Astronomy & Celestial Mechanics
the sky, computed
Concept

Mass ratio — where it appears

The ratio of the two masses in a binary or a restricted problem, which is what most dynamical results actually depend on. The Lagrange points' stability, the Hill radius and the depth of a barycentre's offset are all functions of it alone.

Named by 23 essays across 5 fields — each of them below, with the objects they name alongside it.

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.

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

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

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

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

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

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

gravitation · Lagrange points The two-body problem, and the one-body problem it is. Left: two bodies with mass ratio 0.4 on ellipses of eccentricity 0.5 about their common barycentre, the heavier one on the smaller orbit. Right: the same system as one body of the reduced mass on a single ellipse of the same eccentricity about a fixed centre, at the separation of the two. The right-hand curve is the point-by-point difference of the two left-hand curves, so the substitution is drawn rather than asserted.

Two bodies replaced by one that does not exist

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

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

The distance at which a moon stops holding together

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

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

Three bodies, and what "no solution" actually means

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

gravitation · The three-body problem The surface a star stops at. The equipotential through L₁ — the Roche lobe — at mass fractions 0.50, 0.20, 0.05, in the frame that rotates with the pair. Each is a level set of exactly the same function the zero-velocity curves are level sets of, at exactly the critical value, so nothing here is a new construction: the Roche lobe is the last closed equipotential, and it is closed only because the two lobes touch at a single point. Material that reaches that point is no longer bound to the star it came from, and it leaves through an opening of zero area. The lobes are drawn in the orbital plane; in three dimensions each is a teardrop, and its volume-equivalent radius is what "the size of a Roche lobe" means. As the mass ratio becomes extreme the smaller star's lobe shrinks towards it, which is why a white dwarf accreting from a companion has a lobe smaller than the Sun.

The surface a star stops at

Around each star of a close pair there is a last closed equipotential, and the two touch at a single point. A star that swells to reach it hands its outer layers to its companion through an opening of zero area — and the transfer, once started, makes itself worse.

gravitation · Lagrange points 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.

gravitation · Gravitational constant The Hill radius against the mass ratio, and where four planets keep their outermost moon. The Hill radius R_H = a(μ/3)^⅓ and the exact L₁ distance from the same quintic, both in units of the planet's own semi-major axis, against the mass ratio μ = m/(M+m) from 10⁻⁸ to 0.12. The two are the same line to the width of the stroke across the whole planetary range — the cube-root formula runs 0.33% short of the exact L₁ distance at the Earth's mass ratio and 10.0% short at the right-hand edge — so the departure between them is drawn on its own percentage scale at the right, which is the only place it can be seen. Below the curves are the numerically determined stability limits, 0.5 R_H for a prograde satellite and 0.7 R_H for a retrograde one, and the outermost known satellite of each of four planets, each at a Hill fraction computed from that planet's own mass and orbit: Moon 0.257, Sinope 0.451, Phoebe 0.198, Neso 0.424. Of the four, three are retrograde — Sinope, Phoebe, Neso — and they reach 0.451 of their planet's Hill radius against 0.257 for the one prograde satellite: 1.8 times as far out, in a diagram whose two stability limits stand in the ratio 0.7 to 0.5. No equilibrium argument predicts that, and it is what the drawing makes plain.

The region a planet may keep a moon in

A satellite is not held by orbiting but by orbiting inside the radius at which the Sun's tidal field would take it away. That radius is the same distance the innermost Lagrange point sits at, asked a different question — and the boundary the sky actually respects is smaller, by a factor that depends on which way the moon is going round.

gravitation · Hill sphere One measured number, and every pair of masses that produces it. The plane of the two component masses of an inspiralling binary, with three curves of constant chirp mass across it. The middle one is GW150914's value of 28.7 solar masses, and the chirp mass recovered from the coordinates of the drawn curve varies along its whole length by 7.4e-14 per cent — which is the point: every binary on that line radiates the same frequency sweep at leading order, so the early inspiral cannot tell them apart. Two of them are marked. An equal pair of 33.0 and 33.0 solar masses and a lopsided pair of 63.6 and 18.4 sit on the same contour, and their total masses differ by a factor of 1.24. What separates them is the mass ratio, which enters the phasing only at the first post-Newtonian order, suppressed by the square of the orbital speed in units of the speed of light — small through the hundreds of cycles that carry most of the signal, and appreciable only in the last few, where that speed approaches a third of c. So the chirp mass is a measurement and the individual masses are an inference from the end of the signal, which is exactly the part a detector's high-frequency noise eats first.

