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

The ratio that is not √2

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

What is left at infinity, and what it costs from 200 km. The velocity change needed to leave Earth from a circular orbit 200 kilometres up, against the speed the vehicle still has when the pull has run out. Escaping with nothing left over costs 3.22 km/s. Arriving at infinity with three kilometres a second in hand costs 3.63 — an extra 0.40 km/s to buy three, a discount of 87 per cent, because the two speeds do not add and their squares do — the departure speed is √(2μ/r + v∞²) — and adding a small square to a large one barely moves the root. The curve is therefore almost flat where the planetary missions are and straightens to slope 0.79 at the right-hand edge, where v∞ has overtaken escape speed and each metre a second bought costs a metre a second spent. The practical reading is that the expensive part of an interplanetary departure is getting to escape, and the destination is a surcharge on it.

The speed that is left over

No mission is designed to sit on the escape threshold. What an interplanetary departure is specified by is the speed still remaining when the planet's pull has run out — and because energies add while speeds do not, the first three kilometres a second of it cost four hundred metres a second to buy.

The relative orbit when the target's own is not circular. Free relative motion of a chaser about a target, over one revolution, for four eccentricities of the target's orbit — drawn by propagating two exact Keplerian orbits and differencing them in the target's own rotating frame rather than by any linearised solution. The chaser differs from the target by 0.0035 in eccentricity, so the two semi-major axes are identical. Along-track distance runs across the page. At e = 0 the path is the familiar closed ellipse, 2.00 times as long along the track as across it — recovered here from an exact propagation, which is the check that the calculation is right. Raise the target's eccentricity and the path stops being an ellipse and starts breathing: the separation varies by a factor of 2.00 over one revolution at e = 0 and 2.26 at e = 0.7, because the frame itself is now turning at a rate that changes through the orbit and the chaser is swept around it fastest at periapsis. Every one of them still closes exactly, and that is the one statement of the circular theory that survives — two orbits of equal semi-major axis have equal periods and cannot drift apart, whatever else differs between them.

The relative orbit that breathes

Every result in the close-approach frame — the two-to-one ellipse, the burn that moves the wrong way, the eight kilometres behind — assumes the target's own orbit is circular. Around an eccentric target none of the geometry survives. One statement does, and it is about energy rather than shape.

A camera that cannot see how far

The equations that make a close approach tractable are linear and homogeneous, so scaling an entire relative trajectory gives another valid one with every bearing identical. A vehicle navigating on angles alone is therefore not merely imprecise about range — it is exactly blind to it, and the only thing that restores the scale is spending propellant.

The weak spot in the field. Contours of the strength of the Earth's magnetic field at the surface, in microtesla, over longitude and latitude, from an eccentric dipole: a dipole tilted to pass through the geomagnetic poles and displaced about 570 km from the Earth's centre towards the western Pacific. The field is strongest near the poles and weakest near the equator, as any dipole's is, but the displacement makes the side of the Earth furthest from the dipole's centre weaker still: the minimum, 22.9 µT, lies at 8° S, 40° W, over the South Atlantic. The measured anomaly sits further south and west, near 26° S and 55° W, and is slightly deeper, because the real field has higher harmonics that no dipole carries; the eccentric dipole gets the side of the planet and the size right and the exact place approximately. Trapped particles mirror where the field reaches a fixed strength, so wherever the field is weak they mirror closer to the ground — and over this region the inner belt reaches down into the upper atmosphere.

A belt emptied over one ocean

The Earth's magnetic field is not centred on the Earth. Its dipole sits five or six hundred kilometres towards the western Pacific, so the opposite side of the planet — the South Atlantic — has the weakest field at any altitude, and a trapped particle that mirrors safely above the atmosphere everywhere else dips into it there. The inner edge of the inner radiation belt is drawn by that one region, and it is moving.

The cheapest circular orbit to stop in at Mars is not the lowest. The speed change needed to capture into a circular orbit round Mars, in km/s, against the orbit's radius in Mars radii, for arrival excesses of 2.65, 4, 6 km/s — the first being a minimum-energy transfer from the Earth. One burn at periapsis, where the spacecraft is fastest, turns the hyperbola into the circle. The cost has a minimum, at a radius of 2μ/v∞² from the planet's centre, where it is exactly v∞/√2: 3.6 radii and 1.87 km/s for 2.65 km/s; 1.6 radii and 2.83 km/s for 4 km/s; 0.7 radii and 4.24 km/s for 6 km/s. Lower orbits cost more because the circular speed there is higher than the gain from arriving deeper in the well; higher orbits cost more because the burn is made where the spacecraft is moving slowly and the Oberth effect is lost. A faster arrival moves the optimum inward, and at a large enough excess it moves below the planet's surface, where every circular orbit is on the expensive side of the minimum.

The cheapest circle to stop in

A spacecraft arriving at a planet has to lose speed to stay, and the cheapest circular orbit to lose it into is not the lowest. It sits at a radius set only by the arrival speed — twice the planet's gravitational parameter divided by the square of the excess — where the braking costs exactly the excess divided by the square root of two. Most orbiters want something else, and the arithmetic of what they do instead decides which planets are cheap to stop at and which are not.

