pass through into the secondary's realm and 15 turn back, and they are not interleaved — sweeping the launch direction through 65° finds one changeover and nothing in between. Bisecting the first of them pins it to 2.6e-12 radians, and the integrator runs out of digits before the boundary runs out of sharpness. That surface is the tube. It is the stable manifold of the periodic orbit about L₂, it separates transit from non-transit everywhere and not only in this fan, and a mission that wants to arrive for nothing has to be put inside it.

The tube that leads out of a neck

Below a certain energy the forbidden region opens at a Lagrange point, and a trajectory may pass. Which ones do is decided by a surface with no width at all — and two tubes that meet give a transfer that costs nothing at the join.

A burn along the track moves the chaser 8330 m backwards. Three 0.5 m/s impulses from rest alongside a target in a 400 km circular orbit, followed for 2 revolutions in the frame riding on the target. Along-track distance runs across the page with the direction of travel to the left, and radial distance up. The prograde burn ends 8330 metres behind after one revolution — exactly 6πΔv/n, and it is behind rather than ahead because the burn raised the orbit and a higher orbit takes longer. The retrograde burn ends 8330 metres ahead by the same arithmetic with the sign reversed. The radial burn is the third case and the strange one: it opens a closed loop and returns exactly to where it started after a revolution, having gone nowhere at a cost of 0.5 m/s. That is not a curiosity but the basis of the R-bar approach, in which a vehicle closes on a station from below along a path that costs nothing to abandon.

A burn that moves the wrong way

In the frame riding on an orbiting target, a thrust along the direction of travel leaves a chaser eight kilometres behind after one lap, a radial thrust returns it exactly to where it started, and every free relative orbit is the same ellipse — twice as long along the track as it is across.

Along a contour is free; across one costs, and 6 resonances sit on this one. Perihelion against aphelion, in units of Jupiter's orbit, with contours of constant v∞ — which is the Tisserand parameter through v∞² = (3 − T)v_p². Every contour crosses the line r_p = r_a = 1, because a spacecraft has to be at the planet's orbit to have an encounter there at all, and every point on one contour is reachable from every other point on it with no propellant: a flyby rotates the v∞ vector without changing its length, which moves the pump angle and slides the spacecraft along the curve. A burn is the only thing that moves it between curves, and that is the whole economy of a gravity-assist tour. The diagonals are resonant orbits — 1:1, 3:2, 2:1, 5:2, 3:1, 4:1 with the planet — and each is a straight line of slope −1 because a resonance fixes the semi-major axis and r_p + r_a = 2a. They matter because a spacecraft on one comes back to the same place at the same time as the planet, which is what makes a second encounter possible without waiting for a chance alignment. The marked points are where the v∞ = 0.3 contour meets each: a tour is a walk along the highlighted curve from one to the next, and the arithmetic that decides whether it can be walked is the turn one flyby delivers. At 1.35 Jupiter radii and 3.9 km s⁻¹ that turn is 163°, so the longest step drawn here needs 0.1 encounters — which is why a real tour has dozens of flybys and why the ones with the largest steps are the ones that need a deep-space manoeuvre in between.

A map of the transfers that are free

Drawn as contours of perihelion against aphelion, the invariant that survives an encounter becomes a map. A flyby slides a spacecraft along its own contour and costs nothing; a burn is the only thing that moves it between contours — so tour design is reading a graph.

$An eight-hour pass, and a 351 m s⁻¹ sinusoid that is the whole of the angle. Above: the range rate a two-way Doppler measurement returns over one pass from Goldstone, for a spacecraft receding at 14.6 km s⁻¹. Nothing here is an angle. The measurement is the fractional shift of a carrier the spacecraft coherently turned around and sent back, and its interpretation is that the distance is changing at some rate. Below: the same data with the spacecraft's own smooth signature removed. What is left is a sinusoid of exactly one cycle per day — the station's own motion, carried east at 379 metres a second by the rotation of the Earth, projected onto the line of sight. Its amplitude is that speed times cos δ and returns a declination of 22.0°; its zero crossing is the moment the spacecraft passed the meridian and returns the right ascension. The Earth's rotation is the interferometer. With Doppler good to 0.05 mm s⁻¹ at a 60-second cadence, 480 samples fit that amplitude to 0.003 mm s⁻¹ and the declination to 23 nanoradians — which is 4.7 milliarcseconds, from an instrument with no image plane and no angular resolution of any kind. What the picture cannot show is the part that makes this hard in practice: the spacecraft's own signature is not a straight line but a trajectory with unmodelled accelerations in it, and separating a slow non-gravitational force from a slow drift in the angles is the whole art of the fit.$ Spaceflight

A position measured from a frequency

A spacecraft is unresolvable and unreachable, and everything known about where it is comes from two scalars — a round-trip light time and a Doppler shift. Neither is an angle. The orbit solution returns two angles anyway, because the antenna is bolted to a rotating planet.

Four contact points, and what they fix

The depth of a transit gives a radius ratio. The shape gives the impact parameter, and then — through nothing but Kepler's third law — the mean density of the star being crossed.

Pluto and Triton sit on the nitrogen line, and the Earth sits between helium and nitrogen. Escape speed against exospheric temperature, with a criterion line for each molecular species at v_esc = 6 v_th — the speed at which the Jeans loss time is comparable to the age of the solar system. Every line has slope one half, because the thermal speed goes as √T; a body above a line keeps that gas and a body below it does not. The horizontal axis is the temperature at the exobase, which for the Earth is near 1000 K rather than the 255 K of its equilibrium — the difference is extreme-ultraviolet heating, and using the wrong temperature puts the Earth above the hydrogen line, keeping an atmosphere it observably lost. Three readings are worth making. The Earth falls between helium and nitrogen and does exactly that: it loses helium as fast as radioactive decay supplies it, and keeps nitrogen for ever. Titan sits just above nitrogen and just above methane, which is why it has a thick nitrogen atmosphere and is slowly losing its methane. And Pluto and Triton sit on the nitrogen line, within 1 per cent — which is why both have atmospheres that are marginally bound and measurably escaping. Where it fails it fails in one direction only. Mercury, the Moon and the Galilean satellites all plot above lines for gases they do not have, because retention is necessary and not sufficient: a body also needs a source, and needs to survive the non-thermal losses this criterion says nothing about. Venus is the sharpest case — it sits above the hydrogen line and has still lost an ocean, because that hydrogen left by charge exchange with the solar wind rather than by moving fast enough.

