Astronomy & Celestial Mechanics
the sky, computed
Concept

Escape velocity — where it appears

The speed at which a body's kinetic energy equals its gravitational binding energy, so that it does not return. It is the square root of two times the circular speed at every distance, which is a ratio no property of the body can change.

Named by 20 essays across 7 fields — each of them below, with the objects they name alongside it.

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

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

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

orbits · Conic sections 7 decades of relaxation time, and a Hubble time between them. Crossing time and two-body relaxation time for 5 self-gravitating systems, against the number of bodies in each. The lower marks are the crossing time R/σ, which spans 3 decades; the upper ones are t_relax ≈ 0.1N/ln N times it, which spans 8. The horizontal rule is a Hubble time, and the same expression puts these systems on opposite sides of it. Two of them relax — an open cluster and a globular cluster — so their stars have exchanged energy, their heavy stars have sunk, and their present structure is not the one they were born with. The rest do not: a dwarf spheroidal, an elliptical galaxy, a cluster of galaxies, the elliptical missing it by a factor of 10⁶, which is why a galaxy is modelled as a smooth potential with no stars in it at all. The dependence is on N and hardly at all on anything else, because the Coulomb logarithm makes every decade of impact parameter contribute equally. A cluster that relaxes also evaporates: about a stellar mass in a hundred and forty leaves per relaxation time, so the globular cluster empties in 6.1·10¹¹ years.

A cluster that boils itself away

A star cluster has no thermostat. Encounters between its members push a few of them above escape speed, the cluster loses them, and losing them makes it contract — which makes it hotter, which makes more of them escape. A self-gravitating system heats up as it loses energy, and the process ends by destroying the system.

gravitation · Two-body relaxation 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.

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

The cheapest place to turn

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

spaceflight · Plane change 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.

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.

spaceflight · Rocket equation Hard below 8.3 astronomical units, soft above, and nothing settles at the line. The binding energy of a binary of two 0.7 solar-mass stars against its separation, both axes logarithmic, with the mean kinetic energy of a single cluster star at a velocity dispersion of 5 kilometres a second drawn as a level. Where the curve is above the level the binary is bound more tightly than a passing star's motion, and encounters on average take energy out of the field and put it into the pair; where it is below, they do the reverse. The crossing at 8.3 astronomical units is the hard–soft boundary, and it is the entire content of Heggie's law: hard binaries harden and soft binaries soften. The arrows are the direction each side moves, and they point away from the crossing in both directions rather than toward it. That is not a coincidence but a negative heat capacity, the same property that makes a star contract when it radiates: taking energy out of a bound pair moves it closer together and speeds it up, so a hard binary that gives energy to the cluster becomes harder still and gives more. A cluster with binaries in it therefore has a heat source that turns itself up, and the boundary drawn here is a watershed rather than an equilibrium.

A pair that heats what is trying to cool

A star cluster has a negative heat capacity, so it cannot reach equilibrium — its core contracts and gets hotter without limit. What stops it is a binary, and which way a binary exchanges energy with the stars around it is settled by one comparison of two energies.

gravitation · Binary heating Two bodies at a mass ratio of 2.6 to 1. Both bodies orbit their common centre of mass, on similar ellipses whose sizes are in inverse proportion to the masses — here 2.6 to 1, so the heavier body's path is 2.6 times smaller.

The companion survives, and it is moving

When one star of a close pair explodes, the other is left holding an orbital velocity and no orbit. Whether the pair stays bound turns on a single question — did more than half the total mass leave? — and the stars that answer no are running across the Galaxy at a hundred kilometres a second with nothing visible behind them.

stars · Supernovae The ratio that is √2 only for a point. Escape speed divided by circular speed, against distance in units of the Sun's, for a point mass and for flat-rotation-curve haloes truncated at four different radii. The point-mass value is √2 at every distance and it is a floor rather than a rule. Mass outside an orbit contributes nothing to the circular speed — a spherical shell exerts no force inside itself — and everything to the depth of the well, so any extended distribution has a ratio above √2, by an amount that measures how much lies beyond. For a flat rotation curve the escape speed is not even defined without an outer edge: the potential grows as the logarithm of radius and diverges, so the ratio is √(2 ln(R/r)) with R the truncation radius, and the answer depends on a boundary nobody can see. The measured value near the Sun is about 2.27, which on this curve puts the edge at 13 times the Sun's distance — and the strength of that inference is exactly the weakness of the method, because a ratio of 2.50 would put it at 23.

