A climb out of the ecliptic paid for in orbits

Two spacecraft have flown far out of the plane of the planets, and both had to borrow the turn from a planet. Ulysses was launched in October 1990, flew past Jupiter in February 1992, and came away on an orbit inclined 80 degrees to the Sun’s equator. Solar Orbiter, launched in February 2020, has been raising its inclination a few degrees at a time with flybys of Venus, and after the encounter of 2025 it was seeing the Sun from about 17 degrees above the equator. The arithmetic of those two routes explains why nothing else works: inclination bought directly at the Earth’s orbital speed is the dearest velocity change in the solar system, and the only alternatives are to borrow a planet’s gravity or to fly somewhere slower before turning.

A solar sail is a third way, and it differs in kind. It carries no propellant, so it never pays the rocket equation for its turn; it pays in time. The price is not a velocity change but a number of orbits, and the questions this essay works through follow from that. How much does one orbit buy? Why does a sail that simply leans out of the plane gain nothing? Why is the climb faster closer to the Sun, and what stops it going closer still? How long does a slow sail take, and how light must one be to beat the planets?

A sail that leans one way gains nothing #

An ideal flat sail is pushed along its own normal with a force proportional to $\cos^2\alpha$, where $\alpha$ is the cone angle between the normal and the sunline. On a sail raising or lowering its orbit, the normal is turned within the orbit plane and the useful part is the transverse component. To change the plane instead, the same tilt is turned through a right angle, so that the normal leans north or south of the orbit plane. The force then has a radial part, $\cos^3\alpha$ of the face-on push, and an out-of-plane part, $\cos^2\alpha\sin\alpha$. The second is largest at the same angle that maximises the transverse push, $\alpha = 35.26°$, where it is $k = 2/(3\sqrt3) = 0.385$ of the face-on value.

Gauss’s equation for the inclination says what that push does. For a circular orbit of radius $r$ with specific angular momentum $h = \sqrt{\mu r}$, an out-of-plane acceleration $a_N$ changes the inclination at the rate

$$\frac{di}{dt} = \frac{r\cos u}{h}\,a_N,$$

where $u$ is the argument of latitude, the angle travelled from the ascending node. The cosine is the whole of the difficulty. At the nodes, where the orbit crosses the reference plane, $\cos u = \pm1$ and a push out of the plane rotates the plane directly. A quarter of an orbit later, at the top of the orbit, $\cos u = 0$: the same push no longer tilts the plane and only slides the node along. Past the top the cosine is negative, so a push that was raising the inclination now lowers it.

The dashed curve in the figure is a sail that never changes sides. It holds the optimal cone angle, leans permanently north, and finishes every orbit exactly where it started. Its inclination rises to a peak a quarter of an orbit after the node and falls back to zero by the next node. Leaning one way does not tilt the orbit. It wobbles the plane about a fixed mean, and the wobble is smaller the faster the orbit turns.

The solid curve flips sides at the top and bottom of the orbit, wherever $\cos u$ changes sign. It does not stop at the steps, because every half-orbit is now spent on the side that helps. It does slow there, at the flat stretches where $\cos u$ is near zero and the push is mostly wasted on the node. Averaged over a full orbit, the useful fraction of the push is the mean of $|\cos u|$, which is $2/\pi$. This is the same $2/\pi$ that sits inside Edelbaum’s closed form for a low-thrust plane change with an engine. The steering law is the same; what differs is the size of the push.

Five degrees an orbit, at any distance #

For an engine, the out-of-plane acceleration is whatever the thruster delivers and is the same everywhere. For a sail it is set by the sunlight, $a_N = k\beta\mu/r^2$, with $\beta$ the lightness number: the ratio of the sail’s light pressure to the Sun’s gravity on it, a ratio that does not change with distance because both fall as the inverse square. Putting that in Gauss’s equation, and multiplying by the time per radian of orbit, $\sqrt{r^3/\mu}$, gives

$$\frac{di}{du} = k\beta\cos u.$$

Every $r$ has cancelled. The tilt a sail adds per radian of orbit does not depend on how far it is from the Sun. Integrating $|\cos u|$ over a full orbit gives 4, so each orbit adds

$$\Delta i_{\text{orbit}} = 4k\beta = \frac{8\beta}{3\sqrt3}\ \text{radians} = 88.2\,\beta\ \text{degrees}.$$