The gas a planet cannot keep

Escape is a statement about the tail of a distribution rather than about its mean, so the threshold is not a speed but a dimensionless number near thirty — and past that number the loss rate falls by twelve orders of magnitude. Then, for a hot Jupiter, the whole picture fails and the atmosphere leaves as a wind.

The line beyond which ice counts as rock

A disc of gas around a young star gets colder outwards, and at about a hundred and seventy kelvin water stops being vapour and becomes a building material. Crossing that one line multiplies the solid mass available by roughly three and a half, in a single step, and the architecture of every planetary system is downstream of it.

Where a 2-armed pattern is allowed to exist in the Milky Way. Angular speed against radius, with the two combinations that matter for a pattern of 2 arms. A wave turning at 25 km/s per kpc corotates with the stars at 8.7 kpc, has an inner Lindblad resonance at 1.9 kpc and an outer one at 13.8 kpc. Between the Lindblad radii the wave propagates; at them it is absorbed. Every one of these radii is found by bisection on the curves as drawn, not quoted — and none of them is a property of the pattern alone, because κ comes from the rotation curve, which comes from the mass.

Where a pattern is allowed to turn

A spiral wave has one pattern speed, and that single number decides where in the disc it can exist at all. The boundaries are set by the frequency at which a star rocks radially about its circular orbit — a frequency the rotation curve already contains.

A delay of 81 days, and a sheet nobody can see that moves H₀ to 82.4. Above: the arrival-time surface of a lensed source, along the line through the lens. The curve is the Fermat potential in days — the geometric cost of taking a longer path, minus the gravitational cost of climbing out of the potential — and the images sit at its stationary points, at -1.20″ and 2.04″, which for an isothermal sphere is β ± θ_E. Fermat's principle is doing all of the work here: light does not take the shortest path or the quickest one, it takes every stationary one, and the number of images is the number of stationary points. The vertical distance between the two is 81 days, and it is measurable — the source is a quasar, quasars vary, and the same wiggle appears in one image and then the other. That single number carries an absolute distance: the delay is D_Δt/c times a dimensionless function of the lens model, and D_Δt goes as 1/H₀, so a monitoring campaign gives the Hubble constant with no calibrated distance beneath it. Below: the two light curves, shifted by exactly that delay. The dashed curve is the second image with the delay removed, and the agreement is the measurement. What the picture also shows is the reason the answer keeps moving. The second arrival-time curve is the same lens with a uniform sheet of convergence added and the source moved to compensate: every image sits at the same place, every flux ratio is the same, every image shape is the same, and the delay is λ = 0.85 times as long. A lens model fitted to positions alone cannot see the sheet, and inferring H₀ from the same delay under it gives 82.4 instead of 70 — a 15 per cent shift with no observable attached. Breaking it needs a mass measured some other way: the velocity dispersion of the deflector, or a count of everything else along the line of sight.

A distance measured with a stopwatch

The images of a lensed quasar sit at the stationary points of an arrival-time surface, and the height between two of them is a delay in days. That delay is proportional to a distance, and the distance is proportional to one over the Hubble constant — so a flickering quasar gives H₀ with no calibrated distance under it.

A sphere reconstructed as a spheroid, and two distortions 0.15 apart. Left: the acoustic scale in the plane of separation across the line of sight against separation along it, one quadrant of it. In the cosmology that actually holds, the sound horizon is a sphere of 99.0 h⁻¹ Mpc and its locus here is a quarter circle. That is the whole content of the Alcock–Paczyński test: nothing about the early universe distinguishes the radial direction from the transverse one, so any departure from a circle is a statement about the observer's arithmetic rather than about the ruler. Converting angles into transverse separations needs the transverse comoving distance and converting redshift intervals into radial ones needs H(z), so assuming distances 1.1 times too large and rates 0.94 times too small returns an ellipse of axis ratio 0.855 — and the ellipticity measures that distance times the expansion rate over c, in which the sound horizon has cancelled. A ruler of unknown length still measures a shape. The third curve is the difficulty: peculiar velocities also distort the same correlation function along the same axis, squashing it by 1/(1+β) = 0.704 for β = 0.42, and a squashing is a squashing. Separating a geometric distortion from a dynamical one is the entire art of the measurement, and it is done by using the fact that they have different dependences on scale — the velocities act on the broad-band shape and the ruler is a feature. Right and below: the two numbers the same feature gives at each redshift. Across the line of sight, the transverse distance divided by the sound horizon; along it, c divided by the expansion rate times the sound horizon. Two functions of the expansion history, from one bump in one correlation function, and their agreement with a single model is one of the sharper consistency tests in the subject.

A ruler measured along and across

The sound horizon is a sphere, and a sphere in a redshift survey is measured twice over — across the line of sight it gives an angle, along it a redshift interval. Two different functions of the cosmology out of one feature, and their ratio is a measurement with no ruler in it at all.

Capture is a direction, not a strength

A resonance holds a body that drifts into it from one side and lets go of one that drifts out the other way, and the asymmetry is not about how strong the resonance is. It is the sign of a derivative — whether the trapped region is growing or shrinking — which is why a chain of planets in resonance is direct evidence that they migrated toward each other.

An invariant that is only almost one

The Tisserand parameter survives a close encounter with Jupiter exactly, and comet families are separated by boundaries in it drawn to two decimal places. The exactness holds for a Jupiter on a circle. Jupiter's eccentricity is 0.0489, and integrating the same encounter twice shows what that costs.

A duration that measures an eccentricity

A transit's length is a measurement of how fast the planet was moving when it crossed, and that speed depends on where it was on its orbit. For a circular orbit the duration gives the star's density; for an eccentric one it gives the density times a factor of up to four — and if the density is known independently, the factor is the eccentricity.

Windows every 780 days, costing between 5 and 13. The cheapest departure energy available in each of 8 consecutive launch opportunities, each found by solving a grid of Lambert problems around the window and taking the minimum. Opportunities recur every 780 days — the synodic period of Earth and Mars, which is exact and which is why the interval between missions is always about twenty-six months. Their cost is not periodic on that interval. The cheapest here is 5.0 km²/s² and the dearest 12.9, a factor of 2.59, and the pattern repeats on a period of about fifteen years rather than on the synodic one. The cause is Mars's eccentricity of 0.093: a transfer that arrives near Mars's perihelion has less distance to cover and meets a faster-moving planet, and whether an opportunity does that depends on where Mars is in its own orbit — which drifts relative to the synodic cycle by a fixed amount each time and comes back into phase after seven windows. In launch mass the factor is larger than it looks: departure energy enters the rocket equation through an exponential, so a C₃ of 13 rather than 5 costs roughly 1.08 times the propellant at departure.