The ratio that is not √2

Escape speed is exactly √2 times circular speed at every distance from a point mass, and the Sun's own escape speed from the Galaxy is 2.3 times its orbital speed. The excess is not an error in either number. It is a measurement of the mass that lies outside the Sun's orbit, which pulls on nothing and holds everything.

spaceflight · Escape 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.

spaceflight · Escape A push that depends on how fast the wind is already going. The force multiplier — the radiative push on a line-driven wind as a multiple of the push on its free electrons — against t, the electron-scattering optical depth across one thermal Doppler width, which is small where the wind accelerates steeply. The solid curve is an ensemble of spectral lines whose strengths follow a power law, the number of lines stronger than a given strength falling as that strength to the power 1 − α, with α = 0.6. If every line were optically thin the push would be the flat line at 2000, independent of the flow; if every line were thick it would fall as 1/t, set only by the velocity gradient. The ensemble lies between: over most of the range it falls as t to the power −0.59, because as t rises lines cross from thin to thick one after another. That single exponent is the whole of the theory — it fixes the terminal speed, the mass-loss rate and how both scale with the star — and it is a property of the atomic physics of the ions in the wind, not of the star.

A rate the wind has no choice about

A hot star's wind is driven by starlight caught in hundreds of thousands of spectral lines, and the way those lines are distributed in strength fixes everything else. The mass-loss rate turns out not to be a free parameter at all but the one value for which a smooth flow can leave the surface — and multiplying it by the wind's speed and the square root of the star's radius cancels the star's mass, leaving a relation with the luminosity alone.

starlight · Stellar winds 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.

spaceflight · Escape How often a merged black hole is kicked faster than a given speed. The fraction of black-hole mergers whose remnant recoils faster than the speed on the horizontal axis, for mass ratios spread evenly in logarithm between 0.1 and 1 and spins of 0.9 of the maximum, spins within 10° of the orbit; spins within 30°; spins in random directions, from a fitting formula to numerical relativity over 4000 random mergers each. The median kicks are 113, 234, 534 km/s. Faster than 100 km/s: 62%, 85%, 96%. Faster than 300 km/s: 5%, 40%, 74%. Faster than 1000 km/s: 0%, 2%, 25%. Gas accreted before the merger is expected to drag both spins into line with the orbit, which is why a gas-rich merger keeps its black hole far more often than a gas-poor one.

A black hole thrown out by its own merger

When two black holes merge, the gravitational waves they radiate carry momentum as well as energy, and unless the two are identical and spinning in line the waves leave more in one direction than the other. The remnant recoils, at anything from a few tens of kilometres a second to several thousand — and whether a galaxy keeps its central black hole comes down to how its spins were pointing in the last orbit before the merger.

galaxies · Galactic nuclei The binaries in a fed core are spread evenly in energy. The steady distribution of binding energies among the hard binaries of a core where new binaries enter at x = 5 and leave by ejection from a cluster whose escape speed is 4 times its dispersion: the number per unit binding energy, against binding energy in units of mσ², from 20,000 simulated binaries weighted by the time each spends at each energy. Between twice the entry energy and a third of the typical ejection energy (about 120) it is flat to within 5 per cent, and it falls away smoothly near ejection because a single encounter can throw a binary out from anywhere above about 24. A flat distribution is what a population with a constant flux through energy and a constant drift must have: binaries pass every energy at the same rate, and spend the same time at each.

The binaries a core keeps are spread evenly in energy

A hard binary in a star cluster gains binding energy at the same average rate however hard it already is. So in a core that is fed new binaries and loses hardened ones, every binding energy is passed through at the same speed, and the steady population is flat in energy — which, counted per factor of energy, means most binaries are within a factor of ten of being thrown out. Where they were born hardly matters; where they leave decides everything.

gravitation · Binary heating Ejected before touching, or touching before ejected. The separation at which a hardening pair of 0.7-solar-mass main-sequence stars is typically ejected — where one encounter's recoil exceeds an escape speed 4 times the dispersion, at binding energy about 120 mσ² — against the cluster's velocity dispersion, both axes logarithmic, beside the separation at which the two stars fill their Roche lobes and begin to merge, 0.0082 au (1.8 solar radii). The ejection separation falls as the inverse square of the dispersion, because a hotter cluster needs a harder binary to throw it out. The lines cross at a dispersion of about 18 km/s. In clusters slower than that — open clusters and nearly all globulars — a binary is ejected while its stars are still apart. In faster ones — the most massive globulars and the nuclear star clusters at galaxies' centres — it is driven into contact first, and its two stars become one.