$The same porkchop with Mars's orbit tilted, and the ridge that appears. Contours of departure energy C₃ for transfers from the Earth to Mars, over departure date and flight time, with Mars's orbit inclined by its real 1.85° to the ecliptic and every point a solved three-dimensional Lambert problem. The dashed line is where the transfer angle is exactly 180°. Along it the contours pile up into a ridge reaching C₃ of 346 km²/s² and more, against 7.9 at the cheapest point: a transfer half-way round the Sun has to lie in a plane through the Sun and two nearly opposite points, and reaching a target a fraction of a degree out of the ecliptic tilts that plane towards the pole. The ridge splits the plot into two basins — Type I transfers, less than half-way round, below and to the left, and Type II, more than half-way, above — and the cheapest transfers sit in one basin or the other, never on the line between them.$

A wall across the porkchop

Draw the cost of going to Mars with Mars's orbit in the Earth's plane and the porkchop plot is a smooth pair of basins. Tilt Mars's orbit by its real 1.85 degrees and a wall appears between them, hundreds of times higher than the valley floor, along the line where the transfer goes exactly half-way round the Sun. The wall is a plane change forced by geometry, its height changes from one launch window to the next, and a single burn half-way to Mars removes it.

What it costs to reach Mars faster than the Hohmann transfer. The total velocity change to go from a 400 km orbit round the Earth to a 400 km circular orbit round Mars (solid), and the heliocentric part alone (dashed), against flight time, for transfers that leave the Earth's orbit tangentially and cross Mars's before reaching their far point; both planets are on circular, coplanar orbits. The right-hand end is the Hohmann transfer: 259 days and 5.65 km/s. Shortening it to 200 days costs 0.82 km/s more; to 150 days, 3.4 km/s more; below a hundred days the cost climbs past twenty. The curve leaves the Hohmann point with zero slope — the cheapest transfer is a minimum, so the first days saved cost almost nothing — and turns steeply upward only once the transfer ellipse has become long and thin. For the fastest transfers the orbit-to-orbit cost is below the heliocentric one, because a burn made deep in a planet's gravity well adds more to the speed at which the craft leaves it than the burn itself.

The first weeks off a Hohmann transfer are nearly free

The Hohmann transfer to Mars takes 259 days and is the cheapest route there is. Because it is a minimum, the cost of going faster starts flat — ten days saved costs twenty metres a second — and only climbs steeply once weeks have become months. For a crew, the time is itself a cost, paid in radiation at about two millisieverts a day.

How far a small error at departure moves the arrival at Mars. The distance by which a craft on the Hohmann transfer from the Earth's orbit to Mars's misses its aim point at the nominal arrival time, against an error in its departure speed, on logarithmic axes, integrated numerically for the 259-day flight. An error along the direction of motion (solid) moves the arrival by 76 km for every millimetre a second; one across it (dashed) by 23 km. The relation is linear over the whole range drawn: a metre a second at departure — the precision of a good launch — is 76 thousand kilometres at Mars, several times the planet's own radius and a thousand times what a landing needs. The ratio is the corresponding element of the state transition matrix, and it is the number every navigation plan is built around.

A millimetre a second at launch, seventy-six kilometres at Mars

A craft on its way to Mars arrives seventy-six kilometres from where it was aimed for every millimetre a second its departure speed was wrong. No launch is that good, so no trajectory is flown as designed; it is flown as corrected, by Newton's method run on the matrix that says how the arrival depends on the departure — which doubles its correct digits at every step and makes a two-body sketch into a real trajectory in two iterations.

Two clocks for a radiation-belt electron, at Kp 2. Two times against drift shell L, on a logarithmic scale, for electrons of about 1 MeV. The first is how long radial diffusion takes to move an electron by one Earth radius, the inverse of an empirical diffusion coefficient that grows as the tenth power of L — at Kp 2, 550 years at L = 2, 10 years at L = 3, 21 days at L = 5. The second is how long the electron survives before it is scattered into the loss cone. Inside the plasmapause, here at L = 4.7, that is set by plasmaspheric hiss: years at L = 1.3, 14 days at L = 2.5, 5.2 days at L = 3; outside it a single lifetime of 30 days stands for every other loss. Inside L = 4.8 the electron is lost before diffusion can carry it further in, which is the shaded region; outside, diffusion wins and the belt is fed. The two curves cross at a single place, and that crossing is the inner edge of the outer belt.

The gap between the belts is a race between two clocks

Electrons reach the inner magnetosphere by diffusing inward from the outer edge, and they are lost there to waves that scatter them into the atmosphere. The diffusion slows as the tenth power of distance and the loss does not, so the two clocks cross at one sharp place. Inside it lies the slot, and at megaelectronvolt energies the inner electron belt is not a steady population at all but what the last great storm left behind.

