Field

Spaceflight

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

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

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

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

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

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

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

A gravity assist with a 70° turn. The velocity triangle of a flyby. In the planet's frame the spacecraft's speed is unchanged and only its direction turns; adding the planet's own velocity converts that turn into a gain in speed measured from the Sun.

Stealing speed from a planet, which does not notice

A flyby cannot change a spacecraft's speed relative to the planet. It changes its direction — and adding the planet's own motion back turns that into free velocity.

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

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.

Catching a target 40° ahead. A phasing manoeuvre. Dropping into an orbit 6% lower shortens the period to 0.9553 of the target's, so the chaser gains 16.1° each lap and closes 40° in 3 revolutions. Speeding up would have lost ground instead.

Catching up by slowing down, which cost Gemini 4 its fuel

To reach something ahead in the same orbit, a spacecraft must fire backwards. Pointing at the target and thrusting makes the gap grow, and a crew found that out in orbit before anyone had flown the correct manoeuvre.

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

Going too far in order to arrive cheaply

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

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

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.

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

The same planet, three times

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

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

The same burn is worth more when moving fast

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

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

Two dates decide a mission

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

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

The cheapest place to turn

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

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

The stage that has to be thrown away

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

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

The transfer that costs more the gentler it is

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

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

An orbit that speeds up as it is slowed down

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

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

A corridor a degree and a half wide

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

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

The line under a satellite

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

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

The tube that leads out of a neck

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

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.

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.

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.

A map of the transfers that are free

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

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

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.

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.

A gravity assist with a 68° turn. The velocity triangle of a flyby. In the planet's frame the spacecraft's speed is unchanged and only its direction turns; adding the planet's own velocity converts that turn into a gain in speed measured from the Sun.

The planet pays, and it shows

A gravity assist takes energy from a planet and gives it to a spacecraft, and the planet's loss is exactly the spacecraft's gain. For a two-tonne probe past Jupiter that loss is unmeasurable. Do it with a hundred Earth masses of icy debris and the same bookkeeping moves Neptune outward by several astronomical units.

$Two interiors, 2.59 mm/s apart, against a floor of 0.02. What a radio link measures when a spacecraft flies past a moon. The horizontal axis is time from closest approach in minutes and the vertical axis is the accumulated change in the line-of-sight velocity in millimetres per second, after the pull of the moon as a point mass has been fitted and removed. What is left is the part of the field that is not spherically symmetric, and the two curves are what two interiors predict for it. The upper one is a body whose tidal Love number is 0.616 — a shell floating on a global liquid layer, free to deform almost as a fluid would. The lower one is a body solid throughout, at 0.03. They differ by 2.59 millimetres per second in the accumulated deflection, against a floor of 0.02 for a coherent two-way X-band link integrated over tens of seconds: 130 standard deviations in a single pass. That ratio is the whole reason the measurement is possible, and it is worth stating what is being compared. The point-mass deflection itself is 837 metres per second — five orders of magnitude larger than the signature of interest — so the interior is not read off the Doppler curve but off the residual left after a model of everything larger has been subtracted: the moon's mass, the planet's, the spacecraft's own thrusting and outgassing, the plasma along the path, the station's motion, and relativity. Every one of those has to be right to a part in a hundred thousand before the last curve here means anything, which is why gravity science needs many passes and a global fit rather than one flyby and a subtraction. The published uncertainty on Titan's k₂ is eleven per cent rather than the fraction of a per cent this signal-to-noise would suggest, and the difference is entirely correlations with the other parameters in that fit — a reminder that a formal error on a curve is not the error on the number extracted from it. The straight-line path assumed here is exact only in the limit of a fast flyby; a slow one bends, and the bending is solved for rather than approximated.$

An ocean found in a Doppler residual

A spacecraft flying past a moon is deflected by hundreds of metres a second, and the part of that deflection which says whether the moon has an ocean is two millimetres a second. Everything larger has to be modelled and removed first, exactly, and what is left over is an interior.

