The cheapest place to turn

A manoeuvre that changes nothing about an orbit except the plane it lies in is the most expensive routine operation in spaceflight, and its price has remarkably little to do with the angle. The two-burn climb between circular orbits is priced by the ratio of the two radii and by nothing else. A rotation is priced by the speed at which it is attempted.

That is stranger than it looks, because the orbit before the burn and the orbit after it are the same size, the same shape and the same energy. Nothing has been bought. The vehicle is going equally fast in a slightly different direction, and the bill arrives in kilometres per second.

Three prices for one rotation, spanning a factor of 4.7.

What the proportionality is claiming #

The claim is that the cost of a plane change is

$$\Delta v = 2v\sin\!\left(\frac{\Delta i}{2}\right),$$

in which $v$ is the orbital speed at the moment of the burn and $\Delta i$ the angle between the old plane and the new. The angle enters as a sine of its half, the speed as a plain multiplier, and nothing else enters at all — not the eccentricity, not the vehicle’s mass, not which way round the turn goes.

Stated plainly enough to be wrong: two identical rotations of the same orbit differ in price by exactly the ratio of the speeds at which they are performed, and by no other factor. A turn is not cheap or expensive; a place is.

Three facts that look unrelated fall out of that one proportionality: the turn is deferred to the far end of a transfer, a launch site’s latitude is a permanent tax, and a full reversal costs twice the orbital speed — which from low orbit is more than leaving the planet.

The chord of an isoceles triangle #

The derivation is four lines, and all three consequences are visible in it. A pure rotation leaves the speed unchanged, so the velocity before the burn and the velocity after it are two vectors of the same length $v$ with an angle $\Delta i$ between them. The impulse the engine must supply is their difference, and the law of cosines gives its length:

$$|\Delta \mathbf{v}|^2 = v^2 + v^2 - 2v^2\cos\Delta i = 2v^2(1 - \cos\Delta i) = 4v^2\sin^2\!\left(\frac{\Delta i}{2}\right).$$

The two equal sides make the triangle isoceles, and the impulse is the chord subtending $\Delta i$ on a circle of radius $v$. That is the whole formula, and its shape is worth reading off: for small angles $\Delta v \approx v\,\Delta i$ in radians, so the first degree is cheap; at $\Delta i = 60°$ the triangle is equilateral and the burn costs the entire orbital speed; at 180° it costs $2v$.

The speed itself comes from the vis-viva relation, which for a circular orbit is $v = \sqrt{\mu/r}$. So the price of a fixed rotation falls as $r^{-1/2}$, and falls further if the burn is made at the apoapsis of an ellipse rather than on a circle, an eccentric orbit’s apoapsis speed being below the circular speed there. The geostationary radius is 6.2 times the low orbit’s, and the square root of that is 2.5 — the factor between 3.775 and 1.514 km/s. Arriving on an ellipse rather than a circle supplies another 1.9, since 1.618 km/s at apogee is well under the 3.075 km/s a circular orbit there would have. The two discounts multiply to 4.7, the whole spread in the opening figure, and both are the same instruction read twice.

One burn instead of two #

Deferring the turn to apogee is the obvious half of the optimisation. The figures found the half that is not.

At the top of a transfer ellipse two things need doing. The orbit must be circularised, which adds speed along the direction of travel, and the plane must be rotated, which adds speed across it. Done one after another they cost the sum of two lengths; done in one burn they cost the length of the sum, which is the third side of a triangle rather than the two sides walked in turn.

A saving of 1.146 km/s on a 5.368 km/s budget is not an optimisation, it is a fifth of the mission. And it holds at every non-zero angle by the triangle inequality, which is to say it is not a numerical result at all: two sides exceed the third unless the angle between them is zero, and at zero angle the two routes are the same manoeuvre.

In its useful form, a plane change made alongside a burn happening anyway costs only its own perpendicular part, because $\sqrt{a^2 + b^2}$ is much less than $a + b$. So the rotation out of a launch site’s inclination is never a manoeuvre of its own in a well-planned mission, and a commercial satellite’s apogee burn is aimed a dozen degrees or more off the flight direction.

The part of the turn that belongs at perigee #

Standard practice puts the whole rotation at apogee, for the reason the first figure gives. That answer is very nearly right and it is not the cheapest, and the shortfall is of a kind only a scan finds.

The mechanism is the triangle again, differentiated. Adding $b$ across a burn of length $a$ raises the cost by $b^2/2a$ to leading order, which is quadratic in the angle moved, while removing that angle from the apogee turn saves an amount linear in it. A quadratic penalty against a linear saving balances strictly inside the interval, so the optimum can never sit at either end.

That is worth more than the 25 m/s. A comparison of two options — all of the turn low down, or all of it high up — returns the better of two answers and gives no hint that a third lies between them; the interior optimum is invisible to it by construction. The generator scans the interval instead, and then refuses to report an answer sitting at either end of its own bracket, because a search that has run to the edge of its range returns a number that looks perfectly reasonable.