The window that comes back and the cost that does not

Launch opportunities to Mars recur every 780 days exactly, because that is the synodic period and a synodic period is arithmetic. What they cost does not repeat on that interval at all — the cheapest window is a factor of two and a half below the dearest, and the pattern comes back every fifteen years rather than every two.

$The best split of a plane change between perigee and apogee, for three turns. How much is saved by moving part of the plane change into the perigee burn, against how much is moved, for turns of 15°, 28.5°, 51.6°. Doing the whole rotation at apogee is the standard answer and it is not the cheapest one: at perigee the turn is bought as a small correction to a burn that is happening anyway, so the first fraction of a degree is nearly free while the apogee saving is linear. Each curve therefore rises to an interior maximum — 10 m/s at 1.35° for a 15° turn, 25 m/s at 2.23° for a 28.5° turn, 40 m/s at 2.88° for a 51.6° turn — and falls back through zero at about twice that split. The saving is small against a 4.78 km/s budget, and it is free.$ Spaceflight

A rotation split between two burns

The standard answer is to do the whole plane change at apogee, where the vehicle is slowest. It is not the cheapest answer, and the reason is that a small part of the turn bought at perigee is a second-order correction to a burn that is happening anyway.

Rotating an orbit by flying away from it first. Total Δv against the angle turned, in units of the circular speed of the starting orbit, for two ways of rotating an orbital plane at a fixed radius. The single combined burn does everything at once and costs √(v₁² + v₂² − 2v₁v₂cos Δi); the three-burn route raises the apoapsis to 200 starting radii, turns there where the speed is only 0.7 per cent of what it was, and comes back down. Below 48.9° the single burn is cheaper and the two extra burns are not worth paying for. Above it the three-burn route wins, and it wins by more the larger the angle: at 90° it costs 0.8314 against 1.4142, a saving of 41 per cent. The mechanism is the one thing worth carrying away. A plane change costs 2v sin(Δi/2) and is therefore proportional to the speed at which it is done, so the cheapest place to turn is the slowest place available — and a vehicle can make a slow place by climbing, at a cost that does not grow with the angle while the rotation's cost does. That is why the crossover is an angle rather than a distance, and why it exists at all. What the figure does not price is time: the round trip to 200 starting radii takes 2015 times the period of the starting orbit, which for a low Earth orbit is months.

Flying further away in order to turn

A plane change costs the speed at which it is done, so the cheapest place to turn is the slowest place available — and a vehicle can make a slow place by climbing. Above about thirty-nine degrees the round trip pays for itself, and the crossover is an angle rather than a distance.

Which inclinations a launch site can reach, and which it cannot. Orbital inclination against launch azimuth for three sites, from cos i = sin A cos φ. Due east is the only azimuth that gives the minimum, and that minimum is the latitude itself: Kourou 5.2°, Kennedy 28.5°, Baikonur 45.6°. Everything below the shaded line is unreachable from the highest-latitude site by any azimuth at all, and getting there costs a plane change afterwards — 5.94 km/s from a 400 km orbit to reach the equator from 45.6°. Baikonur in fact flies no lower than 51.6° rather than its 45.6°, and the extra 6.0° is overflight constraint rather than mechanics: the azimuth that would give 45.6° sends the spent stages over places they may not fall on.

A plane change paid at the worst speed there is

A launch reaches an inclination fixed by its latitude and its azimuth, and no azimuth reaches an inclination below the latitude. Getting there afterwards means turning at orbital speed, which is the most expensive place available — so a site's latitude is a floor no trajectory removes.

The observed sky

Two fictitious suns in sequence

The equation of time is usually described as one quantity with two causes. It is better read as two separate reductions, each with its own imaginary body — one that moves uniformly along the ecliptic and one that moves uniformly along the equator — and the second is not the first.

Only two force laws let an orbit come back

That a planet returns to the same point of its own path after one lap is not a fact about orbits. It is a fact about the exponent in the force, and out of the whole continuum of attractions only two — the inverse square, and a spring — bring every bound orbit back to where it started.

$The solid mass available, as a staircase of 5 fronts. The share of the condensable material that is solid, against distance from a 1 solar-mass star, on a logarithmic radius axis. Each riser is one species freezing out, at the radius where the disc's temperature — falling as the inverse square root of the distance — reaches that species' condensation point. silicates and iron at 1400 K and 0.04 AU; water ice at 170 K and 2.71 AU; carbon dioxide at 70 K and 16.00 AU; methane and ammonia at 30 K and 87.11 AU; carbon monoxide at 20 K and 196.00 AU. Inside every front the solid surface density is 22 per cent of what is available, which is the refractories alone; water alone contributes 53 per cent, more than twice everything else combined, which is why one of these steps is called the snow line and the others are not. A body's composition is decided by which pair of risers it formed between, and the steps are narrow because a vapour pressure is exponential in the inverse temperature.$ Exoplanets

There is not one line, there is a staircase

Water freezing is the biggest step in a protoplanetary disc and it is one of five. Each condensable species has its own temperature and therefore its own radius, and what a body is made of is decided not by which side of a line it formed on but by which pair of risers it formed between.

Every method prefers a circle, and not for the same reason

A transit is more likely on an eccentric orbit and shorter when it happens, and the two very nearly cancel. A velocity curve loses amplitude to harmonics no sinusoidal search is looking at, and that one does not cancel at all.

The sunniest place on Earth is the summer pole. Daily-mean insolation against latitude, on the June solstice, an equinox, the December solstice, computed from Q = (S₀/π)(ā/r)²[H₀ sin φ sin δ + cos φ cos δ sin H₀] with the half-day angle clipped to a whole day inside the polar circle. On the June solstice the north pole receives 524 W/m² and the equator receives 385 — a third more, at a place where the Sun never climbs above 23.4°. Both halves of the product move: the Sun is low, so each square metre catches little; and it never sets, so the catching goes on for twenty-four hours. The second wins. That is why the daily total and the noon altitude disagree about where summer is strongest, and the daily total is the one an ice sheet answers to. The two hemispheres are not equal either. The Earth is nearest the Sun in early January, so the December solstice delivers 559 W/m² to the south pole against the north pole's 524 in June — 6.7 per cent more. What this cannot show is any temperature. Antarctica is the coldest place on Earth and receives the most sunlight of anywhere on any day; between the insolation and the temperature sit an albedo, an altitude and an ocean.

The observed sky

The sunniest place is the summer pole

On the June solstice the north pole receives 524 watts per square metre averaged over the day, and the equator 385. The Sun there never climbs above 23.4° and never sets, and the second wins.

