Astronomy & Celestial Mechanics
the sky, computed
The collection

Every essay — page 21

One idea per essay, ordered so that the earlier ones set up the later ones — but nothing here depends on being read in sequence. Essays 401–420 of 534.

Cosmology

One object, seen once, from inside — and every number in it the output of a model.

The same history, on the clock that straightens light. Conformal time upward against comoving distance sideways, for the Planck 2018 cosmology. Conformal time is ∫dt/a, which is exactly the comoving distance light covers, so on these axes every photon moves at forty-five degrees — at every epoch, whatever the expansion is doing. That single property turns every curved thing in the ordinary space-time diagram into a straight one. The universe began at η = 0 and is now at η = 46.1 Gly of conformal time; it will ever accumulate only 62.8, because the integral ∫dt/a converges once Λ dominates, and that finite ceiling is the whole reason an event horizon exists. The particle horizon is the 45° line from the origin and the event horizon is the 45° line back from the ceiling, so the two horizons that were curves are now the two edges of one light cone drawn twice. The shaded wedges are the past light cones of two points on the last scattering surface, at conformal time 0.914 Gly and comoving distance 45.2 Gly from here. They do not overlap. Two points on that surface separated by more than 2η_rec were never in causal contact, which subtends 2.31° on the sky, and the microwave sky therefore contains about 9,805 patches that have no common past and the same temperature to one part in a hundred thousand. That is the horizon problem, and in these coordinates it is a statement about whether two triangles intersect.

The clock on which light travels in straight lines

Cosmic time makes light cones bulge and horizons curve. There is another time coordinate on which a photon's worldline is a forty-five degree line at every epoch, and on it the horizon problem stops being a piece of arithmetic and becomes a question about whether two triangles overlap.

8 figures · Horizons Whether an event horizon exists at all, against one number. The comoving event horizon today — the distance a signal sent now will ever cover — against the equation of state of the dark energy, for a flat universe with the measured matter density. The curve runs away at w = −1/3 and does not exist above it: that is where the expansion stops accelerating, and in a universe that does not accelerate the integral ∫da/a²E diverges and every galaxy is eventually reachable, however far away. Below −1/3 the horizon is finite and shrinks as w falls, because a more negative equation of state makes the dark energy density grow with time rather than stay constant. At the cosmological constant's w = −1 the horizon is 16.7 billion light years against a particle horizon of 46.1, so 4.7 per cent of the volume now observable is still reachable. The band is the measured −1.03 ± 0.03. What the figure is for is the asymmetry in what the measurement still allows: two sigma toward zero puts the horizon at 17.5 Gly and two sigma the other way at 14.7, and the shape of the curve means that the closer the true value sits to −1/3 the more violently the answer moves. The reachable fraction is not a robust number in the way the particle horizon is.

Whether there is a horizon at all

An event horizon exists precisely when the expansion accelerates, and its size is not a smooth function of how much. The integral that defines it runs away as the equation of state approaches minus a third, so two sigma either way on a measured number are two very different futures.

8 figures · Horizons Two horizons, one formula, and a factor of two. Horizon temperature against horizon radius, on logarithmic axes, for the two kinds of horizon this collection has. The upper line is a de Sitter horizon at T = ħc/2πk_BR and the lower is a Schwarzschild horizon at T = ħc/4πk_BR — the same expression with the same constants, differing by exactly two, and both falling as one over the radius so that a bigger horizon is a colder one. The cosmological horizon today has a radius of 4,451 megaparsecs and a temperature of 2.65e-30 K, which is thirty orders of magnitude below the microwave background and will never be measured by anything. A solar-mass black hole sits at 6.2e-8 K, and a black hole as cold as the sky would weigh 2.32e+22 solar masses — of the order of the mass inside the observable universe, which is not a coincidence, since both numbers are c³/GH up to factors of order one. The factor of two between the two lines is the one place the analogy is not exact, and it is not a convention: it comes from the periodicity of the Euclidean time coordinate, which is 8πGM/c³ for a black hole and 2π/H for de Sitter space. The entropies, by contrast, agree exactly.

Two horizons that differ only in who is inside

A black hole's horizon and the cosmological one share an entropy formula exactly and differ in temperature by precisely a factor of two. The factor of two is the whole of the difference, and what it encodes is which side of the surface the observer stands on.