Fifty objects that hold most of the future

Low Earth orbit holds thousands of dead satellites and spent rocket stages, and the debris of the future will come mostly from their collisions. But not in equal shares. A derelict's contribution is the chance it is hit, times the fragments it would make, times how long they would stay up — and all three favour the heaviest objects in the most crowded, highest orbits. Ranked that way, a few tens of rocket bodies near 850 kilometres carry nearly half the risk, and the arithmetic of cleaning space becomes the arithmetic of choosing them.

Braking on a moon before braking with the engine

A gravity assist is usually told as a way to gain speed, but the same exchange runs the other way. A spacecraft falling toward Jupiter can pass behind one of its moons on the way in, lose a few hundred metres per second of its speed to the moon's orbit, and arrive at its capture burn needing a fifth less propellant. Which moon brakes best depends on three things at once — its mass, how fast it is met and how fast it moves — and the best one sits in radiation no spacecraft wants to fly through.

The three terms of the minimum disposal height. The minimum height above the geostationary ring to which a satellite should be raised at the end of its life, by the international guideline ΔH = 235 km + 1000 · Cᵣ · A/m, broken into its three terms, for reflectivity times area-to-mass ratios of 0.005, 0.01, 0.02, 0.03, 0.05 square metres per kilogram. The first 200 km is the protected zone itself; the next 35 km is the most that the Moon, the Sun and the Earth's uneven gravity lower an orbit's perigee over time; the rest is the eccentricity that sunlight raises every year, which a derivation from first principles puts at 942 km per unit of Cᵣ·A/m for an orbit started circular — the guideline's round thousand. Totals: 240 km, 245 km, 255 km, 265 km, 285 km. The term that depends on the satellite is the one sunlight sets, and for a light satellite with large solar arrays it is the largest.

A graveyard height with a term for sunlight

A geostationary satellite at the end of its life is raised about three hundred kilometres and switched off. The height is written as a formula — 235 kilometres plus a thousand times its reflectivity and area-to-mass ratio — and each term is a separate piece of orbital mechanics, the last one sunlight bending the orbit a little every year.

Phase-space density during a storm fed from inside. Electron phase-space density at fixed first and second adiabatic invariants, against the drift-shell parameter L, before a two-day storm (grey) and 0.5, 1, 2 days after it begins, from a model of radial diffusion with a diffusion rate rising as L¹⁰ and a loss lifetime of 4 days. The boundary is held fixed and a source centred at L = 4.5, standing for acceleration by whistler-mode chorus waves, acts for the storm's two days; the density grows a peak at L = 4.5, well inside the boundary, and diffusion then spreads it both inward and outward. A peak in phase-space density inside the outer boundary cannot be produced by diffusion from outside while the boundary is rising — diffusion only smooths gradients — so its presence then is the signature of local acceleration, and its absence is consistent with transport.

A peak that says where the electrons were made

The outer radiation belt refills with electrons of millions of electronvolts within hours of a storm. Whether they were carried inward from the edge of the magnetosphere or accelerated where they are found shows in one place only — the density of electrons in the coordinates their motion conserves, which a detector does not measure directly.

Each kilometre of aim error, turned into a velocity error. The error in the outgoing velocity per kilometre of error in where a spacecraft passes a moon, in metres per second per kilometre, against the flyby altitude on a logarithmic scale, for the four large moons of Jupiter at an encounter speed of 6 km/s. The flyby turns the velocity through an angle δ with tan(δ/2) = aₕ/b, where b is the aim point's distance from the moon's centre and aₕ = μ/v∞²; missing the aim by a kilometre along b changes δ, and missing it by a kilometre across b rotates the plane of the turn, and both move the outgoing velocity by about 2v∞aₕ/b² per kilometre. At 300 km above Ganymede that is 0.32 m/s per km along and 0.32 across. The closer and more massive the flyby, the more it turns — and the more it amplifies whatever error the approach carried.

A flyby that multiplies its own aiming error

A close flyby of a moon turns a spacecraft's path by ten degrees for nothing, and it turns every error in the aim with it. A kilometre of miss at three hundred kilometres above Ganymede becomes a third of a metre per second of velocity error, and a week later two hundred kilometres of miss at the next moon. A tour of flybys is an amplifier that must be re-zeroed after every pass, and whether a small spacecraft can use one in place of an engine burn depends on how many kilometres it can aim to.

What it costs to follow a turning grapple point. The velocity a chaser must spend each minute to hold station on a grapple point carried round by a flat spin, ω²r × 60 s, against the spin rate in degrees per second, for points 1, 3, 10 m from the spin axis, both axes logarithmic. The dashed line is what holding station 10 m from a target in a 400 km orbit costs in the same units, from the orbital dynamics alone: 2.3 mm/s per minute. At 1° per second, a point 3 m out costs 5 cm/s per minute — 24 times the orbital term; at 5° per second, 1.4 m/s per minute, 595 times. The cost rises as the square of the rate, so a target spinning twice as fast is four times as expensive to follow, and above a few tenths of a degree per second the orbit — the whole subject of the approach up to this point — has become the small correction.