An oscillation that does not matter, on a ramp that does. Stored angular momentum in a reaction wheel over 160 days at 550 kilometres, with a capacity of 25 newton metre seconds. The total environmental torque is 1.75e-4 newton metres, of which 35 per cent is taken to survive averaging over an orbit. The fast oscillation is the part that does not survive: it has the 95.6-minute orbital period, reaches 0.10 newton metre seconds, and returns to where it started every revolution, so it consumes capacity and nothing else. The ramp under it is the secular part, and its slope measured between two instants a whole number of orbits apart is 6.117e-5 newton metres, which is the secular torque and is how the figure checks itself. The wheel fills in 4.7 days and has to be emptied 33 times in the span drawn. Every attitude-controlled spacecraft in the collection lives on this sawtooth, and the vertical drops are the only part of it that costs anything: the store can be moved between wheels for nothing, and taken out of the vehicle only by pushing against something outside it.

The spin that has to be put somewhere

A spacecraft holding an attitude is not resisting a force. It is absorbing a slow, one-directional trickle of angular momentum from the gradient of gravity across its own body, from sunlight, from the last of the atmosphere — and every store it has for that trickle fills up.

Three motions, three decades apart, and that is the point. The three periods of a trapped 1-MeV electron's motion against the shell it is trapped on, on a logarithmic time axis. It gyrates about a field line in 0.22 milliseconds, bounces between its mirror points in 0.33 seconds, and drifts right round the planet in 16 minutes — ratios of 1498 and 3015 at L = 4. Each motion carries a conserved quantity: the magnetic moment, the longitudinal invariant, and the magnetic flux the drift shell encloses. The separation is what makes them conserved. An invariant survives anything that changes slowly compared with its own period, so a disturbance lasting minutes destroys the third and leaves the first two untouched — and a particle that keeps its magnetic moment while being moved inward to a stronger field must gain energy. That is not a loophole; it is how the belts are filled.

Three clocks and nothing to fall onto

A charged particle in a dipole field gyrates, bounces and drifts, on timescales a millisecond, a second and a quarter of an hour. The three periods are three decades apart, and that separation is not a curiosity — it is the reason each motion has a conserved quantity, and the reason a magnetic storm can accelerate particles rather than merely stir them.

Two longitudes a satellite falls towards, and 0.5 m/s a year to stay elsewhere. The along-track potential a geostationary satellite feels, against longitude. The Earth's equator is slightly elliptical — about seventy metres between its long and short axes — and that one harmonic of the gravity field gives the geostationary ring two minima and two maxima. A satellite left unattended drifts towards the nearer minimum at 75° or 255° east, overshoots, and librates about it with a period of a couple of years. The peak acceleration accumulates 0.5 metres a second of velocity change a year, and an operational east–west budget is a small multiple of that once the correction cycle is accounted for — a fixed cost of every commercial slot in the ring, paid forever. The two minima are the graveyard of the geostationary population: uncontrolled satellites accumulate there, which is why they are the most crowded longitudes in the belt and why an uncontrolled object is most likely to be found near one.

A satellite that drifts to one of two longitudes

The Earth's equator is elliptical by about seventy metres. That one harmonic of the gravity field gives geostationary orbit a potential with two minima, so an unattended satellite slides towards the nearer one and stays — and every operational slot in the ring is paid for with fuel, every year, forever.

A density that spans a factor of 25 at 400 kilometres. Thermospheric density against altitude, for three levels of solar activity, with the model's own uncertainty band drawn around the middle curve. The extreme ultraviolet output of the Sun heats the upper atmosphere, so the scale height rises with activity and the density at a fixed altitude rises with it — by a factor of 25 at 400 kilometres between solar minimum and maximum. Superposed on that are a diurnal bulge of about a factor of two, semiannual variations, and geomagnetic storms that raise the density by tens of per cent within hours. The best empirical models reproduce past conditions to about 15 per cent, and orbital lifetime is inversely proportional to density, so a re-entry predicted a year ahead carries that error and the far larger one of not knowing what the Sun will do.

A density model wrong by a factor of two

Everything about a low orbit's future depends on the density of the air at four hundred kilometres, and that density varies by a factor of twenty-five over the solar cycle, by two within a day, and by tens of per cent during a storm nobody predicted. Every model of it is an empirical fit, and re-entry dates are quoted with the honesty that implies.