One number where two masses were

The hundreds of orbits an inspiralling binary completes inside a detector's band depend on its two masses only through one combination of them. Every pair on that contour radiates an identical sweep, so the early signal — which carries nearly all the signal-to-noise — cannot say which pair it was.

gravitation · Gravitational waves Luminosity against mass, against a slope of 3.5. Main-sequence luminosity against mass, both in solar units, on logarithmic axes, over the range 0.079 to 63 solar masses. The measured curve comes from the eclipsing binaries and is the same in every drawing of it; what changes here is what it is compared against. The dashed line is a pure power law of exponent 3.5, and the curve crosses it rather than following it — the local slope runs from about 2.3 at the bottom of the range, where the interiors are convective, through nearly 4 near a solar mass where bound-free opacity dominates, to about 3 among the massive stars where electron scattering does. Quoting one exponent across the whole sequence is a convenience and the places it fails are the places the interior physics changes. Because the slope is between three and four across most of the range, a small spread in mass becomes an enormous spread in output: the 63-solar-mass end is 3.0e+9 times brighter than the 0.079-solar-mass end.

Mass decides everything, by a power of three and a half

Two stars of the same mass are almost the same star. Double the mass and the output multiplies by eleven — which is why a modest range of masses produces a colossal range of stars.

stars · The mass–luminosity relation Two radial-velocity curves, and one mass ratio. The line-of-sight velocity of each star through one orbit of AI Phoenicis. Both curves are computed from the two masses and the period; what a spectrograph delivers is the reverse. The ratio of the amplitudes is the inverse ratio of the masses — 48.2 to 50.3 kilometres a second, so the heavier star moves more slowly — and the sum of the amplitudes with the period gives the mass sum, 2.437 solar masses, once the inclination is known from the eclipses.

The only stars whose masses are known

A star's mass cannot be measured by looking at it. It can be measured by watching two stars pull on each other, and if the pair also eclipses, the same observations give both radii as well — with no stellar model anywhere in the chain. A few hundred such systems calibrate everything else.

stars · Binary stars 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.

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.

spaceflight · Rocket equation How small a sphere of influence is. Each planet's sphere of influence as a fraction of its own orbital radius, against that radius, both logarithmic. The largest belongs to Jupiter at 6.19% and the smallest to Mercury at 0.194%. The patched-conic method treats a trajectory as heliocentric everywhere outside these, and the figure is the argument for why that costs so little: they are thousandths of the journey.

One trajectory, stitched from three two-body problems

An interplanetary flight is a problem with no closed solution. It is flown by cutting it into pieces that each have one, and the seams are places where the model is knowingly false.

spaceflight · Patched conics 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 A fold 0.06 Einstein times wide, and two configurations that make the same one. A binary lens of mass ratio 0.003 at a projected separation of 1.5 Einstein radii. Top left: the source plane, with the caustic — the set of source positions at which the magnification is formally infinite — and the track of a background star across it. A single lens has no such curve; it magnifies smoothly and diverges only at one point. A second mass makes the lens mapping fold, and the fold has edges: crossing one, the number of images changes from 3 to 5, because a pair is created out of nothing on the critical curve. The small closed curve near the origin is the central caustic, always there; the larger one at 0.83 Einstein radii is the planetary caustic, and its distance from the origin is s − 1/s, which is where the planet's own image lies. Top right: the central caustic drawn twice, once for s = 1.5 and once for s = 0.667. They are 0.0184 and 0.0169 Einstein radii across and they lie on top of each other. That is not a coincidence of these numbers: to the order that a central-caustic anomaly is measured, a close binary and a wide one with the reciprocal separation produce the same perturbation, so an event with only a central anomaly returns two separations and no way to choose. Below: the light curve along the track. The smooth part is what a single lens of the same total mass would do; the spikes are the two crossings, 0.06 Einstein times apart, so a few hours inside an event lasting a month. The two curves differ in one thing only — the size of the source. A point source diverges at each fold and reaches 27; a source of angular radius 0.006 Einstein radii averages over its own disc and reaches 9 — an eighth of a source radius inside the fold the two are 8 and 4, with the divergence replaced by a rounded shoulder whose width is the source's own diameter. Everywhere else in this collection the finite size of a star is a nuisance that degrades a measurement. Here it is the ruler: the fold is a straight edge of known sharpness sweeping across a disc, so the shape of that shoulder gives the source's angular radius, and dividing it by the crossing time gives the angular Einstein radius — which is the one quantity a light curve otherwise cannot supply.