A grapple point that moves faster than the orbit

Every rendezvous problem up to the last few metres is orbital mechanics: two bodies in nearly the same orbit, drifting apart and together under gravity. A derelict satellite tumbling at a few degrees a second changes the problem entirely. Holding station on a point of its surface costs more each minute than a week of orbital station-keeping, the moment to grab it comes and goes in seconds, and a free tumble never repeats — so the orbit becomes a small correction to a problem in rotation.

The wall sells inclination at the dearest price

Transfers to Mars that go almost exactly half-way round the Sun are forced into planes tilted steeply out of the ecliptic, which is why they form a wall of expensive dates across every launch chart. A mission that wanted to leave the ecliptic could use the wall on purpose. It would be buying inclination at the Earth's orbital speed — the dearest place to buy it — and the routes that reach the Sun's poles have always gone somewhere slower to make the turn.

A plane change borrowed from the Moon

Turning an orbit's plane by the engine costs a fixed fraction of the spacecraft's speed, which is why it is done as far from the Earth as possible. Farther still, at the Moon's distance, a spacecraft is moving so slowly that the Moon can turn its velocity through tens of degrees for nothing — but a flyby close enough to turn it hard also flings it outward, and the useful flyby is a gentle one, in a window of heights the arithmetic picks out.

Two ways out of a transfer orbit. The eccentricity against the semi-major axis of a spacecraft spiralling out from a geostationary transfer orbit — perigee 300 km up, apogee at geostationary height — under a continuous thrust of 0.2 millimetres a second squared, integrated through hundreds of revolutions, for two steering laws. Thrust always along the velocity (red) raises the orbit efficiently but leaves its shape almost unchanged, reaching the geostationary semi-major axis after 64 days still with an eccentricity of 0.59 — an orbit swinging between 17,000 and 67,000 km. A law that points the thrust to reduce a weighted distance from the target in both elements (blue) circularises as it raises, and reaches a circular geostationary orbit in 129 days.

The thrust that rounds an orbit turns once a revolution

An electric engine pushes gently for months, and the obvious way to point it — along the velocity — raises an orbit as fast as possible and leaves its shape almost untouched. Starting from an elongated transfer orbit, that is the wrong job. Rounding the orbit off needs the thrust forward at apogee, backward at perigee and sideways in between, a full turn of the thrust vector every revolution, and the law that does it can be written down from the equations for how each element changes.

How fast an error in the departure grows, near a neck and on a conic. How much a small error in a spacecraft's velocity is magnified in its later position — the largest singular value of the block of the state-transition matrix that maps velocity errors to position errors, in the Earth–Moon system's units — against time in days, integrated from the variational equations of the restricted three-body problem. A trajectory held at the L1 point between the two bodies (blue) amplifies errors exponentially, by a factor of e every 1.48 days — the rate the linearised dynamics there predict, 2.932 in the system's units, which the integration reproduces to within two per cent. A two-body circular orbit at half the separation (green), integrated the same way with the second body's mass switched off, amplifies them only secularly — an error in speed is an error in period, and that grows along the track in proportion to time. Over 14 days the neck has multiplied an error by 2,359; the conic by 9.

A trajectory with no good first guess

A trajectory designed from conics is corrected onto the real one by Newton's method, and the method works because the arrival depends on the departure only polynomially. Near a Lagrange point the dependence is exponential: every day and a half near the Earth–Moon neck multiplies an error by e. The first guess has to be exponentially good, the digits of the departure run out in weeks, and the only cure is to stop treating the trajectory as one piece.

$Removals a year, set almost entirely by one percentage. The number of derelicts that must be removed each year to hold a shell at 50 per cent of its critical count, against the fraction of disposals that fail, for a 650-satellite constellation at 800, 1000, 1200 km. Because a dead satellite at these heights stays up for centuries, almost every failure has to be removed by hand, and the removal rate is very nearly the failure rate: at 1200 km, 13 a year if ninety per cent of disposals succeed and 1.3 if ninety-nine do. The removal budget is therefore set by an engineering reliability rather than by the orbital environment, and a percentage point of reliability is worth a fixed number of removal missions a year for as long as the constellation flies. Five removals a year is enough for this shell only above 96.2 per cent reliability.$

A removal rate set by one percentage

How many dead satellites have to be taken out of a high orbit each year is decided less by the orbit than by how often a satellite fails to dispose of itself. Above eight hundred kilometres a derelict stays for centuries, so the removal rate needed to keep a shell stable is very nearly the failure rate, and moving disposal reliability from ninety per cent to ninety-nine cuts the work tenfold.

What a departure loses to taking time. The gravity loss of a departure burn — the velocity a rocket spends beyond what an instantaneous impulse would need — against its initial thrust-to-weight ratio, on logarithmic axes, for departures from a 300-km orbit to hyperbolic excess speeds of 0.5, 3, 6 km/s, an exhaust speed of 4.41 km/s and thrust held along the velocity. The impulsive costs are 3.21, 3.60, 4.74 km/s. At a thrust-to-weight of 1.2 the losses are 4, 5, 11 m/s; at 0.1, 298, 396, 708; at 0.03, 1485, 1785, 2546. For a stage of high thrust the loss falls as about the inverse square of the thrust (slope -1.98 at the high end), because the burn arc shrinks in proportion and the loss grows as the square of the arc.