Four media, and only two of them care about the frequency. The four propagation delays between a tracking station and a spacecraft, in metres of round-trip range, against the angle between the target and the Sun. The troposphere contributes a few metres and depends only on the elevation; the ionosphere is smaller at X band and scales as the inverse square of the frequency; the solar plasma rises steeply towards conjunction and scales the same way; and the Sun's gravitational delay rises as a logarithm and has no frequency dependence at all. That last distinction is the whole of the calibration strategy: transmitting and receiving at two widely separated frequencies measures the two plasma terms and removes them, leaving the troposphere to a weather model and the relativistic term to theory. Near conjunction the plasma exceeds everything else by an order of magnitude, which is why a spacecraft passing behind the Sun is both untrackable and the best available laboratory for measuring the relativistic term.

Four media between the antenna and the spacecraft

A deep-space range measurement is a round-trip time, and the signal spends that time passing through a troposphere, an ionosphere, the solar wind and a curved region of spacetime. All four delay it, none of them is the orbit, and the whole of the navigation depends on removing them.

A patch that throws away 2.5 to 436 metres a second. The velocity error a patched-conic approximation makes at each planet's sphere of influence, on a logarithmic scale, for an approach at 3 kilometres a second. The sphere of influence is where the two ways of writing the problem — planet-centred with the Sun as a perturber, or Sun-centred with the planet as one — become equally bad, and at that radius the neglected solar tide is exactly two times the fifth root of the planet's mass ratio times the planet's own pull. That is drawn beside each planet, it runs from 0.09 to 0.50, and it is not the same for all of them — a factor of 5.6 that the definition does not remove. Because both neglected terms are largest exactly at the surface where the switch is made, the trajectory has a discontinuity in its acceleration and an accumulated velocity error of metres a second. That is negligible for a mission design and enormous for a navigation solution, which is why patched conics are used to find a trajectory and never to fly one: the real trajectory is obtained by numerically integrating the full n-body problem, differentially corrected onto the patched-conic solution as a starting guess.

The discontinuity a patched conic hides

An interplanetary trajectory is designed as two exact solutions glued along a surface where neither is valid. At that surface both neglected forces are at their largest, so the stitched path has a kink no real trajectory has — and the size of the kink is metres a second, which is a rounding error for a mission design and a catastrophe for a navigation solution.

The fragments nobody can see and cannot shield against

A catastrophic collision in low orbit makes about a thousand pieces big enough to track and sixty thousand big enough to destroy a spacecraft. The catalogue is under two per cent of the hazard, the gap is not one better radars will close, and the fragments' lifetimes span a factor of twenty-five within a single event.

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

The window that comes back and the cost that does not

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

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

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.

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.

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.

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

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.

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

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.

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

Flying further away in order to turn

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

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

A plane change paid at the worst speed there is

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

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.

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.

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.

A 5/5/1 constellation at one instant: where no satellite is above 0°. The points beneath the 5 satellites of a 5/5/1 Walker delta pattern at 43.5°, at 11,605 km, at one instant, with each satellite's 0° footprint — a circle 69.2° in radius at the Earth's centre — drawn round it, on an equirectangular map that swells the circles towards the poles. The shaded cells are ground with no satellite above the mask: none at this instant. Everywhere else is seen by between 1 and 3 at once. The largest empty circle at this instant is 67.6° in radius, computed exactly from the satellites' directions, inside the footprint, which is why there are none.

Five satellites and not four

No single orbit keeps a satellite above every point on the Earth. How many are needed is a question about the largest empty circle among their directions at the worst instant, and it has sharp answers. Four can never do it, five can from 11,605 kilometres up, and past that the arrangement of the orbits matters as much as their number.