Ejected from a globular, merged in a nucleus

A hard binary in a star cluster hardens until something ends it: an encounter that throws it out, or its own stars touching. Which comes first is decided by one comparison — the separation at which the cluster's escape speed is exceeded against the separation at which the stars meet — and the first shrinks as the square of the cluster's velocity dispersion. In a globular cluster binaries leave while their stars are still apart; in the nucleus of a galaxy they merge first, and the heating that holds the cluster up becomes a factory of stars more massive than they should be.

gravitation · Binary heating The velocities that decide whether Jupiter keeps a visitor. A velocity diagram in Jupiter's orbital plane, in km/s, with the Sun's frame at the origin. Jupiter moves at 13.06 km/s (the arrow along the axis). The dashed circle is the escape speed from the Sun at Jupiter's distance, 18.47 km/s: a body whose heliocentric velocity ends inside it is bound. A visitor arriving with 3 km/s to spare reaches Jupiter's distance at 18.71 km/s, just outside the circle; here its velocity makes 150° with Jupiter's, so it meets Jupiter at 30.7 km/s. A flyby keeps that relative speed and only turns it, by at most 81° for a pass grazing the cloud tops — so the possible outcomes lie on the drawn arc of a circle of radius 30.7 km/s about Jupiter's velocity. The part of the arc inside the escape circle (darker) is the set of flybys that leave the visitor bound to the Sun; the slowest outcome is 17.7 km/s. Capture needs the visitor to arrive barely faster than escape, so that the turn of its velocity relative to Jupiter can take the small excess away.

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.

orbits · Hyperbolic orbits The propellant a long departure burn wastes, under three ways of pointing it. The gravity loss of a departure from a 300-km orbit to a hyperbolic excess speed of 3 km/s with an exhaust speed of 4.41 km/s, against the stage's initial thrust-to-weight ratio, both logarithmic, for thrust held along the velocity (orange), held fixed in space at the best direction (gold) and turned at the best constant rate (blue). At a thrust-to-weight of 0.05: 1264 m/s, 3597 m/s and 1521 m/s. At 0.2: 166 m/s, 328 m/s and 154 m/s. At 0.8: 12 m/s, 20 m/s and 10 m/s. No one of the three is best everywhere. For burns of a few minutes to half an hour the constant-rate turn beats following the velocity by up to a tenth; for the longest burns, which sweep most of an orbit, following the velocity wins, because the direction that does most good turns faster as the vehicle gains speed and no constant rate keeps up. A fixed attitude, the simplest to fly, costs 88 per cent more than the best turn even for the strongest stage drawn, and several times more for weak ones. The true optimum, which no simple family contains, lies below all three and is computed numerically for each mission.

No single way to point a long burn

A departure burn that lasts tens of minutes wastes propellant because it cannot all happen at the lowest, fastest point of the orbit, and how much it wastes depends on where the engine points while it runs. Three simple rules — follow the velocity, hold one direction in space, turn at a steady rate — are each best somewhere and none everywhere. A steady turn saves up to a tenth for burns of minutes; following the velocity wins for burns that last most of an orbit; holding still costs twice as much as either.

spaceflight · Orbital transfer Two arrivals with the same weak thruster. Two spacecraft entering Mars's sphere of influence (outer circle, 577 thousand km) from the right, each thrusting continuously against its velocity at 0.3 mm s⁻² — about what an electric thruster gives a spacecraft of a few tonnes — and aimed to pass 6 planetary radii from the centre. Arriving at 0.40 km/s (blue), the thrust removes the excess while the craft crosses the sphere: it is captured 10 days after entry, having spent 268 m/s, on an orbit whose apoapsis lies inside the sphere, where the planet rather than the Sun keeps it, and the drawn path follows the same thrust winding it inward until it is within one and a half planetary radii. Arriving at 0.80 km/s (orange), it passes through on a bent hyperbola and leaves again: the thrust had too little time, and too little leverage, to remove an excess twice as large.