A burn that cannot happen at one point

Every transfer cost in orbital mechanics is computed as though the engine fired for an instant. A real departure burn runs for minutes or hours, and the vehicle travels a long arc while it does — climbing, slowing, and delivering thrust ever further from the point where it would have done the most good. The extra velocity that costs is the gravity loss, and it grows as the square of the arc: negligible for a strong stage, ruinous for a weak one, and recoverable by splitting the burn at the price of time.

$The atmosphere's delay, by how high the spacecraft stands. The one-way delay of a radio signal through the troposphere, in metres of range, against the elevation of the line of sight, for an atmosphere whose refractivity falls exponentially with a hydrostatic scale height of 8 km and a wet one of 2 km. At the zenith it is 2.42 m. Integrated along a straight ray through a curved atmosphere (solid), it is 5.0 m at 30°, 13.4 m at 10° and 24.6 m at 5°. The flat-atmosphere rule, zenith delay over the sine of the elevation (dashed), agrees above about twenty degrees and then runs away, giving 46.2 m at 3° where the curved atmosphere gives 35.9: a ray at a low angle leaves the atmosphere through the top of a shell, not through the side of a slab. Ray bending, ignored here, adds centimetres below about five degrees.$

Two metres overhead and thirty at the horizon

A radio signal crossing the lower atmosphere is delayed by about two and a half metres of range straight up and by more than ten times that near the horizon. Tracking a spacecraft means converting the first number into the second, and the conversion — the mapping function — is where almost all of the troposphere's error lives: under a millimetre above thirty degrees, centimetres to decimetres in the last ten. And the last ten degrees are exactly the observations the geometry cannot do without.

Three sails against the two routes that have flown. Inclination reached against years since launch. The Venus staircase and the Jupiter step are the arithmetic of the launch-window figures — repeated Venus flybys at a v∞ of 15 km/s reaching 23°, and a single Jupiter flyby after 1.2 years reaching 90° on a heavy launch. The sails, of lightness number 0.020, 0.040, 0.080, spiral in to 0.48 AU — flat lines, 4.6, 2.3, 1.1 years — and then climb at a constant rate until they are polar. They pass the Venus route after 9.0, 4.5, 2.2 years; only the lightest catches Jupiter inside the decade, and none ever stops climbing.

A climb out of the ecliptic paid for in orbits

A solar sail can tilt its orbit without propellant by leaning against sunlight and changing which way it leans twice an orbit. Each orbit then adds the same few degrees whatever its size, so the climb runs faster the closer to the Sun the sail goes. Nothing in the orbit fixes how close that can be. The sail film's temperature does.

The one axis the field will not take

A satellite that keeps one face towards the Earth can empty its reaction wheels into the Earth's magnetic field with coils, using no propellant. Its own rotation, once an orbit, cancels every steady torque except those about one axis, the pitch axis normal to the orbit. In an orbit near the equator the field lies along that same axis, and a coil cannot push about the direction of the field.

How far inside the door a day of compression reaches. The innermost drift shell at which phase-space density has halved after a day of compression, against where the compression put the last closed drift shell, for diffusion at Kp = 4, 5, 6, 7. With the door at L = 5 and Kp = 6 the loss reaches L = 3.8; at Kp = 4 it reaches only 4.3. The diagonal is the door itself: everything above the curves and below the diagonal was emptied without ever being opened. A strong storm both brings the door in and speeds the diffusion that drains the shells behind it, and the two together carry the loss a full Earth radius or more inside the shells that were opened.

A belt drained through its outer door

When a storm's solar wind squeezes the magnetopause inward, the outermost drift shells of the radiation belt are cut open and their electrons leave within minutes. The shells inside are untouched by the boundary and by the atmosphere alike, and they empty anyway: the opened region is a sink, and the same diffusion that fills the belt from outside carries electrons out to it. A day of compression drains the belt a full Earth radius inside the shells that were actually opened.

A dropout that partly comes back. The flux of 1 MeV electrons at L = 4.5, relative to before the storm, through a storm whose ring current deepens to Dst = −150 nT over eight hours and decays with a one-day e-folding. Dashed: the adiabatic part alone, with phase-space density unchanged — the flux falls to 18 per cent at the storm's peak and returns completely. Solid: the same with a real loss of 60 per cent of the phase-space density during the main phase, as compression and outward diffusion would cause. The flux falls to 7 per cent and recovers only to 39. Seen in flux at fixed energy, the two causes look the same while the storm is on, and only the recovery — days later — tells how much of the dropout was electrons lost and how much was the field.

A dropout that is only the field

During a storm the ring current weakens the field the radiation belt sits in, and every electron, keeping its adiabatic invariants, loses energy. A detector counting electrons at a fixed energy and place sees the flux fall by half or more — and not one electron has gone anywhere. When the ring current decays, the electrons regain their energy and the flux comes back. Only in the coordinates the electrons themselves conserve can a storm's real losses be told from the field's sleight of hand, and the bigger the storm, the larger the sleight.