One square root that raises the orbit and turns it. The velocity budget from a 300 km circular orbit to geostationary, against the plane change carried out along the way. Edelbaum's closed form — √(v₁² + v₂² − 2v₁v₂cos(½πΔi)) with Δi in radians — puts the whole continuous manoeuvre in one square root, and at Δi = 0 it collapses to |v₁ − v₂| = 4.651 km/s, which is the spiral's cost with no plane change in it. The two-impulse curve puts its rotation into the circularisation burn at apoapsis, where the vehicle is moving at 1.608 km/s and a rotation is cheap. At 28.5° the continuous transfer costs 5.951 km/s against the impulsive 4.256 — the plane change adds 1.300 to one and 0.363 to the other. That is the opposite of the usual claim that low thrust turns for free. It turns continuously, which is not the same thing: the gain is that the propellant is not the budget, and the Δv is worse here as it is everywhere else.

One square root that raises the orbit and turns it

Edelbaum put the plane change inside the same radical as the raise, and the half-pi in its cosine is the whole result — a continuous turn costs π/2 times an impulsive one below 140° and less above it.

$There is a best exhaust speed, and the calendar picks it. Payload fraction against exhaust speed for a 11 km/s mission, at three thrusting durations, with a power plant of 0.025 kilograms per watt of jet power. The vehicle is payload plus power plant plus propellant and the arithmetic is one line: λ = e^(−Δv/c) − (αc²/2t)(1 − e^(−Δv/c)), the first term the rocket equation and the second the mass of the machine that makes the jet. They pull opposite ways. A slow exhaust burns propellant; a fast one needs power, and the power per newton rises in proportion to c, so the plant's mass rises as c². Neither end is where anybody builds, and the optimum is 3,211 s over 200 days, 6,523 s over 700 days, 11,424 s over 2000 days — the same mission, the same Δv, and the best engine for it changes by a factor of 3.6 depending only on how long there is to do it. A gridded ion engine at 3,100 seconds sits at 30.4 km/s, which suits the shortest of these and is slow for the longest. What the curve cannot show is that α is not a constant either: a solar array's mass per watt falls as the fourth power of the distance from the Sun, so the same vehicle is a different point on this plot at Mars and at Jupiter.$ Spaceflight

The engine is chosen by the calendar

A chemical stage's exhaust speed is fixed by chemistry. An electric one's is a dial, and turning it up costs power — so there is a best setting, and it is decided by how long the mission has rather than by how far it is going.

A sail has to be tilted, and tilting it throws most of it away. The thrust on an ideal flat sail, resolved into the orbit frame, against the angle between the sail's normal and the sunline. The force is along the normal and goes as cos²α — one cosine for the area the sail presents to the light, one for the momentum the reflection returns along the normal — so the radial component goes as cos³α and the transverse one as cos²α sin α. A sun-facing sail has no transverse push at all. Its thrust is purely outward and falls as 1/r² exactly as solar gravity does, so it merely replaces μ with μ(1 − β): the orbit stays the same conic with a smaller central mass, and the vehicle raises nothing. Every manoeuvre a sail makes it makes by tilting, and the transverse push peaks at 35.26° — arctan(1/√2), differentiated rather than tabulated — where it is 0.385 of the face-on force, or 2/(3√3). Two thirds of the thrust is the price of pointing any of it somewhere useful. The lightness number β is the sail's whole specification, radiation pressure and gravity both falling as 1/r² so their ratio is a constant: IKAROS, 2010 at 1607 g/m² gives β = 9.5e-4; LightSail 2, 2019 at 156 g/m² gives β = 9.8e-3; a 5 µm film with no structure at 7 g/m² gives β = 0.219, against the 1.53 g/m² at which the Sun would push as hard as it pulls. What no figure here can show is the thing a sail actually has instead of a rocket equation, which is nothing: the exponential that limits every other vehicle is absent, and what limits this one is a structure that has to hold a square kilometre of film flat.

A drive with no rocket equation

Radiation pressure and solar gravity both fall as the inverse square, so their ratio is a constant of the vehicle. A sun-facing sail therefore only rescales the central mass — it has to be tilted to do anything, and the best tilt throws away sixty-two per cent of the thrust.

Venus is brightest 51 days from inferior conjunction, 38 per cent lit. The brightness of Venus through one synodic period of 584 days, in magnitudes below its brightest, against days from superior conjunction — inferior conjunction at the two ends of the axis. The planet is treated as a matte, Lambert-scattering sphere on a circular orbit of 0.723 AU, so its flux is its phase function divided by the square of its distance from the Earth. The two factors fight: near inferior conjunction the planet is closest but shows only a thin crescent, and near superior conjunction it is fully lit but 1.72 AU away. For this orbit the contest has an interior winner. The brightest moment is 51 days either side of inferior conjunction, at an elongation of 44.6° from the Sun, with 38 per cent of the disc lit and the planet 0.539 AU away; greatest elongation, at 46.3°, comes 71 days from inferior conjunction, after the brightness peak on the way out from inferior conjunction. At superior conjunction the planet is 0.82 magnitudes fainter than its best. The marked point is the observed greatest brilliancy, about 36 days from inferior conjunction at an elongation near 39° — closer to conjunction and to a thinner crescent than any matte sphere on this orbit can be brightest at. A surface that sends more light forward, towards large phase angles, would move the peak exactly that way. Very near either conjunction the planet is lost in the Sun's glare, and the curve there describes light nobody sees.

The observed sky

Brightest as a crescent, and not as a disc

Venus is fully lit when it is furthest away and nearest when it is barely lit, and it is brightest in between, as a crescent weeks from inferior conjunction. A matte planet only has such a peak if its orbit is wider than about 0.46 of the Earth's — Mercury's is not, and Mercury is brightest full — and Venus's real peak sits closer to conjunction than any matte sphere allows, which is its clouds throwing light forward.

What is left at infinity, and what it costs from 200 km. The velocity change needed to leave Earth from a circular orbit 200 kilometres up, against the speed the vehicle still has when the pull has run out. Escaping with nothing left over costs 3.22 km/s. Arriving at infinity with three kilometres a second in hand costs 3.63 — an extra 0.40 km/s to buy three, a discount of 87 per cent, because the two speeds do not add and their squares do — the departure speed is √(2μ/r + v∞²) — and adding a small square to a large one barely moves the root. The curve is therefore almost flat where the planetary missions are and straightens to slope 0.79 at the right-hand edge, where v∞ has overtaken escape speed and each metre a second bought costs a metre a second spent. The practical reading is that the expensive part of an interplanetary departure is getting to escape, and the destination is a surcharge on it.

