Astronomy & Celestial Mechanics
the sky, computed
Concept

Hohmann transfer — where it appears

The two-burn ellipse tangent to both of two circular orbits, which is the cheapest transfer between them in the two-body problem. It is optimal only below a radius ratio of 11.94, above which going further out than the destination and coming back is cheaper.

Named by 11 essays across 2 fields — each of them below, with the objects they name alongside it.

Flight time against semi-major axis, for a fixed 135° sweep. Lambert's theorem drawn: the time to fly between two points 1 and 1.524 AU out and 135° apart, against the semi-major axis of the orbit that does it. Nothing else about the orbit enters — not its eccentricity, not where its periapsis is, not how it is oriented — which is the content of the theorem and the reason a two-point transfer is a one-dimensional search rather than a six-dimensional one. Two branches: the lower one is the ellipse whose arc stays short of apoapsis, falling towards the parabolic floor at 103.2 days as a grows without limit; the upper one is the ellipse of the same size whose arc runs through apoapsis, rising without limit. They meet at a = s/2 = 1.2161 AU, 244.2 days, which is the minimum-energy transfer and the slowest ellipse available — every faster one is bigger. Each branch is monotone, checked point by point across the drawn range, so a horizontal line cuts each at most once: for a given pair of points and a given time there is exactly one ellipse, and at 260 days it is a = 1.2189 AU on the upper branch. The freedom a mission designer has is not in this picture: it is the choice of the two points, which is what a porkchop plot sweeps.

Two places and a clock decide the path

The time to fly between two points depends on the semi-major axis, the chord between them, and the sum of their distances — and on nothing else about the orbit. Not the eccentricity, not where periapsis is, not the orientation. Lambert's theorem is why an interplanetary launch date is the root of one equation.

orbits · Lambert's problem 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.

spaceflight · Orbital transfer 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.

spaceflight · Orbital transfer 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.

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

The cheapest place to turn

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

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

The stage that has to be thrown away

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

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

The transfer that costs more the gentler it is

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

spaceflight · Low-thrust transfer 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.

spaceflight · Launch windows 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.

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

Flying further away in order to turn

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

spaceflight · Plane change One square root that raises the orbit and turns it. The velocity budget from a 300 km circular orbit to geostationary, against the plane change carried out along the way. Edelbaum's closed form — √(v₁² + v₂² − 2v₁v₂cos(½πΔi)) with Δi in radians — puts the whole continuous manoeuvre in one square root, and at Δi = 0 it collapses to |v₁ − v₂| = 4.651 km/s, which is the spiral's cost with no plane change in it. The two-impulse curve puts its rotation into the circularisation burn at apoapsis, where the vehicle is moving at 1.608 km/s and a rotation is cheap. At 28.5° the continuous transfer costs 5.951 km/s against the impulsive 4.256 — the plane change adds 1.300 to one and 0.363 to the other. That is the opposite of the usual claim that low thrust turns for free. It turns continuously, which is not the same thing: the gain is that the propellant is not the budget, and the Δv is worse here as it is everywhere else.

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.

spaceflight · Low-thrust transfer

Named alongside it

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

ΔvVis-vivaLaunch windowOberth effectPlane changeApoapsisCharacteristic energyCircular velocityOrbital inclinationOrbital transferSpecific impulseSynodic period

All concepts