Turning round costs more than leaving #

The extreme case of the proportionality is a reversal, and it is where the pricing stops being an inconvenience and starts forbidding things.

The escape burn is the circular speed times $\sqrt2 - 1$, the subject of the speed that does not come back. That the crossing sits as low as 23.9° is what turns the pricing into a constraint: a launch site’s latitude is routinely larger than that, so the inclination it imposes is never removed low down.

Nor is a low orbit ever reversed. A retrograde orbit is achieved at launch, by flying west and throwing away the Earth’s rotation instead of using it — and a mission needing a large heliocentric inclination borrows its angle from a planet, which the closing section returns to.

The tax a launch site cannot pay off #

A launch delivers a vehicle into a plane, and which planes are on offer is fixed before any engine lights. The orbital plane must contain the launch site’s position vector at insertion and the velocity that was flown, and spherical trigonometry on those two facts gives $\cos i = \sin A \cos\varphi$, with $A$ the launch azimuth measured east of north and $\varphi$ the site’s latitude. Due east makes $\sin A = 1$ and returns $i = \varphi$ exactly; every other azimuth returns more. A site cannot reach an inclination below its own latitude by any azimuth whatsoever, and the deficit is paid later, at orbital speed.

The honest reading of that figure includes the part it had to be told rather than derived. The identity says 45.6° is reachable from Baikonur due east. Baikonur flies no lower than 51.6°, because the azimuth that would give 45.6° drops spent stages on territory they are not permitted to fall on — a constraint that is a map of who lives where, and no orbital mechanics contains it. A figure deriving 45.6° and presenting it as the flown minimum would be quietly wrong in exactly the way this collection exists to avoid.

That 51.6° then propagates. When Russia joined the space station programme in 1993 the station’s inclination was raised from the 28.5° of the earlier design so that Baikonur could reach it — and Kennedy, at 28.5°, reaches 51.6° only by flying an azimuth of 45.0°, surrendering about 120 m/s of the head start the Earth’s rotation supplies. The station’s orbit is a compromise between two latitudes and a debris footprint, and every shuttle flight to it paid the difference in cargo.

The identity is not really about rockets. It is the same spherical right triangle that says a star passes through the zenith only at the latitude equal to its declination, which is why the same sky looks different from different places. A launch site reaches the great circles that pass over it, and so does a telescope.

The turn that costs nothing but time #

There is one way of changing an orbital plane that appears on no budget at all, and it is used routinely.

The Earth’s equatorial bulge exerts a torque on any orbit not in the equatorial plane, and the effect is to make the node regress — the orbit’s plane rotates about the Earth’s axis, westward for a prograde orbit, at a rate that depends on the altitude and on the cosine of the inclination.

That is a plane change, delivered continuously and free. What it cannot do is change the inclination: the plane rotates about the polar axis, so the angle to the equator is preserved and only the orientation of the line of nodes moves.

For a mission that needs a particular node rather than a particular inclination, waiting is therefore an alternative to burning. Two satellites launched into the same inclination at slightly different altitudes regress at slightly different rates, so their planes drift apart at a rate the altitude difference controls, and a constellation can be deployed by putting everything into one plane and letting the drift separate it over months.

The economics are stark. Separating two planes by ninety degrees of node costs several kilometres per second as a burn and costs a few tens of metres per second as a pair of small altitude changes plus a wait — which is why a constellation of satellites in many planes is deployed by a small number of launches rather than one launch per plane.

The same drift is what makes a Sun-synchronous orbit possible, and it is worth noticing that this is the one case in the essay where the perturbation is the point rather than the nuisance: an inclination near ninety-eight degrees makes the node regress at exactly one turn per year, so the orbit’s plane keeps a fixed angle to the Sun without any propellant at all.

The expensive manoeuvre is the one that changes the angle to the equator, and the free one is the one that changes where the crossing is.

What is actually measured #

Nothing here is an observation of the sky, so the honest question is which of these numbers were measured, and by what.

Every speed above is $\sqrt{\mu/r}$ with $\mu_\oplus = 398{,}600$ km³/s², and that product is measured directly from spacecraft tracking rather than assembled from $G$ and a mass — it is known to something like a part in $10^9$ while $G$ itself is known to a few parts in $10^5$, which is why nothing in the sky is weighed in kilograms.

An inclination is not observed either; it is a fitted element. Ground stations measure range by the round-trip time of a coded signal, to a few metres, and range-rate by Doppler on the carrier, to something like 0.05 mm/s. An orbit is the six-parameter conic reproducing a run of those numbers, the inclination is one output of that fit, and $\Delta i$ is a difference of two of them.