The speed that is left over

No mission is designed to sit on the escape threshold. What an interplanetary departure is specified by is the speed still remaining when the planet's pull has run out — and because energies add while speeds do not, the first three kilometres a second of it cost four hundred metres a second to buy.

The relative orbit when the target's own is not circular. Free relative motion of a chaser about a target, over one revolution, for four eccentricities of the target's orbit — drawn by propagating two exact Keplerian orbits and differencing them in the target's own rotating frame rather than by any linearised solution. The chaser differs from the target by 0.0035 in eccentricity, so the two semi-major axes are identical. Along-track distance runs across the page. At e = 0 the path is the familiar closed ellipse, 2.00 times as long along the track as across it — recovered here from an exact propagation, which is the check that the calculation is right. Raise the target's eccentricity and the path stops being an ellipse and starts breathing: the separation varies by a factor of 2.00 over one revolution at e = 0 and 2.26 at e = 0.7, because the frame itself is now turning at a rate that changes through the orbit and the chaser is swept around it fastest at periapsis. Every one of them still closes exactly, and that is the one statement of the circular theory that survives — two orbits of equal semi-major axis have equal periods and cannot drift apart, whatever else differs between them.

The relative orbit that breathes

Every result in the close-approach frame — the two-to-one ellipse, the burn that moves the wrong way, the eight kilometres behind — assumes the target's own orbit is circular. Around an eccentric target none of the geometry survives. One statement does, and it is about energy rather than shape.

The spike, and where one passage through the planets sends it. The distribution of the reciprocal semi-major axis 1/a of long-period comets, in units of 10⁻⁴ AU⁻¹ and each curve scaled to its own peak. The narrow shaded peak is the Oort spike: comets arriving from the Oort cloud for the first time, with 1/a of about 4·10⁻⁵ AU⁻¹ and a spread of a few times 10⁻⁵. The broad curve is where those same comets go after one passage through the planetary region, whose perturbations change 1/a by a typical 5·10⁻⁴ AU⁻¹ — ten times the spike's width. 47 per cent end with negative 1/a, on hyperbolic orbits that leave the Solar System; only 4 per cent remain inside the spike's width. A comet in the spike is therefore on its first passage, almost by definition: the planets cannot put a comet back into the spike once it has been through them.

A spike only first-time visitors can be in

Long-period comets arriving from the Oort cloud have orbital energies packed into a spike a few hundred-thousandths of an inverse astronomical unit wide. One passage through the planets changes that energy by ten times as much, so no comet can be in the spike twice — and the spike's position, near thirty thousand astronomical units, is where the Galaxy's own tide is just strong enough to lift a comet's perihelion over Jupiter and Saturn in a single orbit.

The cheapest circular orbit to stop in at Mars is not the lowest. The speed change needed to capture into a circular orbit round Mars, in km/s, against the orbit's radius in Mars radii, for arrival excesses of 2.65, 4, 6 km/s — the first being a minimum-energy transfer from the Earth. One burn at periapsis, where the spacecraft is fastest, turns the hyperbola into the circle. The cost has a minimum, at a radius of 2μ/v∞² from the planet's centre, where it is exactly v∞/√2: 3.6 radii and 1.87 km/s for 2.65 km/s; 1.6 radii and 2.83 km/s for 4 km/s; 0.7 radii and 4.24 km/s for 6 km/s. Lower orbits cost more because the circular speed there is higher than the gain from arriving deeper in the well; higher orbits cost more because the burn is made where the spacecraft is moving slowly and the Oberth effect is lost. A faster arrival moves the optimum inward, and at a large enough excess it moves below the planet's surface, where every circular orbit is on the expensive side of the minimum.

The cheapest circle to stop in

A spacecraft arriving at a planet has to lose speed to stay, and the cheapest circular orbit to lose it into is not the lowest. It sits at a radius set only by the arrival speed — twice the planet's gravitational parameter divided by the square of the excess — where the braking costs exactly the excess divided by the square root of two. Most orbiters want something else, and the arithmetic of what they do instead decides which planets are cheap to stop at and which are not.

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

A wall across the porkchop

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

What it costs to reach Mars faster than the Hohmann transfer. The total velocity change to go from a 400 km orbit round the Earth to a 400 km circular orbit round Mars (solid), and the heliocentric part alone (dashed), against flight time, for transfers that leave the Earth's orbit tangentially and cross Mars's before reaching their far point; both planets are on circular, coplanar orbits. The right-hand end is the Hohmann transfer: 259 days and 5.65 km/s. Shortening it to 200 days costs 0.82 km/s more; to 150 days, 3.4 km/s more; below a hundred days the cost climbs past twenty. The curve leaves the Hohmann point with zero slope — the cheapest transfer is a minimum, so the first days saved cost almost nothing — and turns steeply upward only once the transfer ellipse has become long and thin. For the fastest transfers the orbit-to-orbit cost is below the heliocentric one, because a burn made deep in a planet's gravity well adds more to the speed at which the craft leaves it than the burn itself.

The first weeks off a Hohmann transfer are nearly free

The Hohmann transfer to Mars takes 259 days and is the cheapest route there is. Because it is a minimum, the cost of going faster starts flat — ten days saved costs twenty metres a second — and only climbs steeply once weeks have become months. For a crew, the time is itself a cost, paid in radiation at about two millisieverts a day.

How far a small error at departure moves the arrival at Mars. The distance by which a craft on the Hohmann transfer from the Earth's orbit to Mars's misses its aim point at the nominal arrival time, against an error in its departure speed, on logarithmic axes, integrated numerically for the 259-day flight. An error along the direction of motion (solid) moves the arrival by 76 km for every millimetre a second; one across it (dashed) by 23 km. The relation is linear over the whole range drawn: a metre a second at departure — the precision of a good launch — is 76 thousand kilometres at Mars, several times the planet's own radius and a thousand times what a landing needs. The ratio is the corresponding element of the state transition matrix, and it is the number every navigation plan is built around.

A millimetre a second at launch, seventy-six kilometres at Mars

A craft on its way to Mars arrives seventy-six kilometres from where it was aimed for every millimetre a second its departure speed was wrong. No launch is that good, so no trajectory is flown as designed; it is flown as corrected, by Newton's method run on the matrix that says how the arrival depends on the departure — which doubles its correct digits at every step and makes a two-body sketch into a real trajectory in two iterations.