7 figures · Horizons

Spaceflight

Celestial mechanics used forwards — where to burn, and what it costs.

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.

7 figures · 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.

8 figures · Rocket equation Where the missing kilometres a second go. The two losses along a gravity-turn ascent, against the vehicle's thrust-to-weight ratio at lift-off, from an integration of the trajectory rather than from a table. Every run spends the same 9400 m/s of ideal Δv and every one is flown as the same manoeuvre: one pitch kick, solved by bisection so that the vehicle is horizontal at burnout, and thereafter zero angle of attack so that gravity alone turns it — which makes the steering loss identically zero and the comparison a fair one. What differs is how much of the Δv survives as speed. The gravity loss is ∫g sin γ dt, the part of the thrust spent holding the vehicle up rather than accelerating it, and it falls as the thrust rises because a vehicle that leaves quickly spends less time doing it: 1263 m/s at T/W = 1.15 against 381 at 2.2. The drag loss is ∫(D/m) dt and it rises, because the same haste means reaching high speed lower down where the air is: 119 m/s against 1874. The sum is least at T/W ≈ 1.5, at 1198 m/s, and the minimum is shallow — which is why real vehicles cluster between 1.2 and 1.5 and none of them is there because of this curve. At the marked 1.5 the 9400 m/s of ideal Δv leaves the vehicle at 8202 m/s and 42 km, against a circular speed of 7884 m/s there, so the 1198 m/s of loss is the whole of the answer to why orbit costs about 9.4 km/s when orbital speed is under 7.9.

Orbit costs 7.8 and a launch buys 9.4

The gap between orbital speed and the velocity change a launcher spends is not overhead. It is three integrals along the ascent, only one of which can be reduced by flying better, and the two that can be traded move in opposite directions.

6 figures · 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.

6 figures · Rocket equation 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.

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.

8 figures · Plane change 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.

7 figures · Plane change 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.

7 figures · Plane change

Exoplanets

Planets nobody has seen, weighed and measured from a dip, a wobble and a delay.

A light curve pulled out of shape by the Earth's own orbit. Magnification against time for a 90-day event at impact parameter 0.3, computed with the observer's orbital motion included at three values of the microlens parallax. The dashed curve is π_E = 0 and is exactly symmetric in time, because a straight-line track past a point lens has to be. The others are not. The Earth's displacement over the months the event lasts adds a term π_E times its projected orbital motion to the lens–source trajectory, so one wing is pushed closer to the lens and the other further away, and the curve acquires an asymmetry of 16 per cent at π_E = 0.15 and 34 per cent at π_E = 0.35. That asymmetry is the whole measurement. An ordinary event gives one dimensioned number, t_E, which mixes the lens mass with two distances and a proper motion and therefore weighs nothing; the parallax gives a second, and two constraints on the same lens are what a mass requires. It is only available on long events — the Earth has to move appreciably while the magnification is changing — which is why parallaxes are measured for the timescales above about fifty days and not for the short ones that a low-mass lens produces.

An asymmetry that the Earth's own orbit puts in

A microlensing event delivers one number with dimensions, and one is not enough to weigh anything. The Earth's motion over a long event distorts the light curve, and the distortion is the second constraint — after which the mass follows with no distance in it at all.

7 figures · Microlensing Two signals that peak at different separations. The astrometric centroid shift and the photometric magnification against the source–lens separation in Einstein radii, on a common horizontal axis and their own vertical ones, for θ_E = 1 milliarcseconds. The shift is u/(u²+2) times θ_E: it vanishes at u = 0, because the two images are then symmetric about the lens and their centroid is the lens itself; it vanishes as u grows, because the minor image fades and the major one approaches the source; and it is largest in between, at u = √2 exactly, where it is θ_E/2√2 = 0.3536 mas. The magnification at that separation is only 1.15, which is a nineteen per cent brightening — a signal a survey would barely flag. The two observables are therefore complementary rather than redundant. The photometric event is short, bright and centred on the closest approach; the astrometric one is broad, largest on the wings, and lasts several times longer, falling only as 1/u once the source is well away. That slow decline is why astrometric microlensing needs years of monitoring and why it was a prediction for sixty years before it was a measurement.

