Astronomy & Celestial Mechanics
the sky, computed
Concept

Orbital transfer — where it appears

A manoeuvre that moves a spacecraft from one orbit to another, whose cost is the sum of the velocity changes rather than the distance covered. The cheapest one between two circles is the Hohmann ellipse below a radius ratio of about twelve and the bi-elliptic above it.

Named by 6 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 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 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 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.

spaceflight · Low-thrust transfer 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.

spaceflight · Low-thrust transfer 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.

spaceflight · Orbital transfer

Named alongside it

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

ΔvLow-thrust transferSpecific impulseElectric propulsionHohmann transferRocket equationOberth effectPropellant fractionVis-vivaApsidal precessionCharacteristic energyConic section

All concepts