A light curve with a fold in it

A single lens magnifies smoothly. A second mass makes the lens mapping fold, and a fold has an edge — a curve across which two images appear out of nothing and the magnification formally diverges. Crossing it turns the finite size of the source star from a nuisance into a ruler.

exoplanets · Microlensing A magnification, and a spike inside it. The brightness of a background star as a foreground one passes in front of it. The smooth curve is exact: a point mass magnifies a point source by (u² + 2)/(u√(u² + 4)), which peaks at 6.7 for this track. The spike is the planet, of mass ratio 0.001, lensing one of the two images. Its duration is the Einstein time scaled by √q — about 23 hours against 30 days — so the whole planetary signal is a few hours in an event lasting a month, and it never repeats. The deviation is drawn in the approximation that the planet lenses the image in isolation; a real caustic crossing has structure this smooths over, and its true height is set by the source's size rather than by the geometry.

A star magnified by a planet nobody will see again

Microlensing weighs a planet by the way its gravity bends light around it. The measurement lasts a few hours, cannot be repeated, and is the only one that does not require the planet's star to be visible at all.

exoplanets · Microlensing The same equation, run away at q = 0.8 and settled at q = 0.5. The donor's overfill of its own Roche lobe against time, integrated for 3 mass ratios at a donor adiabatic response of −0.33 — a star with a deep convective envelope, which expands as it loses mass. The overfill sets the transfer rate and the transfer rate changes the overfill, and the sign of that feedback is the stability criterion: where the lobe shrinks faster than the star does, ζ_L > ζ_ad, the overfill grows and the growth is exponential. At q = 0.8 the lobe responds at 0.03 and the overfill runs away; At q = 1.4 the lobe responds at 1.32 and the overfill runs away. At q = 0.5 it responds at -0.62 and the transfer throttles itself back. The critical ratio for this donor is 0.63 — below equal masses, which is the result the whole subject turns on: a giant transferring to a lighter companion is already unstable before the mass ratio has reversed, and what follows is not accretion but a common envelope. Nothing here is a fitted rate. The vertical scale is logarithmic and the runaway is a straight line on it, which is what an exponential is.

Three clocks and a runaway

Whether mass transfer between two stars is stable is a comparison of two logarithmic derivatives. How fast it runs is a separate question with three possible answers fourteen orders of magnitude apart — and the answer decides whether the companion accretes, is buried, or is swallowed.

stars · Mass transfer Where the 9.4 km/s should be divided between two unequal stages. Overall payload fraction against the share of the 9.4 km/s given to the first stage, for a vehicle whose two stages are not alike: kerosene and oxygen, first stage at ε = 0.06 and I_sp 300 s, so vₑ = 2.942 km/s; hydrogen and oxygen, second stage at ε = 0.09 and I_sp 450 s, so vₑ = 4.413 km/s. The curve falls to zero at both ends, because a stage asked for too much Δv has a negative payload fraction and the vehicle does not close. The maximum is at 26.8 per cent — 2.52 km/s in the first stage and 6.88 in the second — and it delivers 5.130 per cent of the lift-off mass as payload against 4.240 per cent for an equal division. The gain from optimising is 21.0 per cent of the payload — worth having on a vehicle whose payload is five per cent of its mass, and small next to the difference the second stage's exhaust speed makes. What the shape says is more useful than where the peak is: moving ten points either side of the optimum still delivers 4.905 per cent, so the penalty for getting the split wrong is 4.4 per cent of the payload. A penalty that small is why launch vehicles are staged on structural and operational grounds — where the tank domes go, which engines exist, what can be transported by road — and the calculus is run afterwards to check that nothing has been left on the table. Note also which way the optimum leans: the better stage is the second, and it is given more of the work, because a share of Δv bought at a higher exhaust speed costs less mass.

The split that is not an equal split

Two stages sharing a velocity budget do not share it evenly, and the calculus that divides it hands more of the work to whichever stage has the better exhaust speed. The optimum is interior, it is worth about a fifth of the payload, and at this mission it is flat enough that nobody designs to it.