A centroid that moves when the brightness does not

The two images a lens makes are never resolved, but their centre of light is displaced from where the source would be — by an amount that is largest at a separation where the magnification is only 1.34, long after the photometric event is over.

7 figures · Microlensing How long an event lasts, against the mass that causes it. Einstein crossing time against lens mass, on logarithmic axes, for a lens at 6.5 kpc in front of a source at 8 moving at 6 milliarcseconds a year — a typical bulge geometry. The slope is a half, exactly, because θ_E goes as the square root of the mass and t_E is θ_E divided by a proper motion: an M dwarf gives 21 days, a Jupiter gives 22 hours, a Neptune gives 4 hours, an Earth gives 1 hours. Nothing else about the event changes with the mass. The peak magnification is set by the impact parameter alone, so a free-floating Earth passing close enough produces exactly the light curve a star would, at exactly the same height, and lasts 1.2 hours instead of a month. That is the entire difficulty of detecting them: the events are ordinary and brief, so what decides whether they are found is the survey's cadence rather than its sensitivity. The dashed line is where the source becomes larger than the Einstein ring — 1.5e-4 solar masses for a 6-microarcsecond source — and below it every event is finite-source-dominated, which flattens the peak and, in the same stroke, measures θ_E.

An event of a few hours and no host

The Einstein crossing time is the square root of the lens mass and nothing else changes, so a free-floating Earth produces exactly the light curve a star does, at exactly the same height, and is over in ninety minutes. What decides whether it is found is cadence rather than sensitivity.

6 figures · Microlensing 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.

6 figures · Microlensing

The observed sky

Coordinates, seasons, phases and shadows — geometry seen from inside it.

The equation of time, and its two causes. The difference between a sundial and a clock over the year, in minutes, computed from Kepler's equation and the tilt. The eccentricity term has one cycle a year and the obliquity term has two; their sum runs from −14.2 to 16.4 minutes.

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.

8 figures · Equation of time The extreme sunset and sunrise, neither of them on the December solstice. Sunset and sunrise in local mean solar time through the ninety days around the December solstice, at latitude 52°, computed from Kepler's equation and the tilt. Each is the solar noon — which is 12:00 minus the equation of time — plus or minus half the day arc, and the two ingredients have different stationary points. The day length is stationary at the solstice, exactly, because the declination is. The equation of time is not: it is changing at about 0.50 minutes a day there, moving solar noon steadily. So the sum is still changing after the day length has stopped, and the earliest sunset comes on 14 Dec — 9 days before the solstice on 23 Dec — while the latest sunrise comes on 31 Dec, 9 days after it. The shortest day is the solstice, and neither of its ends is extreme there. Nothing in this is an approximation or a correction: it is a sum of two functions with different stationary points, and the stationary point of a sum is not the stationary point of either.

The earliest sunset is not the shortest day

A sunset time is solar noon plus half the day length, and the two have different stationary points. The day length stops shortening at the solstice; the equation of time does not stop moving, so the sum keeps falling — and the earliest sunset comes days before the shortest day.

7 figures · Equation of time

Spaceflight

Celestial mechanics used forwards — where to burn, and what it costs.

15 revolutions of a 185 km swath. The band an instrument 185 km wide sweeps over 15 revolutions of a 233/16 repeat orbit at 98.2° and 699.6 km, drawn on an equirectangular map. Each band's edges are placed 92.5 km either side of the track on the sphere, so the bands really are the same width everywhere and only look wider towards the poles because this projection stretches longitude by 1/cos φ. Successive passes are 24.72° of longitude apart, and across the track at the equator that is 2691 km — so neighbouring passes leave strips about 2,506 km wide unimaged between them at the equator, while near ±82° the same bands lie on top of one another many times over. Filling the tropical strips is what the rest of the 16-day cycle is for.

A swath is sized at the equator

An imaging satellite's tracks crowd together towards the poles and spread apart towards the equator, so whether a swath leaves gaps is settled at the equator and nowhere else. The order in which a repeat cycle then closes those gaps is not orbital mechanics at all. It is a theorem about points on a circle.