A sail with $\beta = 0.06$ gains 5.3 degrees an orbit, and that holds at the Earth’s distance, at Mercury’s, or anywhere else. The cancellation has a physical reading. Closer to the Sun the push is stronger as $1/r^2$, but the orbit is faster and the spacecraft carries more angular momentum per unit mass, so a given push turns the plane less. The two effects balance exactly, because both follow from the inverse square.

Time is where the distance comes back. An orbit at radius $r$ astronomical units takes $r^{3/2}$ years, so the rate per year is

$$\frac{di}{dt} = \frac{88.2\,\beta}{r^{3/2}}\ \text{degrees per year}.$$

The vertical scale is logarithmic, so each sail’s curve has the same shape and a change of lightness only moves it up or down. The curve that matters is the bottom one, $\beta = 0.01$, near the value of the sails that have actually flown. At the Earth’s distance it gains under one degree a year, a fifth of the Venus sequence’s average. Only inside about a third of an astronomical unit, close to Mercury’s orbit, does it match what Venus provides for free. A sail of $\beta = 0.06$, the value a solar polar mission has been designed around, gains 15.6 degrees a year at 0.48 AU and 5.3 at 1 AU.

Every curve also gets steep at the left of the figure. Halving the distance raises the rate by a factor of $2^{3/2} = 2.8$, and that is what decides where a sail should crank.

The orbit never says stop #

A sail starts at the Earth’s distance and has to get closer before cranking pays. Spiralling inward with the transverse component at its best angle, the orbit shrinks at the rate set by Gauss’s equation for the semi-major axis. For a near-circular spiral that rate is $da/dt = -2k\beta\sqrt{\mu}\,a^{-1/2}$, and integrating from 1 AU to $r$ gives the time in years:

$$t_{\text{spiral}} = \frac{1 - r^{3/2}}{6\pi k\beta}.$$

The time to crank to a target inclination $i$ at that radius is $i\,r^{3/2}/(88.2\beta)$. Both terms have $\beta$ underneath, so a lighter sail is faster everywhere in proportion. As functions of $r$ they pull opposite ways: the spiral costs more the further in it goes, and the cranking costs less. Differentiating the sum and setting it to zero, $\beta$ cancels and so does $r$, leaving one number:

$$i_{\text{turn}} = \frac{4 \times 180°}{6\pi^2} = 12.2°.$$

For any target above 12.2 degrees the total time falls all the way in, whatever the sail. The two derivatives have the same dependence on $r$, so whichever is larger is larger at every radius. Above 12 degrees the cranking always wins. Since every reason for leaving the ecliptic wants far more than 12 degrees, the orbital dynamics never set an optimum radius.

The figure shows that in practice. Each curve falls steadily from right to left with no minimum, and the spiral’s cost never catches up with the saving. At 0.48 AU a sail of $\beta = 0.06$ reaches 60 degrees in 5.3 years; at 0.3 AU in 3.8; at 0.2 AU in 3.1. The improvement slows towards the left because the cranking time is shrinking towards zero and the spiral’s fixed cost takes over. It slows, but it never turns round.

So the radius is chosen by something outside the orbit, and the first candidate is the sail itself. A thin film in sunlight reaches the temperature at which it radiates what it absorbs. An aluminised front reflects about 88 per cent of the light and absorbs the rest. A chromium-coated back has an emissivity near 0.55, against 0.05 for the aluminium front, so the film sheds heat mostly from behind. At a cone angle of 35 degrees such a sail sits near 250 K at the Earth’s distance. Its temperature rises as $r^{-1/2}$: 361 K at 0.48 AU, and 520 K just inside a quarter of an astronomical unit. The polyimide films proposed for sails survive higher temperatures than that. The bonds, seams, coatings and deployment hardware are what limit a real design, and their limits are engineering numbers, not orbital ones. The figure’s 520 K is an assumption made to draw the boundary, and nothing in the argument depends on it except where the shading starts.