A cloud of fragments that pinches where it was born. 240 fragments thrown from a satellite on a circular orbit 800 kilometres up, with kicks of about 100 metres a second in random directions in the orbital plane, each then following its own Kepler ellipse, drawn at 0.5, 1, 4.5, 20 orbital periods of the parent, with the Earth's centre marked and the break-up point at the top. Radial offsets from the parent's circle are exaggerated 2.5 times. Half a period after the break-up the fragments are spread 28 degrees along the orbit and 1374 kilometres in altitude, the kicks having carried some to higher apogees and others to lower perigees. After one period the along-track spread has grown to 53 degrees but the whole cloud is only 112 kilometres thick, and the 83 fragments within eight degrees of the break-up point span 35: every fragment's orbit passes through the break-up point, and the cloud pinches there once a period. By 4.5 periods it is 231 degrees long and 863 kilometres thick overall; by 20 it is 352 degrees long, still pinched to 11 kilometres where it began.

Every fragment comes back through the same point

When a satellite breaks up, its pieces fly off in every direction and every one of them returns, an orbit later, to the place where it broke. The cloud pinches there once a period, stretches along the orbit at a rate set by one component of each kick, and becomes a ring in hours.

Lambert's time of flight in the semi-major axis and in Lancaster and Blanchard's variable. The non-dimensional time of flight T = t√(2μ/s³) of the transfer between radii 1 and 1.524 through 135 degrees, against two variables that label the same family of orbits. Left: against the semi-major axis in units of the minimum-energy ellipse's. The family is two curves — the fast branch, which passes the empty focus on one side, and the slow branch, on the other — joined at a = aₘ with T = 1.566, where both have a vertical tangent: the time changes infinitely fast with the semi-major axis there, and every a above aₘ corresponds to two times. Right: against x, the variable of Lancaster and Blanchard, in which a = aₘ/(1 − x²). The same family is one smooth, decreasing curve through T = 1.566 at x = 0 to the parabolic time 0.662 at x = 1, with no tangent vertical and every time given by exactly one x. Nothing about the orbits differs between the panels; only the label on them does.

The variable that makes Lambert's problem easy

Written in the semi-major axis, the time-of-flight equation has two branches joined by a vertical tangent, and a solver near the join struggles. Written in a variable Lancaster and Blanchard chose in 1969, it is one smooth curve — and every trajectory library now solves it that way.

The three terms of the minimum disposal height. The minimum height above the geostationary ring to which a satellite should be raised at the end of its life, by the international guideline ΔH = 235 km + 1000 · Cᵣ · A/m, broken into its three terms, for reflectivity times area-to-mass ratios of 0.005, 0.01, 0.02, 0.03, 0.05 square metres per kilogram. The first 200 km is the protected zone itself; the next 35 km is the most that the Moon, the Sun and the Earth's uneven gravity lower an orbit's perigee over time; the rest is the eccentricity that sunlight raises every year, which a derivation from first principles puts at 942 km per unit of Cᵣ·A/m for an orbit started circular — the guideline's round thousand. Totals: 240 km, 245 km, 255 km, 265 km, 285 km. The term that depends on the satellite is the one sunlight sets, and for a light satellite with large solar arrays it is the largest.

A graveyard height with a term for sunlight

A geostationary satellite at the end of its life is raised about three hundred kilometres and switched off. The height is written as a formula — 235 kilometres plus a thousand times its reflectivity and area-to-mass ratio — and each term is a separate piece of orbital mechanics, the last one sunlight bending the orbit a little every year.

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

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.

One detection, and every density it could mean. The number density of interstellar objects implied by one detection of a given size in 8 years, against the distance from the Sun inside which a survey could have seen it, for objects arriving at 26.3 km/s, if the survey caught 100 per cent, 10 per cent, 1 per cent of those passing inside that distance. Each estimate is the count divided by the rate of passages per unit density, πq²v∞ with gravitational focusing, times the efficiency. If ʻOumuamua-sized bodies could be seen inside 0.5 AU at 10 per cent efficiency, there are about 0.05 per cubic AU; if inside 2 AU with every one caught, 7.9 × 10^−4. The inferred density is a statement about the survey as much as about the visitors.

A census of visitors is a census of the survey

Three bodies from other stars have passed through the inner solar system in eight years of looking. Turning that count into a number of such bodies per cubic astronomical unit needs the volume in which they could have been seen, and that volume is set by the Sun's gravity, by each body's size and by how many weeks a survey had to catch it — so the first number the census gives is a measurement of the survey.

A plane change borrowed from the Moon

Turning an orbit's plane by the engine costs a fixed fraction of the spacecraft's speed, which is why it is done as far from the Earth as possible. Farther still, at the Moon's distance, a spacecraft is moving so slowly that the Moon can turn its velocity through tens of degrees for nothing — but a flyby close enough to turn it hard also flings it outward, and the useful flyby is a gentle one, in a window of heights the arithmetic picks out.

Two ways out of a transfer orbit. The eccentricity against the semi-major axis of a spacecraft spiralling out from a geostationary transfer orbit — perigee 300 km up, apogee at geostationary height — under a continuous thrust of 0.2 millimetres a second squared, integrated through hundreds of revolutions, for two steering laws. Thrust always along the velocity (red) raises the orbit efficiently but leaves its shape almost unchanged, reaching the geostationary semi-major axis after 64 days still with an eccentricity of 0.59 — an orbit swinging between 17,000 and 67,000 km. A law that points the thrust to reduce a weighted distance from the target in both elements (blue) circularises as it raises, and reaches a circular geostationary orbit in 129 days.

The thrust that rounds an orbit turns once a revolution

An electric engine pushes gently for months, and the obvious way to point it — along the velocity — raises an orbit as fast as possible and leaves its shape almost untouched. Starting from an elongated transfer orbit, that is the wrong job. Rounding the orbit off needs the thrust forward at apogee, backward at perigee and sideways in between, a full turn of the thrust vector every revolution, and the law that does it can be written down from the equations for how each element changes.

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

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.

A distant planet felt only as a stretch

An undiscovered planet far beyond Neptune pulls the Sun and Saturn almost equally, and an ephemeris, which measures only where Saturn is relative to the Sun, cannot see a pull that moves both together. What it can see is the difference — a tidal stretch falling as the mass over the cube of the distance — and most even of that is absorbed by refitting Saturn's orbit. What is left decides which planets the solar system can be hiding.

The kick speeds read back out of the periods: median 100 m/s against the 100 m/s put in. The distribution of kick speeds recovered from the fragments' periods alone, by p(v) = −2v g′(v) applied to the along-track kicks, for 300 and 3000 fragments (points), against the log-normal distribution actually used, median 100 m/s and width 0.3 dex (line), on a logarithmic speed axis with the density per logarithmic interval. With 3000 fragments the recovery follows the shape of the truth and puts the median at 100 m/s, though differentiating a measured distribution twice smooths its peak and roughens its tails; with 300 it is noisy but already in the right place. The inversion uses one assumption, that the kicks were isotropic, and one measurement per fragment, its period — the quantity tracking networks determine first and best.