The path of a spin across the sphere of fixed momentum. A body with principal moments 1, 8, 8.6, started about its axis of least inertia with a 3° wobble and an internal energy sink strong enough that the whole motion fits in 80 turns and each circuit of the path can be seen. The disc is the near hemisphere of the sphere of fixed angular momentum, seen from a direction between all three body axes, with the near end of each axis marked. Each thin curve is a contour of kinetic energy on that sphere — a polhode, one of the paths the angular momentum can follow in the body with no dissipation — and the thick curve is the separatrix through the intermediate axis, which divides motions that circle the axis of least inertia from motions that circle the axis of greatest. The coloured path is what the integration did: solid on the near hemisphere, dashed where it passes behind. It starts at the open dot and ends at the filled one.

A spin that left the axis it was given

The first American satellite was spun about its long axis, like a rifle bullet, and soon after launch it was tumbling end over end. Nothing outside it had pushed. A body that cannot change its angular momentum but can lose energy has exactly one place to end up, and a long body spun about its length is as far from that place as a spin can be.

$A spin about the middle axis of a 1:2:3 body, turning over every 2.9 turns. A body with principal moments of inertia 1, 2, 3, spun about its intermediate axis with a hundredth of its angular momentum knocked onto the axis of least inertia, integrated with no dissipation and no external torque. The curves are the components of the angular momentum along the three body axes, as fractions of its fixed size. The intermediate component stays near one for 1.5 spin periods, then swings through zero to minus one — the body turns over, end for end — and keeps doing so every 2.9 periods, 14 times in the span drawn. Energy and angular momentum are both conserved throughout, the energy to better than one part in a billion; nothing is being lost and nothing drives the flips. A spin about the intermediate axis is an equilibrium like a pencil balanced on its point, and the smallest disturbance grows exponentially, here by a factor of e every 0.28 spin periods, until it carries the body to the opposite equilibrium and back.$

A wingnut that turns over on its own

Spin a rigid body about the axis whose moment of inertia is neither the largest nor the smallest and it turns end over end, again and again, with nothing pushing it and nothing lost. The flip was noticed aboard a space station in 1985 and was already implicit in equations written in 1765. How long it waits is a logarithm, and no care in setting up the spin can make the logarithm infinite.

Where the gradient of gravity holds a spacecraft still. The plane of the two inertia ratios that decide whether a spacecraft pointing at the Earth is held there by the gravity gradient: k₁ — the pitch moment of inertia less the yaw moment, divided by the roll moment — across, and k₃ — the pitch moment less the roll moment, divided by the yaw moment — up, with roll along the velocity, pitch normal to the orbit and yaw towards the Earth. Shaded points satisfy all three conditions of the linear theory — pitch is stable when k₁ > k₃, and roll and yaw together when k₁k₃ > 0 and 1 + 3k₁ + k₁k₃ > 4√(k₁k₃). The large region at upper right, 12.4 per cent of the square, is the one in which the pitch moment is the largest and the yaw moment the smallest, the arrangement of a long boom hanging towards the Earth. The small region just left of the vertical axis and below the horizontal one, 2.0 per cent, is a second, narrow island of stability with the moments in a different order, found by DeBra and Delp in 1961. There the orientation is a maximum of the potential in roll and yaw rather than a minimum, held only by the gyroscopic coupling of the two, and a damper — the very thing the long-boom region needs — destroys it: with damping of 0.05 of the orbital rate, a swing of a hundredth of a radian at (−0.10, −0.21) grows to a full radian within 11 orbits, while the same swing on the long boom shrinks 435-fold in 40. A boom along the vertical, pitch moment largest sits at (0.97, 0.40) and is stable; the same boom with roll and pitch moments swapped sits at (0.93, −0.40) and is unstable: the same boom, with two nearly equal moments exchanged, crosses from one side of an axis to the other.

A boom held upright by a difference in gravity

A long spacecraft in orbit is pulled into line with the vertical for nothing — its near end feels slightly more gravity than its far end, and the difference is a torque. The torque restores and never dissipates, so the vehicle swings like a pendulum whose clock is the orbit. Whether it is held at all comes down to three inequalities between its moments of inertia, and one region that satisfies all three is destroyed by the damper every such spacecraft needs.