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

No single way to point a long burn

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

The cost of turning an orbit, against how much of each revolution the engine runs. The velocity spent per unit of inclination change, relative to an instantaneous turn at a node, when a small out-of-plane thrust is applied only in arcs centred on the two nodes, against the half-width of each arc (blue). Thrusting all the time costs π/2, 1.571; arcs of ±60° cost 1.209; ±30°, 1.047; ±10°, 1.0051. The price is time: the engine is off for the rest of each orbit, and the turn takes longer by the inverse of the duty times the efficiency (orange, right-hand scale) — relative to a push that was always fully effective, 1.57 times as long for continuous thrust, 3.14 for ±30°, 9.0 for ±10°. Most of Edelbaum's penalty is spent in the quarter-orbits near the points where a push turns the node instead of the plane.

A plane turned only where turning is cheap

A low-thrust engine that turns an orbit's plane continuously pays π/2 times what an impulse at the node would cost, because for half of every revolution it is pushing where a push mostly swings the node instead of tilting the plane. Thrusting only in arcs about the nodes recovers almost all of that: arcs of ±30 degrees cost five per cent more than an impulse, at the price of taking twice as long. It is how geostationary satellites hold their planes, and it stops working for large turns, where a chord is shorter than any arc.

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

A thruster that must arrive slow

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

The last correction: accurate if late, cheap if early. For a spacecraft approaching a moon at 6 km/s with a 30 km aim error left by navigation from the Earth, which locates the moon by camera to 20 microradians and acts an hour after its last image: the error remaining at the flyby after one final correction (blue, km) and that correction's cost (orange, m/s), against how long before the flyby it is made, all on logarithmic scales. The error falls in proportion to the time-to-go, because the moon is seen from closer, until it meets a floor: 3.1 km from how well the limb of a Europa-sized moon can be fitted once it fills the camera's view, with the burn's own two-per-cent error, 0.6 km, beneath it. The cost rises as one over the time-to-go, because the same error has to be removed in less time. The floor is reached 6.2 hours out, and a correction later than that costs more and buys nothing. A correction a day out costs 0.35 m/s and delivers 10.8 km; one six hours out, 1.4 m/s for 3.2 km. With 1 m/s to spend, the best is 4.2 km, at 8.7 hours out.

An aim bought with the last day of the approach

A spacecraft that photographs the moon it is approaching can aim its flyby far more precisely than the ground can tell it to, because it sees the target from closer every hour. But each hour closer leaves less time for a correction to act, so every kilometre of aim is bought with propellant — and below a few kilometres the camera stops helping at all, because a moon that fills the view can be located only as well as its limb can be fitted. For a small craft with a metre a second to spend, the aim comes out at four kilometres, set by geometry and not by the antenna.

Four satellites, and which four. Two sets of four navigation satellites seen on the sky from below, zenith at the centre and the horizon at the rim, with circles at 30° and 60° elevation. Left: one overhead and three low and spread around the horizon. Right: four bunched in one part of the sky. Each set fixes a position, and the error it leaves is the range error multiplied by the dilution of precision — 2.0 for the spread set, 62.0 for the bunched one. The same four range measurements, made with the same precision, give positions 31 times more uncertain when the satellites are bunched.

Four satellites in view, and which four

A satellite navigation receiver needs four satellites to fix its position, and a constellation is designed to keep at least four above every place. Four is the count, not the answer. The same four ranges, measured equally well, give a position thirty times worse when the satellites are bunched in one part of the sky than when they are spread around it, and the constellations carry their extra satellites to pay for the moments when the geometry, not the count, fails.

Too slow to shatter. The kinetic energy a projectile brings per gram of a 2,000 kg satellite, against the impact speed on logarithmic axes, for projectiles of 1, 10, 100 kg. Above 40 joules per gram (dashed) the standard breakup model counts the collision as catastrophic — the whole target is shattered; below it, only a crater is made. The speeds needed are 12.6, 4.0, 1.3 km/s respectively. In the geostationary ring, where nothing crosses faster than about 0.8 km/s (shaded), only a projectile of hundreds of kilograms — another satellite — can shatter a satellite; in low orbit, at ten kilometres a second, a kilogram can.

A ring where nothing leaves and little shatters

Every argument about orbital debris in low orbit ends in the atmosphere, which pulls fragments down in years or centuries. The geostationary ring has no atmosphere, so its removal term is exactly zero and anything left there stays for millions of years. It is saved by a second fact of the same geometry. Everything in the ring moves at nearly the same speed in nearly the same direction, so collisions there are slow, and a slow collision usually leaves a crater where a fast one leaves a cloud.

A turn spread where the climb leaves room

An electric spacecraft spiralling from low orbit to geostationary has to remove its launch inclination on the way, and the classic steering law tilts its thrust out of the plane by the same angle all the way round each orbit. But out-of-plane thrust turns the plane only near the nodes, while in-plane thrust raises the orbit equally well everywhere. Letting the tilt follow the nodes — leaning hard where turning works and lying flat where it does not — saves 150 metres a second on a transfer from Cape Canaveral, with no coasting, because the climb fills exactly the parts of each orbit where the turn would be wasted.