8 figures · Ground tracks How far across the ground a satellite can be seen from, at 0, 10, 30° masks. The footprint of a satellite — the largest angle at the Earth's centre between the point beneath it and a station that can still see it above an elevation mask — against altitude, on a logarithmic axis. λ = arccos(R cos ε / (R + h)) − ε. At low altitude it grows as the square root of the height, λ ≈ √(2h/R), drawn dashed for the horizon mask, so doubling a low orbit's altitude widens its footprint by only about forty per cent; far out it saturates, and no altitude sees past 90° − ε. At 420 km (a space station) the 0° footprint is 20.2°, a circle 2,254 km in radius holding 3.1 per cent of the Earth's surface. At 20,180 km (a navigation satellite) the 0° footprint is 76.1°, a circle 8,472 km in radius holding 38.0 per cent of the Earth's surface. At 35,786 km (geostationary) the 0° footprint is 81.3°, a circle 9,050 km in radius holding 42.4 per cent of the Earth's surface. Raising the mask costs most at low altitude: at 420 km a 10° mask takes 38 per cent off the footprint's radius and 62 per cent off its area, because most of what a low satellite could see is near the horizon, while at 35,786 km the same mask takes 20 per cent of the area.

The circle a station can see

A ground station can talk to a satellite only while the satellite is above its horizon, and the region it can do that from is a circle drawn round the point beneath the spacecraft. The circle grows as the square root of the altitude and then stops growing, the time spent inside it diverges at one altitude, and the stations that get the most contact are not where anyone would first put them.

8 figures · Ground tracks The figure of eight a geosynchronous orbit draws at 5°, 15°, 30° of inclination. The ground track over one sidereal day of a circular orbit whose period is exactly a sidereal day, at inclinations of 5, 15, 30°, centred on its own mean longitude. It is not a point. The latitude swings to ±i and back twice a day, and the longitude falls behind and then runs ahead of the Earth's rotation, because the rate at which an inclined orbit gains longitude is u̇ cos i / cos²φ: slowest at the nodes, where part of the motion is north–south, and fastest at the extremes of latitude, where all of it is eastward and a degree of longitude is shorter — so the track closes as a figure of eight. Its half-width in longitude is ±0.109° at 5°, ±0.993° at 15°, ±4.117° at 30° — exactly arcsin(tan²(i/2)) — against the small-inclination form i²/4 = ±0.109°, ±0.982°, ±3.927°: quadratic in the inclination, so the eight is tall and very thin. The longitude axis is stretched 8 times relative to the latitude axis, and without that stretch every one of these curves would be drawn as a vertical line. This is why a few degrees of inclination, which a geostationary operator spends most of its propellant preventing, moves the satellite a long way north and south and almost not at all east and west.

A stationary satellite that draws a figure of eight

A satellite with a period of exactly one sidereal day returns over the same ground every day, but only an orbit in the equator with no eccentricity returns over a single point. A few degrees of tilt draw a figure of eight, a little eccentricity a swing in longitude, and the two together draw the figure the Sun draws in the sky over a year.

9 figures · Ground tracks

The observed sky

Coordinates, seasons, phases and shadows — geometry seen from inside it.

A sidereal clock that runs up to 1.15 seconds ahead or behind, with the Moon's node. The equation of the equinoxes from 1990 to 2030: apparent sidereal time, counted from the true equinox, minus mean sidereal time, counted from an equinox that only precesses. It is the nutation in longitude times the cosine of the obliquity, and it is drawn here from the four largest nutation terms, which carry it to about a hundredth of a second. The dominant term follows the Moon's node round its 18.61-year cycle with an amplitude of ±1.052 s; riding on it are a half-yearly term of ±0.081 s from the Sun and a fortnightly one of ±0.014 s from the Moon. Over this span the sum runs from −1.147 to 1.146 s. None of it is the Earth's rotation: it is the zero point of the clock moving, because the zero point is the intersection of the equator with the ecliptic and the equator nods.

A clock whose zero is moving

Sidereal time is counted from the equinox, and the equinox does not stay put. Its steady drift makes the sidereal day eight milliseconds short, its acceleration puts a quadratic term into the formula for sidereal time, and the Moon makes it nod by a second every nineteen years. The Earth rotation angle removes all three by counting from a point defined not to move along the equator.

7 figures · Sidereal time

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