A kick speed read off the slope of a histogram

Throw fragments at one speed in random directions and the component of each kick along the orbit is spread evenly between plus and minus that speed — Archimedes' theorem about the area of a sphere, applied to debris. Each fragment's period records that component, so the histogram of periods is a stack of flat bands, one per speed, and its slope gives the speeds back: a break-up weighed by its own violence from a catalogue of orbits, if the event was isotropic, and fragment by fragment if it was not.

Two clouds with the same kicks: the explosion's centre at −2.3 m/s, the collision's at 10.9. The along-track kicks read from the periods of 1500 fragments, for an explosion (blue) and for a collision (orange) whose fragments received the same random kicks plus a shared velocity change of 12 m/s along the track — the centre-of-mass motion a projectile's momentum leaves behind. The two histograms have the same width and shape; the collision's is displaced. The number-weighted mean of the explosion's kicks is −2.3 ± 2.6 m/s, consistent with zero, and the collision's 10.9 ± 2.5 m/s. The width of a cloud says how violent the event was; its centre says whether anything arrived from outside.

A collision moves the cloud's centre, and an explosion cannot

However violently a satellite blows itself apart, its fragments' momenta add up to the parent's: the centre of the cloud keeps moving exactly as the satellite did. A collision is different, because the projectile brings momentum in, and a collision energetic enough to shatter its target needs a projectile heavy enough to move the centre by at least 80,000 joules per kilogram divided by the impact speed — eight metres a second in low orbit. The width of a debris cloud measures how violent the event was. Its centre measures whether anything arrived from outside.

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

A climb out of the ecliptic paid for in orbits

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

Jupiter keeps only the visitors already nearly bound

An interstellar object passing Jupiter can lose enough speed to stay, but one flyby can only turn the visitor's velocity relative to the planet, never shorten it, and the turn removes at most a few kilometres a second from its speed around the Sun. So capture is common for visitors that arrive with less than a kilometre a second to spare and almost impossible for the speeds the three known visitors had. A census of the slow tail of the interstellar population, and the orbits a capture leaves — more than a third of them backwards — follow from the same arithmetic.

An orbit seen at an angle keeps everything but its focus

No orbit outside the solar system is seen face on. It is seen as a shadow on the sky, and the shadow of an ellipse is an ellipse, swept at equal areas in equal times, indistinguishable in shape and timing from a real orbit. Only one thing gives the tilt away. The star the companion circles projects to where the true focus projects, and that is not where the apparent ellipse has its focus.

A two-body arc flown in the real field. How far from its target a spacecraft arrives when it flies the velocity Lambert's two-body problem gives, through the Earth's real field with its equatorial bulge, against the angle of the transfer, from a 400 km orbit to a point on a 420 km orbit, on a logarithmic scale. In a pure two-body field every solution arrives within a metre of the target; exactly 180° is left out, where the two points and the Earth's centre are in line and the transfer's plane is not determined. In the real field the miss grows with the transfer: at 51.6° inclination, 4.2 km after 60° and 43.0 km after 300°. An equatorial orbit misses most, because the bulge's extra pull is strongest over the equator and acts on it all the way round; an inclined orbit feels it only part of the time but also has its plane turned.

An answer from two bodies, flown through a flattened Earth

Lambert's problem gives the exact velocity that carries a spacecraft between two points in a given time — exactly, for a spherical Earth. The real Earth bulges at the equator, by a thousandth of its central pull, and a spacecraft flying the exact two-body answer between two points in low orbit arrives tens of kilometres from where it was sent. The repair is to keep the two-body solver and lie to it about the target, and the lie converges by a factor of a hundred each time it is told.

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

A correction that tilts the arc it meant to aim

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

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.

A turn spread where the climb leaves room

An electric spacecraft spiralling from low orbit to geostationary has to remove its launch inclination on the way, and the classic steering law tilts its thrust out of the plane by the same angle all the way round each orbit. But out-of-plane thrust turns the plane only near the nodes, while in-plane thrust raises the orbit equally well everywhere. Letting the tilt follow the nodes — leaning hard where turning works and lying flat where it does not — saves 150 metres a second on a transfer from Cape Canaveral, with no coasting, because the climb fills exactly the parts of each orbit where the turn would be wasted.

A height the map already knows

A navigation service for the Moon's south pole needs only a handful of satellites, if they are put in elliptical orbits that hang for hours over the pole at the high point of each lap. But satellites hanging over a pole all stand at much the same height in the sky, on a cone, and four ranges from a cone cannot tell whether the receiver is lower or its clock is late. On the surface the receiver knows its height already, from a map of the terrain — and that one number, used as a measurement, does more for the service than two more satellites.

Two satellites moved apart by the attitude they hold. The along-track separation of two identical 5 kg satellites at 475 km, starting together, when one flies broadside to the airflow and the other edge-on for 15 days and then they swap, so that their relative drift ends at zero. The separation grows quadratically and then levels off. At solar minimum the manoeuvre moves the pair 2,797 km apart; mid-cycle 8,299; at solar maximum 24,620 — 206° of the orbit's 43,060 km circumference. No propellant is spent; the same command produces a tenfold different result depending on the state of the Sun.

A rendezvous paid for in air

Two satellites in the same low orbit can be moved along it relative to each other with no propellant at all, by turning one broadside to the thin air and the other edge-on. The draggier one sinks and speeds up, the gap grows as the square of the time, and swapping attitudes halfway brings the drift back to zero. The cost moves from fuel to time, to altitude, and to knowing an atmosphere that changes tenfold with the Sun.

Captures counted over the Sun's life, cluster and field together. The cumulative number of ʻOumuamua-sized bodies Jupiter has captured since the Sun formed, against time, for birth clusters of 300, 1000, 3000 stars per cubic parsec whose bodies were captured at 4.3 a year per unit density for 3·10⁷ years, followed by today's visitors at 7.2 × 10⁻⁴ a year (dotted). The cluster's 3·10⁷ years supply 99.5%, 99.9%, 100.0% of all the captures the Sun has made: 7.2·10⁸, 2.4·10⁹, 7.2·10⁹ bodies, against 3.3·10⁶ from four and a half billion years of fast strangers. Almost everything Jupiter has ever captured, it captured from its siblings.