A height the map already knows

A navigation service for the Moon's south pole needs only a handful of satellites, if they are put in elliptical orbits that hang for hours over the pole at the high point of each lap. But satellites hanging over a pole all stand at much the same height in the sky, on a cone, and four ranges from a cone cannot tell whether the receiver is lower or its clock is late. On the surface the receiver knows its height already, from a map of the terrain — and that one number, used as a measurement, does more for the service than two more satellites.

Two satellites moved apart by the attitude they hold. The along-track separation of two identical 5 kg satellites at 475 km, starting together, when one flies broadside to the airflow and the other edge-on for 15 days and then they swap, so that their relative drift ends at zero. The separation grows quadratically and then levels off. At solar minimum the manoeuvre moves the pair 2,797 km apart; mid-cycle 8,299; at solar maximum 24,620 — 206° of the orbit's 43,060 km circumference. No propellant is spent; the same command produces a tenfold different result depending on the state of the Sun.

A rendezvous paid for in air

Two satellites in the same low orbit can be moved along it relative to each other with no propellant at all, by turning one broadside to the thin air and the other edge-on. The draggier one sinks and speeds up, the gap grows as the square of the time, and swapping attitudes halfway brings the drift back to zero. The cost moves from fuel to time, to altitude, and to knowing an atmosphere that changes tenfold with the Sun.

An arrival held without a burn, and released. A body approaching a planet from a heliocentric orbit just outside the planet's, drawn in the planet's rotating frame (the Sun far to the left, the planet's orbital motion upward), in units in which the Hill radius is 0.69. It drifts in along the shear of the differential rotation, passes through the neck at the Lagrange point L2 into the region the planet controls, loops round the planet 3.3 times and leaves through a neck again. No burn is drawn: the capture and the release are both the Sun's tide working against the planet's pull. For Mars one unit of distance is 1.56 million km and an orbit round the planet near the Hill radius takes about a year.

An arrival the Sun hands over

A spacecraft that reaches a planet with no speed to spare can be captured with no burn at all, passing through the neck between the planet's gravity and the Sun's into a loose, looping orbit. The capture is temporary and chaotic — a few revolutions, and arrivals a few hundred kilometres apart end differently — but while it lasts the spacecraft is already bound, and the burn that keeps it is taken from there.

$Where to stage, if the first stage is coming home. The payload a two-stage rocket puts into low orbit, as a share of its lift-off mass, against how much of the 9.4 km/s it needs is supplied by the first stage, with exhaust speeds of 3 and 3.4 km/s and structural fractions of 0.05 and 0.06. Thrown away, the first stage is best asked for 4.1 km/s and the rocket carries 3.5%. Landed downrange — 0.02 more of its mass in legs and fins, and propellant kept for an entry and a landing burn — the best split moves down to 3.1 km/s and the payload falls to 3.0%. Flown back to the launch site, which first cancels the stage's downrange speed, it moves to 2.5 km/s and 2.7%. A recoverable first stage is asked to do less, and the upper stage does the rest.$

A booster that flies home pays with the stage above it

A first stage that lands carries legs, fins and the propellant to bring itself back, all of which the rocket equation charges at the exponential rate it charges everything else. Recovered from a rocket designed to be thrown away, it costs a fifth of the payload. Recovered from a rocket designed around it, the toll falls by a third — because the best place to separate a stage that is coming home is earlier and slower, and the stage above it does the rest.

Where on the orbit the turning is done, and where the shadow falls. How effectively out-of-plane thrust turns a circular orbit's plane at each point round the orbit, measured from the ascending node: it goes as the absolute value of the cosine of that angle, greatest at the two nodes and zero halfway between them. The shaded bands are the Earth's shadow on an orbit 6,378 km up with the Sun in the orbit's plane, 60 degrees of the circle, placed once over a node and once between the nodes. Over the node the shadow switches the engine off where it would have done the most turning, and the orbit loses 25% of its turning; between the nodes it switches the engine off where it was doing little, and the loss is 7%.

A turn that costs more when the shadow sits on the node

An electric spacecraft turning its orbit's plane does most of the turning near the two nodes, where a push out of the plane tilts it most. Its engine runs on sunlight and stops in the Earth's shadow, and the shadow covers a sixth of the orbit wherever it happens to fall. When it falls on a node it removes a quarter of each orbit's turning, and a law that concentrates its thrust at the nodes loses half. The node drifts one way and the Sun the other, so the shadow slides across the nodes over weeks — and a transfer's length depends on the day it starts.

How far a change of attitude moves a satellite before the encounter. The along-track shift a 5-kg satellite achieves by switching from 0.01 to 0.1 m² of frontal area when warned of a conjunction, against the warning time, at 400, 500, 600 km, between solar minimum (lower edge) and maximum (upper edge), logarithmic. The shift grows as the square of the time. In a day it is 26.1 to 156 km at 400 km; 3.9 to 39 km at 500 km; 0.6 to 10 km at 600 km. The dashed line is the 556 m needed to bring a 300-m position uncertainty's collision risk below one in ten thousand.