A tumble removed with a coil and a compass, at 500 km. The rotation rate of a small spacecraft — principal moments 0.0067, 0.041, 0.043 kg m², the proportions of a three-unit cubesat — tumbling at 8.8° a second after release, against orbits at 500 km, with nothing to control it but magnetic coils driven by the B-dot law: a dipole opposite to the rate of change of the field measured aboard, capped at 0.2 A m². The field is a dipole tilted 9.2° from the Earth's axis and turning with the Earth. In a polar, 97.4°, orbit the rate settles at 0.12° a second over the last orbit drawn, passing 1° a second after 0.58 orbits; in a 51.6° orbit the rate settles at 0.12° a second over the last orbit drawn, passing 1° a second after 0.32 orbits; in an equatorial orbit the rate settles at 2.44° a second over the last orbit drawn. The law needs no knowledge of the spacecraft's attitude: a tumbling body sees the Earth's field swing round in its own frame, and a dipole opposing that swing produces a torque that removes the part of the spin perpendicular to the field. The polar orbit does not reach zero. It settles at 0.94 of twice the orbital rate, 0.13° a second, and twice the orbital rate is how fast the field direction itself turns round a polar orbit: a body turning with the field sees little change to oppose. An equatorial orbit keeps the field pointing nearly the same way all the way round, so the spin about it is reached only through the dipole's tilt and the Earth's turning, and 28 per cent of the starting rate is still there at the end.

A tumble stopped by the field it tumbles through

A small satellite leaves its deployer tumbling, and the first thing most of them do is stop, using nothing but a magnetometer and three coils. The law they run needs no idea where the satellite is pointing. What it cannot do, at any instant, is touch the spin about the local field line — so how much tumble survives is decided by how much the field's direction changes along the orbit, and the stillness it reaches is defined by the field rather than by the stars.

$The drag coefficient of a sphere runs from 2.03 to 2.79, and 2.2 is a convention. The free-molecular drag coefficient of a sphere against the accommodation coefficient — the fraction of striking molecules that thermalise with the surface and leave in a cosine distribution rather than bouncing — at speed ratios 2, 4, 8, with the surface at 0.3 times the flow's temperature. Specular reflection gives 2.469 at the lowest speed ratio drawn and 2.001 in the hypersonic limit, where every molecule delivers exactly twice its own momentum. Accommodation adds the re-emitted flux, which leaves at the wall temperature in a direction the flow did not choose, and it adds most where the speed ratio is smallest — which is high up, where the light species dominate. The conventional 2.2 lies outside this family at both ends: at s = 8 a sphere reaches only 2.112 even at full accommodation, and at s = 2 it is already 2.469 with none. That is not a defect of the arithmetic — 2.2 is a fitted average for satellite shapes, whose flat panels have a higher coefficient than a sphere of the same projected area, and the sphere is drawn because it is the one geometry with a closed form. What survives the shape is the dependence: a satellite's drag coefficient is an assumption about its surface chemistry and its attitude, and every density inferred from drag carries it in inverse proportion.$

A coefficient that belongs to the surface, not the satellite

Every density ever inferred from satellite drag was divided by a drag coefficient, and that coefficient is not a property of the spacecraft. It is a property of what happens when an oxygen atom at eight kilometres a second strikes a surface it has already coated — and the conventional 2.2 is a convention.

A weather forecast made out of orbits

A density model fitted to fifty years of satellite drag is a climatology, and its error does not shrink with more data. Updating it from the drag observed on objects in orbit right now is the manoeuvre a weather forecast makes — and it buys a factor of several for a day and nothing at all for a fortnight.

A corridor 0.2 km wide, and a density known to a factor of 2. The apoapsis a vehicle is left on after a single atmospheric pass, against the periapsis altitude it aimed at, for ballistic coefficients of 60, 130, 300 kg/m² arriving at 3 km/s. The energy removed is the density at periapsis times an effective path length of √(2πrₚH), divided by the ballistic coefficient — so it falls exponentially with altitude and the curve is steep. Hitting a 1000 km apoapsis to ±10 per cent requires a periapsis inside 0.2 kilometres. Getting the atmosphere wrong by a factor of 2 moves the aim point by 7.6 kilometres, which is 30.7 times the corridor's own width — so a ballistic vehicle aiming at a planet whose density is known to a factor of two misses by more than the tolerance allows, and the manoeuvre has to be flown rather than aimed.