spaceflight · Rocket equation Exhaust speed against the molecular weight of the exhaust. The ideal exhaust speed of a converging–diverging nozzle, √(2γ/(γ−1) · R_uT_c/M · [1 − (p_e/p_c)^((γ−1)/γ)]), against the mean molar mass of the exhaust, at three chamber temperatures — 2200, 3000, 3600 K — with γ = 1.2 and an expansion to 1 per cent of chamber pressure. Every choice a propellant makes enters through two symbols, and the speed goes as the square root of their ratio: four times the chamber temperature doubles it, and a quarter of the molar mass doubles it too. That symmetry is the point. A cooler flame with a lighter exhaust beats a hotter one with a heavier, and the marked combinations show it — hydrogen + oxygen at M = 10, T_c = 3500 K, 441 s; methane + oxygen at M = 20.5, T_c = 3550 K, 310 s; kerosene + oxygen at M = 23, T_c = 3670 K, 298 s. The hottest flame drawn belongs to kerosene + oxygen and the fastest exhaust to hydrogen + oxygen, which are not the same entry. Hydrogen's advantage is not that it burns hot; it burns slightly cooler than kerosene. Its advantage is that the mixture is run fuel-rich on purpose, so the exhaust carries unburnt hydrogen and its mean molar mass falls to about 10 rather than water's 18 — buying more in the denominator than it loses in the numerator.

Choosing a propellant is choosing a molecular weight

An exhaust speed is the square root of a chamber temperature divided by a molecular weight, so a cooler flame with a lighter exhaust beats a hotter one with a heavier. Hydrogen wins the rocket equation and loses the tank, and the two cannot be optimised separately.

spaceflight · Rocket equation The mass ratio an interstellar probe needs, and the exhaust that decides it. Mass ratio against final speed, for four exhaust speeds given as fractions of c, on a logarithmic vertical axis. The relativistic rocket equation replaces the velocity change with the rapidity artanh(β), which is the quantity that adds when velocities are combined, so the mass ratio is exp(c·artanh β / vₑ) and the dashed curves are the Newtonian exp(βc/vₑ) for comparison. The two agree wherever the speed is small and part company above about a third of c, with the relativistic answer always the dearer of the two. What the figure is really about is which curve a mission sits on: reaching 0.95c needs a mass ratio of 3.29e+26 at vₑ = 0.03c, 9.02e+7 at vₑ = 0.1c, 2.57e+2 at vₑ = 0.33c, 6.24e+0 at vₑ = 1c. A fusion drive with a realistic exhaust speed sits at the left of that list and the numbers are not engineering numbers. Even the photon rocket — vₑ = c, the fastest exhaust physics permits, and requiring the propellant to be converted entirely to directed radiation — needs a mass ratio of 1.73 to reach half of c and 4.4 to reach nine tenths. The equation never forbids a speed. It prices one, and the price is exponential in a quantity that itself runs to infinity.

An equation that does not break at the speed of light

Relativity replaces the velocity change in the rocket equation with the rapidity, which adds where velocities do not. The equation therefore never forbids a speed — it prices one, and the price is exponential in a quantity that itself runs to infinity.

spaceflight · Rocket equation Two separations that draw the same caustic. The width of the central caustic against the separation, computed from the lens equation for a mass ratio of 0.003, for each separation s and for its reciprocal 1/s. The two curves lie nearly on top of each other. That is the close–wide degeneracy, and it is a theorem rather than a coincidence: expanding the binary lens equation near the primary shows that the central caustic depends on the separation only through s + 1/s to first order in the mass ratio, and that combination is invariant under s → 1/s. It is first order and not exact, which the figure shows rather than hides — the two agree to 0.7 per cent at s = 2.8 and only to 13.1 at s = 1.4, because a smaller separation is closer to the resonant regime where the central and planetary caustics have not yet separated. Reducing the mass ratio to 1.0e-3 brings the worst case to 4.8 per cent, which is the first-order statement being checked rather than quoted. What that means for a measurement is uncomfortable. An event whose planetary signal comes from the source passing near the central caustic — which is most of them, because the central caustic sits where the magnification is already high and the event is already being watched — cannot distinguish a companion at 2.8 Einstein radii from one at 0.357. For a typical lens that is the difference between a planet at four astronomical units and one at less than one. And the disagreement the figure measures is not the way out: at 2.8 Einstein radii the caustics differ in width by 0.7 per cent, which is far below what a light curve sampled through a night's seeing can separate, so the ambiguity is real in the data even where it is not exact in the mathematics.

Two systems that draw the same curve

A binary lens with separation s and one with separation 1/s have central caustics that agree to first order in the mass ratio. The same event is therefore a planet at four astronomical units or one at less than one, and no amount of photometric precision decides between them.

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

A split that decides whether the piece can leave

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

orbits · Rubble piles

Named alongside it

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

ΔvSpecific impulseAngular momentumDegeneracyEccentricityLagrange pointsPropellantReduced massStagingBarycentreBinary starsCaustic

All concepts