Astronomy & Celestial Mechanics
the sky, computed
The collection

Every essay — page 9

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 161–180 of 514.

Spaceflight

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

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.

10 figures · Patched conics 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.

8 figures · Gravity assist 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.

10 figures · Vis-viva 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.

10 figures · Launch windows 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.

10 figures · 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.

8 figures · Rocket equation 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.

8 figures · Low-thrust transfer 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.

8 figures · Atmospheric drag 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.

8 figures · Atmospheric drag 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.

8 figures · Ground tracks 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.

8 figures · Lagrange points Where the fuel goes, and it is not where a satellite points. Left, the orbit pole of a geostationary satellite, in degrees from the Earth's. The Sun and the Moon between them carry it round a circle of radius 7.4° in 53 years, and a satellite launched into the equatorial plane starts on the rim of that circle rather than at its centre — so its inclination climbs from zero at 0.88° a year, reaches 14.8° after 27 years, and comes back. Right, what holding it costs. A plane change of 0.88° at 3.07 km/s is 47.1 m/s a year; holding the longitude against the equatorial bulge, computed from the same resonant term that makes the longitude a pendulum, is 1.8 m/s a year. North–south is 96% of the budget, and a satellite that gives up on it does not fail — it starts tracing a figure of eight on the sky 1.8° tall in the first year, which a fixed dish cannot follow and a steerable one can. Retiring at the end of the propellant is therefore a choice about which service ends first.

The orbit that has to be paid for every year

A geostationary satellite is not in equilibrium in any direction. The Sun and Moon tilt its plane by 0.85 degrees a year, the Earth's equatorial ellipticity makes two longitudes stable and two unstable, and the end of a satellite's life is the end of its propellant.

10 figures · Station-keeping 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.

8 figures · Rendezvous A quadratic and a linear, crossing at 149 objects. The two rates that decide whether a shell at 900 km is stable, against how many objects are in it. Production goes as N² — every collision needs two objects, so the number of collisions is proportional to the square of the population, and each one is taken here to make 1600 trackable fragments. Removal goes as N, because drag acts on each object independently and takes 1,195 years to do it at this altitude. A quadratic and a linear cross exactly once, at 149 objects in this shell, and above that crossing the population grows with nothing launched. The shell presently holds about 2,280, which is 15 times the crossing. Every number on the production side is uncertain by a factor of a few — the fragment yield most of all, and the cross-section is calibrated against an observed collision rate rather than measured — so the position of the crossing carries that uncertainty with it. The shape does not, and the shape is the argument: a quadratic overtakes a linear once and never comes back, the crossing falls as the altitude rises because the lifetime is in the denominator, and what results is a threshold rather than a trend.

A collision rate that needs no collision

The flux through an orbital shell is a gas-kinetic calculation with no orbits in it. Production goes as the square of the population and removal goes as the first power, so a quadratic overtakes a linear once and never comes back — and which side of that a shell is on is decided by its altitude.

8 figures · Orbital debris 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.

9 figures · Tisserand parameter 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.

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.

9 figures · Radiometric navigation A nanosecond across the Earth is 35.69 nanoradians on the sky. The angular accuracy of a differenced-delay measurement against the length of the baseline it is measured on, both axes logarithmic, for three levels of delay precision. The relation is σ_θ = cσ_τ/B and nothing else, so every curve is a straight line of slope −1.00: the only two ways to measure an angle better are a better clock or a wider Earth, and only one of those is available. The three marked baselines are the ones that exist — the deep-space complexes in California, Spain and Australia, 8,400, 10,600, 11,700 kilometres apart. On the longest of them a delay good to 0.05 nanoseconds is 1.28 nanoradians, which at 0.52 astronomical units is 100 metres across the line of sight; a more typical 0.15-nanosecond measurement on the shortest baseline is 5.35 nanoradians. What makes any of this survivable is that the same pair of antennas observes a quasar a few degrees away immediately afterwards. The quasar is at infinity, its position is known better than the measurement, and subtracting its delay from the spacecraft's removes the clock offsets, the water vapour over each dish and the station coordinates in one step — so the number that comes out is not a delay at all but an angular separation from a fixed point in the sky.

An angle measured against a quasar

A tracking station measures how fast a spacecraft is receding, which is one number where three are wanted. The two missing angles come from the Earth's rotation, slowly, and near a planetary encounter there is no time for slowly — so the position is instead measured directly, as a difference of arrival times between two antennas, referred to a quasar a few degrees away.

8 figures · Radiometric navigation

Exoplanets

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

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.

9 figures · Microlensing A transit of a planet 0.103 of its star's radius. The star's brightness through one transit, computed by integrating the uniform stellar disc over the region the planet covers. The depth is 1.05%, which is exactly (Rp/R⋆)² = 0.01055. The four contact points are where the two discs are externally and internally tangent, at separations 1 ± 0.103 stellar radii.

A planet measured by the light it removes

A transit gives a depth, and the depth is a ratio of two radii rather than a size. Everything a transit says about a planet is said in units of a star nobody has visited either.

9 figures · Transits A transit of a planet 0.103 of its star's radius. The star's brightness through one transit, computed by integrating the limb-darkened stellar disc over the region the planet covers. The depth is 1.26%, deeper than (Rp/R⋆)² = 0.01055 because the planet crosses a limb-darkened disc whose centre is brighter than its average. The four contact points are where the two discs are externally and internally tangent, at separations 1 ± 0.103 stellar radii.

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.

9 figures · Transits

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