A thruster that must arrive slow

A rocket arriving at a planet stops by burning hard at the bottom of the gravity well, where every metre a second it spends removes the most energy. An electric thruster cannot burn hard anywhere. It can capture only arrivals slower than a few hundred metres a second, so it has to arrive with almost no excess at all — and then it pays the full circular speed of its final orbit to spiral down, two and a half times what a burn from the same start would cost, because a spiral spends its thrust at every speed on the way rather than at the highest one.

spaceflight · Escape The same worlds, placed by the light that strips. The bodies of the shoreline plane, each moved from its present instellation to the cumulative X-ray and extreme-ultraviolet energy it has received over its star's life, both relative to the Earth's (arrows). The Sun's planets do not move, by construction. A planet of a small red dwarf moves up by a factor of 7 to 16: its star stayed at its saturated activity level — a thousandth of its light in X-rays and extreme ultraviolet — for 2.3 billion years for TRAPPIST-1 against 0.1 for the Sun, and its planets have absorbed that for longer. TRAPPIST-1 e, which receives 0.65 of the Earth's sunlight, has received 10 times the Earth's lifetime XUV and moves from well inside the line to above it. Every planet measured as bare, TRAPPIST-1 b, TRAPPIST-1 c, LHS 3844 b, now lies above the line. The saturation times are a scaling with stellar mass rather than a fit to any one star.

A shoreline measured in light already spent

In the solar system one line of slope four, in the plane of escape velocity against sunlight, separates every world with air from every world without. Drawn against the light a planet receives today it puts the temperate planets of a small red dwarf safely on the side with air. Drawn against the stripping light their star has already spent, it moves them an order of magnitude toward the airless side — and leaves most of them inside a band that the line's unknown height cannot resolve.

exoplanets · Exoplanet atmospheres Two places one tidal dwarf could sit. The radial acceleration relation — observed acceleration against the acceleration the baryons alone produce, logarithmic, in m/s² — with the fitted relation for ordinary disc galaxies (solid) and the line of equality (dashed). The marks are a tidal dwarf galaxy of 10⁹ solar masses of gas and stars in an exponential disc of scale length 1.5 kpc, sampled every half kiloparsec to 6 kpc. If it carries no dark matter, as debris pulled from a disc should not, its points lie on the dashed line (open). If the relation is a law that every galaxy obeys, they lie on the curve (filled), and at the outermost radius the acceleration is 5.6 times the baryons' — a rotation speed 2.37 times higher. Every point is below a₀, where the two predictions are furthest apart.

A galaxy born without a halo

When two galaxies collide, some of the gas flung out along their tidal tails condenses into new dwarf galaxies. Debris launched from a disc cannot take the dark halo with it, so these dwarfs should rotate at the speed their own gas and stars allow and no faster. The relation that predicts every ordinary galaxy's rotation from its baryons says they should rotate twice as fast. One rotation curve decides between a halo and a law — except that the inclination of a lumpy young disc can move the answer by more than the difference.

galaxies · Dark matter One orbit, chosen to spell 1, 3, 1, 4, 1, 5. The height of the particle on the axis of an equal-mass binary of eccentricity 0.3, from launch at phase 210° and speed 1.3105, found by a seeded nested search for the sequence of excursions 1, 3, 1, 4, 1, 5, in whole binary periods. Between each pair of crossings the particle's excursion lasts the listed number of periods, to the nearest whole one. Moser's theorem is about every sequence of large enough integers at small enough eccentricity and is exact; this is one finite word at one finite eccentricity, found numerically to show the kind of thing the theorem says exists.

A particle on a line that can spell any word

Two equal stars orbit each other, and a speck of dust moves up and down on the line through their centre, perpendicular to their orbit. It has one coordinate and cannot leave that line. In 1973 Jürgen Moser proved that for almost any list of whole numbers, as long as they are large enough, there is an orbit whose successive excursions last exactly that many periods of the stars. The list can be chosen at random, so the orbit can do anything a coin can.

gravitation · Chaos

Named alongside it

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

Oberth effectBinary heatingGravity assistCore collapseΔvGlobular clusterHyperbolic excess speedOrbital energyPatched conicsVelocity dispersionBinding energyDark matter halo

All concepts