The dotted line marks the second candidate. An orbit at 0.48 AU has a period of 0.333 years, so the spacecraft goes round exactly three times for each orbit of the Earth. Its geometry relative to the Earth then repeats every year: the same separations, the same communication distances, the same angles from which it sees the parts of the Sun the Earth can also see. A study of a sail-borne solar polar orbiter chose that radius for that reason, giving up some speed for a mission that would be the same every year. The same trade will be faced by any sail that reaches a polar orbit. The dynamics would always go further in; the radius is chosen for the operations.

The resonance is not where Kepler puts it #

The radial part of the push has been left out so far, and at the resonant radius it matters. The sail leans out of the plane at 35 degrees, so $\cos^3\alpha = 0.544$ of its face-on push is directed straight away from the Sun. That part does no work on the plane. It does weaken the Sun’s pull, exactly as a sail facing the Sun does, so the spacecraft orbits a Sun of effective gravitational parameter $\mu(1 - 0.544\beta)$.

For $\beta = 0.06$ that Sun is 3.3 per cent lighter. A circular orbit at 0.48 AU then takes 1.7 per cent longer than Kepler’s law gives, and to keep three orbits per year exactly the sail has to sit at 0.476 AU instead. The difference is small, but it matters to a mission built on repeating geometry: the drift would build up by a full orbit’s worth of phase in about twenty years. It also matters during the climb, since the radial part changes whenever the cone angle does. A sail that varied its lean around the orbit would change its effective gravity as it went and drive its radius up and down. Held at a fixed angle, the radial part is constant and the effective Sun is steady. The plane changes steadily and the orbit stays round.

The flip comes where the orbit cares least #

The sail essay noted that a large sail turns slowly. It has a huge moment of inertia and only weak torques from shifting its own centre of pressure, so a manoeuvre takes days rather than minutes. A law that demands two complete reversals of the lean every orbit might seem to need a vehicle that a sail cannot be.

The cosine answers that worry. The reversal happens at the top and bottom of the orbit, where $\cos u$ is zero and the push turns the plane least. A sail that takes several days to swing from one side to the other is therefore lazy at the moment when being lazy costs least. If the out-of-plane push ramps linearly through zero over a slew of $w$ radians of orbit, centred on the top, the fraction of the ideal rate that survives is close to $1 - w^2/24$ for small $w$. The loss is quadratic, so a slew that is a small part of the orbit loses a smaller part still.

At 0.48 AU, where an orbit takes 121 days, a week-long flip keeps 99.5 per cent of the rate and a month-long one still keeps 90. The cost appears only on short orbits. At 0.2 AU an orbit is 33 days, so a ten-day flip is a third of a half-orbit, and the sail keeps 85 per cent. That is a second limit on going inwards, independent of the temperature. The cranking rate per year rises as $r^{-3/2}$ because the orbits get shorter, and the attitude system has to keep up with them. Twice an orbit, every orbit, it must turn a structure tens of metres across.

The same quadratic shape protects the law from timing errors. A flip made late by an angle $\phi$ gives up only $1 - \cos\phi$ of the gain, so being ten degrees of orbit late, three days at 0.48 AU, costs 1.5 per cent. The precision this manoeuvre needs is in the duration of the mission, not in its individual steps.

Racing the planets #

The figure above sets three sails against the two routes that have flown, drawn with the same arithmetic as the launch-window figures. Each sail starts with a flat stretch while it spirals in to 0.48 AU and then climbs in a straight line at constant rate until it is polar. A sail of $\beta = 0.08$ passes the Venus sequence’s 23 degrees in two and a quarter years and reaches 90 in five and a half. A sail of $\beta = 0.04$ passes the Venus route at four and a half years and is still climbing at ten. A sail of $\beta = 0.02$ takes nearly nine years to pass Venus. The Jupiter route stays out of reach for all of them in the first few years: its single flyby turns the orbit almost to polar a year and a quarter after launch.