What Jupiter kept of the Sun's siblings

The Sun was born among a few thousand other stars, and while their planets formed they flung out small bodies at a few kilometres a second — exactly the speeds at which Jupiter can capture a passing body. Counted honestly, the birth cluster supplied almost every capture Jupiter has ever made: billions of bodies in thirty million years, against a few million in the four and a half billion since. And billions is still a hundred-thousandth of the Sun's own comets. Whatever the siblings gave the Sun, Jupiter was not how it arrived.

Every orbit turns by about the same angle, whatever its size. How far the Earth's equatorial bulge turns an ellipse with a 300-km perigee in one revolution, in degrees, against its semi-major axis, logarithmic: the line of apsides (solid) and the node (dashed), for inclinations of 28.5, 51.6, 98°. The turn per orbit depends on the semi-latus rectum, which for an elliptical orbit lies between one and two perigee radii, so it changes by less than a factor of four from a near-circular orbit to one reaching the Moon's distance: at 28.5°, 0.56° of apsides per orbit at 8,000 km and 0.20° at 300,000 km. Per day the large orbit turns far less, because it completes so few revolutions; per revolution, almost the same.

An orbit lost to a burn that did not fire

A weak stage leaving the Earth splits its departure into several burns at perigee, coasting round a staircase of ever larger ellipses between them. If one burn does not fire, the stage goes round once more. That costs a revolution of time — an hour and a half early in the staircase, more than a day at the top — and a revolution of turning by the Earth's bulge, which comes out at nearly the same angle for every step. The angle is cheap to undo at the slow apogee of a large orbit. The time is not, when the target is the Moon.

The bands of sky from which each planet would be seen to transit. Ecliptic latitude against ecliptic longitude, with the vertical scale stretched 11 times: the band from which each of Mercury, Venus, Earth, Mars, Jupiter would be seen to cross the Sun's disc. The Earth's band lies along the ecliptic, ±0.266° wide. Mercury's is the widest, ±0.69°, and tilted by its 7° inclination; the giants' are narrow, ±0.051° for Jupiter. Each crosses the Earth's at the two nodes of its orbit, and only near those crossings can an observer see two planets transit. From nowhere on the sky can more than 3 be seen.

The stars that could catch the Earth in transit

The geometry that lets one Earth twin in two hundred be seen crossing its star works in reverse: the Earth crosses the Sun for observers in a band of sky half a degree wide, under half a per cent of the sky. About two thousand stars within a hundred parsecs sit in it. The band is fixed to the Earth's orbit and the stars are not, so the list changes on the timescale of recorded history — and from no direction can more than three of the eight planets be seen transiting at once.

How fast a meteor arrives is its orbit's inclination. The speed at which a meteoroid on a near-parabolic orbit with perihelion at 1 AU meets the Earth, against the orbit's inclination to the ecliptic: relative to the Earth far away (dashed) and at the top of the atmosphere after the Earth's gravity has added its pull (solid). A prograde orbit in the ecliptic overtakes the Earth at 12 km/s and enters at 17; a retrograde one meets it head-on at 72 and enters at 73. The points are four showers computed from their parents' elements at the node nearest 1 AU: Draconids 24 km/s at 31.9°, Geminids 38 km/s at 22.3°, Perseids 61 km/s at 113.5°, Leonids 72 km/s at 162.5°. A meteor's speed is a reading of the angle between its orbit and the Earth's.

A shower that is a reading of an orbit

Every August the Earth runs through the dust of comet Swift–Tuttle, and the meteors arrive at sixty kilometres a second from a point in Perseus. Each of those facts is an orbital element in disguise. The date is the longitude of the stream's node, the speed is its inclination, the radiant is the direction it comes from, and the shower exists at all only because one element happens to put a node within a hair of the Earth's orbit.

How much thicker the cable must be at the top. The ratio of an Earth elevator cable's cross-section at synchronous orbit to its cross-section at the ground, as a power of ten, against the specific strength of its material, for safety factors of 1 and 2. The taper is exp(ΔΦ/(σ/ρ)) with ΔΦ = 48.4 MJ/kg, the same exponential as the rocket equation. With a safety factor of 2: high-strength steel (0.25 MJ/kg) beyond 10³⁰; Kevlar (2.5 MJ/kg) about 6.7×10¹⁶; carbon fibre (3.6 MJ/kg) about 4.8×10¹¹; Zylon (3.8 MJ/kg) about 1.2×10¹¹; nanotube yarn, today (4 MJ/kg) about 3.3×10¹⁰; ideal nanotube (50 MJ/kg) about 7. Only a material with tens of megajoules per kilogram — near the ideal strength of a carbon–carbon bond, ten times anything yet made in bulk — gives a cable that is not absurd.

A cable with a rocket equation of its own

A cable hanging from synchronous orbit to the equator would let a climber reach orbit on electricity, with no rocket at all. But the cable has to hold its own weight, and a cable that holds its own weight at constant stress must thicken exponentially with the potential it spans. The exponent is the same 48 megajoules per kilogram a rocket must supply, divided by the material's strength per unit mass. The rocket equation has not been escaped; it has moved from the propellant to the rope.

A way home that needs no engine

A spacecraft can leave the Earth so that the Moon's gravity turns it round and sends it home with no further burn. Drawn in the frame that turns with the Moon the path is a figure of eight. It was flown to the Moon by every early Apollo mission for one reason: if the engine fails, the crew comes back. That safety is exquisitely narrow. A tenth of a metre a second at departure moves the perilune by tens of kilometres and the return by thousands, and the perilune it allows is not one that most landing sites can use.

Seven orbits in fifteen years, each turned from the last. The ellipse of an Aldrin cycler, semi-major axis 1.60 AU and eccentricity 0.390, in the Sun-centred frame, for 7 successive synodic periods, the line of apsides advanced by 48.7° each time. The inner circle is the Earth's orbit and the next is Mars's. Over about fifteen years the pattern has gone once round. The same spacecraft would need to be carried from each ellipse to the next by a flyby of the Earth, or a small burn. The drawing is the circular, coplanar case; the real one has to be recalculated for the actual orbit of Mars.

A timetable that orbits the Sun

Every trip to Mars leaves when the planets line up and accelerates everything it carries. Buzz Aldrin proposed in 1985 that there need be only one departure: a large ship on an orbit that returns to the Earth each synodic period, with crews riding small taxis out to catch it. The orbit almost exists without any propellant. It has a semi-major axis fixed by a single condition, an aphelion that passes Mars with nothing asked for, and one unavoidable bill: it must turn by about fifty degrees every twenty-six months, and the Earth can pay for all but seven of the eighty the turn requires.