The $\Delta v$ is the exception, and a pleasing one. It is among the few quantities in this subject a vehicle measures on itself, because an accelerometer cannot feel gravity — free fall registers nothing — so what an inertial unit integrates during a burn is exactly the non-gravitational velocity change the equations are about. An apogee burn is terminated on that integral rather than on a clock, to about a tenth of a per cent, and the residual is misalignment, slosh and the burn’s finite length rather than any error in the arithmetic.

The latitudes are geodetic positions good to centimetres. The azimuths are commanded, and constrained by range safety rather than by mechanics, which is the whole content of the 6.0° above.

What the picture cannot show #

Where the burn is allowed to happen. Two orbital planes intersect along a line, and a plane change is possible only where the vehicle crosses it. These figures price the manoeuvre and say nothing about scheduling it, yet the whole optimisation assumes a node lying at apogee — and the Earth’s equatorial bulge drags the node round by degrees per day, so that geometry is available only at particular times.

Propellant, which is what is actually spent. Every axis here is in velocity, and velocity converts to mass through an exponential, so the 1.146 km/s the combined burn saves is worth far more than a fifth of the tank. What a stage can carry turns on that exponential, and a plot in kilometres per second understates every saving on it.

Burns of finite length. All of it assumes impulses. A real apogee burn runs for minutes and sweeps tens of degrees of true anomaly, during which speed and direction both drift, so the achieved turn is an integral rather than a chord. Continuous low thrust breaks the picture altogether.

Anything about why a launch site is where it is. The latitude figure derives 45.6° for Baikonur, and Baikonur flies 51.6°. The missing six degrees are drawn as a dotted line because they could not be computed, and the same holds of every other reason a site exists — ocean downrange, existing rail, a treaty.

The same chord, bought from a planet instead of a tank #

The chord formula has a second life, and finding it there is the best argument that the geometry rather than the rocket is doing the work.

A flyby turns a spacecraft’s velocity relative to a planet through an angle $\delta$ without changing its magnitude, which is the definition of a plane change. The change to the heliocentric velocity is therefore $2v_\infty\sin(\delta/2)$ — the identical chord, with the planet supplying it and taking the momentum out of its own orbit. The angle follows from how close the approach is, as $\delta = 2\arcsin(1/e)$, so what a mission buys with navigation is what it would otherwise buy with propellant. The scale of that gift is clearest where it was needed. Ulysses reached an inclination near 79° to the Sun’s equator; at the Earth’s orbital speed of 29.8 km/s the chord for such a turn is $2 \times 29.8 \sin(39.5°) = 38$ km/s, several times the whole budget of any chemical mission ever flown. Ulysses went to Jupiter and turned there for nothing. The same reasoning rescues a manoeuvre that three burns and a detour cannot otherwise justify. Below a radius ratio near twelve a bi-elliptic transfer is simply dearer than a tangent one — and yet high-latitude geostationary launches overshoot the target altitude routinely, because pushing the apogee outwards discounts the waiting turn faster than the extra climb costs. The detour is paid for by the turn, not by the transfer.

Both halves of that trade are the mirror of the Oberth effect. Energy is bought most cheaply where the vehicle is fast, low in the well; direction is bought most cheaply where it is slow, far out. Every manoeuvre in the field is a decision about which of the two it mostly is.

The extreme case makes the size of the gift explicit, and it is worth pricing the turn nobody buys.

That is the sense in which the chord formula is a constraint on mission design rather than a line item in a budget. Below about thirty degrees a plane change is something a spacecraft does; above sixty it is something a mission is built around, by choosing a launch site, by choosing a flyby, or by choosing not to go. There is no intermediate regime in which a larger tank solves it, because the tank’s mass enters the rocket equation exponentially while the chord grows only as a sine.

The consequence is visible in the flown record. Almost every high-inclination mission in the history of spaceflight either launched into its inclination directly or bought it from a planet, and the handful that turned with propellant did so by ten or twenty degrees at most. The formula does not merely say what a turn costs; it says which turns exist. And it does so with no reference to any vehicle. The chord is a statement about two velocity vectors and the angle between them, so it prices a turn identically for a cubesat and for an upper stage, and the only thing a better rocket changes is how much of the resulting number it can afford. That is unusual in this subject: most costs scale with something about the machine, and this one does not. A plane change is priced by geometry alone, and geometry is the one supplier in spaceflight that has never lowered its prices or improved its products in the sixty years anybody has been buying from it. Every other cost in a launch manifest has fallen by an order of magnitude in that time; the sine of half an angle has not moved.

Where this ladder goes next #

Later rungs on this anchor: the combined ascent-and-turn optimisation, where the rotation folds into the launch trajectory and the dog-leg appears. The bi-elliptic plane change, which beats a direct rotation above about 39° and has been flown. Edelbaum’s result for continuous low thrust, whose closed form the chord formula does not resemble. Nodal regression under the Earth’s oblateness, and the inclination near 98° chosen so that the regression matches the year — an inclination doing work rather than costing money. And the plane change as a term in a Lambert solution, where the two ends differ in plane and the split between them is one more interior optimum.