What the sails have that the flybys lack shows in the shape of the curves. A flyby sequence is a staircase that flattens. Venus can only rotate the relative velocity it is given, and once that velocity has been turned as far as it can go the steps stop, at a few tens of degrees. A sail’s line never flattens. It keeps climbing for as long as the sail survives, it can stop at any inclination, and it can go on through the pole into a retrograde orbit if the mission wants one. The only planet involved is the Earth, at launch, and the launch date stops mattering. A sail cranking from 0.48 AU makes its own windows, three a year.

This figure puts a price on it. Each curve gives the lightness needed to reach a target inclination in a given time, and since both legs take a time proportional to $1/\beta$, halving the time doubles the lightness. A polar orbit in five years asks for $\beta \approx 0.086$, a 60-degree orbit in the same time for 0.064. On the right, a decade buys a polar orbit for 0.043.

The horizontal lines show how far off that is. LightSail 2, which raised its orbit around the Earth on sunlight in 2019, carried a whole spacecraft of five kilograms on 32 square metres of film, an areal density of 156 grams per square metre and a lightness number below 0.01. IKAROS, the first sail to fly between the planets, had $\beta \approx 0.001$. A sail at 20 grams per square metre, including the spacecraft it carries, reaches 0.077, and that is a polar orbit in five and a half years. Neither flown sail was built to maximise lightness, and the films themselves are a few grams per square metre. The gap is the structure: the booms, the deployment, the bus, and keeping a sail tens of metres across flat and pointed.

What the arithmetic leaves out #

Every number here is an idealisation, and the idealisations nearly all flatter the sail. The film is taken as a perfect mirror. A real aluminised film reflects about 88 per cent of the light and wrinkles, which both shortens the useful push and tilts it away from the normal. A realistic optical model lowers the out-of-plane push by something like a fifth, and the times above should be read as optimistic by about that much. The cone angle is held fixed at 35 degrees through the orbit, and for pure cranking that is not a simplification: the cosine of the argument of latitude only scales the push, so the angle that maximises it is the same at every point of the orbit. An engine can save propellant by throttling down near the top; a sail has no propellant to save, and leaning less there would buy nothing. The spiral inward is treated as circular throughout, which is accurate for a sail this weak.

Two things are left out deliberately because they are real gains. The sail does not have to spiral in before cranking. It can tilt and fall at once, splitting each orbit’s push between the two jobs, which shortens the total by a fraction of the spiral time. And the figures measure inclination to the ecliptic, while what a solar mission wants is latitude above the Sun’s equator, which is tilted 7.25 degrees to the ecliptic. A sail that starts its climb with the node in the right place begins seven degrees up.

None of this changes the shape of the argument. The gain per orbit is fixed by the sail’s lightness. The gain per year grows as the orbit shrinks. The orbit does not limit how far in to go; the hardware does. The only quantity that moves the whole family of curves is $\beta$, and every factor of two in it is a factor of two in time.

Still open: whether a sail can be made light enough to need no planet #

The case for a sail on this route rests on one number. Above a lightness of about 0.05 a sail reaches high solar latitudes faster than any flyby sequence, with no launch window to wait for and no ceiling on the inclination. Below about 0.02 the planets are faster for a decade. The sails flown so far sit near 0.01 and below. They were demonstrations of deployment and control, built heavy because the spacecraft was the experiment.

The orbital economics are already favourable: a mission that does not need a planet has no calendar, and a mission that never stops climbing has no ceiling. What has not been shown is a sail and spacecraft together near 20 grams per square metre, holding a cone angle steady for years at 0.48 AU. That means flipping twice every four months, with its film at 360 K under a sunlight more than four times as intense as at the Earth. The arithmetic says such a sail would see the Sun’s poles from about a third of the distance at which Ulysses passed over them, and would keep seeing them three times a year. Whether sail structures can reach that areal density without losing the flatness and control the cranking law assumes is still an engineering question.