A dodge made of air

A small satellite with no thruster can still move along its orbit by turning broadside to the thin air, and the question is whether that is enough to get out of the way of debris. Below six hundred kilometres the answer, measured in distance, is easily yes: a day's warning buys kilometres when a few hundred metres would do. Measured in predictability it is no. The density of the air is forecast to about thirty per cent, so the dodge goes somewhere nobody, including the satellite's own operator, can say in advance to the precision the dodge was meant to achieve.

Every orbit turns by about the same angle, whatever its size. How far the Earth's equatorial bulge turns an ellipse with a 300-km perigee in one revolution, in degrees, against its semi-major axis, logarithmic: the line of apsides (solid) and the node (dashed), for inclinations of 28.5, 51.6, 98°. The turn per orbit depends on the semi-latus rectum, which for an elliptical orbit lies between one and two perigee radii, so it changes by less than a factor of four from a near-circular orbit to one reaching the Moon's distance: at 28.5°, 0.56° of apsides per orbit at 8,000 km and 0.20° at 300,000 km. Per day the large orbit turns far less, because it completes so few revolutions; per revolution, almost the same.

An orbit lost to a burn that did not fire

A weak stage leaving the Earth splits its departure into several burns at perigee, coasting round a staircase of ever larger ellipses between them. If one burn does not fire, the stage goes round once more. That costs a revolution of time — an hour and a half early in the staircase, more than a day at the top — and a revolution of turning by the Earth's bulge, which comes out at nearly the same angle for every step. The angle is cheap to undo at the slow apogee of a large orbit. The time is not, when the target is the Moon.

A tow with no rope

A spacecraft parked beside an asteroid pulls it by gravity alone, with a force set by the spacecraft's mass and its distance and nothing else. Twenty tonnes hovering a hundred and fifty metres from the centre of a two-hundred-metre asteroid moves it by two millimetres a second in a year. That is the gentlest engine there is, and its limit is not the pull but the exhaust: the spacecraft must thrust away from the rock to stay put without blasting it, and the angle that takes decides how close it can hover.

How much thicker the cable must be at the top. The ratio of an Earth elevator cable's cross-section at synchronous orbit to its cross-section at the ground, as a power of ten, against the specific strength of its material, for safety factors of 1 and 2. The taper is exp(ΔΦ/(σ/ρ)) with ΔΦ = 48.4 MJ/kg, the same exponential as the rocket equation. With a safety factor of 2: high-strength steel (0.25 MJ/kg) beyond 10³⁰; Kevlar (2.5 MJ/kg) about 6.7×10¹⁶; carbon fibre (3.6 MJ/kg) about 4.8×10¹¹; Zylon (3.8 MJ/kg) about 1.2×10¹¹; nanotube yarn, today (4 MJ/kg) about 3.3×10¹⁰; ideal nanotube (50 MJ/kg) about 7. Only a material with tens of megajoules per kilogram — near the ideal strength of a carbon–carbon bond, ten times anything yet made in bulk — gives a cable that is not absurd.

A cable with a rocket equation of its own

A cable hanging from synchronous orbit to the equator would let a climber reach orbit on electricity, with no rocket at all. But the cable has to hold its own weight, and a cable that holds its own weight at constant stress must thicken exponentially with the potential it spans. The exponent is the same 48 megajoules per kilogram a rocket must supply, divided by the material's strength per unit mass. The rocket equation has not been escaped; it has moved from the propellant to the rope.

A way home that needs no engine

A spacecraft can leave the Earth so that the Moon's gravity turns it round and sends it home with no further burn. Drawn in the frame that turns with the Moon the path is a figure of eight. It was flown to the Moon by every early Apollo mission for one reason: if the engine fails, the crew comes back. That safety is exquisitely narrow. A tenth of a metre a second at departure moves the perilune by tens of kilometres and the return by thousands, and the perilune it allows is not one that most landing sites can use.

Seven orbits in fifteen years, each turned from the last. The ellipse of an Aldrin cycler, semi-major axis 1.60 AU and eccentricity 0.390, in the Sun-centred frame, for 7 successive synodic periods, the line of apsides advanced by 48.7° each time. The inner circle is the Earth's orbit and the next is Mars's. Over about fifteen years the pattern has gone once round. The same spacecraft would need to be carried from each ellipse to the next by a flyby of the Earth, or a small burn. The drawing is the circular, coplanar case; the real one has to be recalculated for the actual orbit of Mars.

A timetable that orbits the Sun

Every trip to Mars leaves when the planets line up and accelerates everything it carries. Buzz Aldrin proposed in 1985 that there need be only one departure: a large ship on an orbit that returns to the Earth each synodic period, with crews riding small taxis out to catch it. The orbit almost exists without any propellant. It has a semi-major axis fixed by a single condition, an aphelion that passes Mars with nothing asked for, and one unavoidable bill: it must turn by about fifty degrees every twenty-six months, and the Earth can pay for all but seven of the eighty the turn requires.

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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