A manoeuvre that has never been flown once

Arrive on a hyperbola, dip once through the atmosphere, leave on a bound orbit having spent no propellant. The saving is a kilometre a second or more, the physics is the same as an entry corridor, and nobody has done it — because the corridor is a tenth of a kilometre wide and the density is known to a factor of two.

One square root that raises the orbit and turns it

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

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

The engine is chosen by the calendar

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

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

A drive with no rocket equation

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

At the Sun–Earth L₂ the cheapest correction is every 23 days, and it costs e σ per e-folding. The annual station-keeping cost at the Sun–Earth L₂ point, against the interval between corrections, on logarithmic axes, for velocity errors of 0.5 cm/s, 2.0 cm/s, 5.0 cm/s along the unstable direction at each correction. Correcting often costs a lot because every correction carries its own error σ; correcting rarely costs a lot because the error has grown by e^(T/τ) in between, with an e-folding time τ = 23.4 days set by the point's growth rate of 2.484 times the orbital mean motion. The product (365.25/T) σ e^(T/τ) has its minimum at exactly T = τ, where the annual cost is 365.25 e σ/τ: 0.21 m/s a year for σ = 0.5 cm/s, 0.85 m/s a year for σ = 2.0 cm/s, 2.12 m/s a year for σ = 5.0 cm/s. The minimum is broad, so an operator can correct at a convenient interval near the e-folding time for little penalty, and the cost scales linearly with how well the spacecraft's velocity is known and executed. This is a one-dimensional caricature: a real halo orbit's correction also removes a stable component it need not, and solar radiation pressure on a large sunshield is a steady error source of its own. The figure's claim is the structure — an unstable equilibrium is cheap to hold if the instability is caught while it is still small, and its cost is a navigation budget rather than a force budget.

An unstable point that costs less to hold than a stable orbit

A spacecraft at the Sun–Earth L₂ point sits on an equilibrium that throws it away, doubling any error every sixteen days. It holds station for a few metres per second a year — a twentieth of what a geostationary satellite pays to stay on an orbit that is stable. The difference is what is being paid for — an instability caught small costs a navigation budget, and a steady torque costs a force budget.

$A gauge good to 7 per cent with a tenth of the load left, and to 23 per cent with three hundredths. The uncertainty in the propellant remaining in a spacecraft tank, as a percentage of what remains, against the fraction of the 450-kilogram load still in the tank, on a logarithmic uncertainty axis with the tank emptying to the right. Bookkeeping — summing every thruster firing through a flow-rate model — carries an error common to all burns of 2 per cent of the mass used, plus an independent 5 per cent per burn that averages down over 2000 firings; its absolute error grows with the mass used. Gauging by pressure and temperature infers the empty volume of the tank from the gas law applied to a known mass of pressurant, with a combined 0.66 per cent uncertainty in n R T / P and a 0.2 per cent uncertainty in the tank's volume; its absolute error grows as the gas fills the tank. The two methods are independent and are combined by inverse variance. With a tenth of the load left the combined estimate is uncertain by 2.9 kg, 7 per cent of what remains; with three per cent left, by 3.1 kg, 23 per cent. Near empty the absolute error barely changes, so halving what is left doubles the relative error — the gauge is at its worst exactly when the last manoeuvre has to be planned from it.$

A fuel gauge that is worst when it is needed

A spacecraft's tank has no float and no dial. The propellant left is estimated by adding up every burn or by reading the pressure and temperature of the gas above the liquid, and both methods' errors grow with the propellant used. Relative to what remains, the error doubles every time what remains halves — so a geostationary satellite has to hold back months of station-keeping as a margin against a gauge that cannot see the last few kilograms.

The ladders in this field

19 anchors · one idea each

Orbital transfer Escape Gravity assist Rocket equation Rendezvous Patched conics Vis-viva Launch windows Plane change Low-thrust transfer Atmospheric drag Ground tracks Lagrange points Station-keeping Orbital debris Tisserand parameter Radiometric navigation Attitude control Radiation belts

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