An unstable point that costs less to hold than a stable orbit

A geostationary satellite pays about fifty metres per second a year to stay where it is, most of it to cancel the tilt the Sun and the Moon give its orbital plane. That orbit is stable in the ordinary sense: left alone, a geostationary satellite does not fly away. It wanders — its inclination grows to fifteen degrees over decades and its longitude slides towards one of two resting places — and holding it in its slot means fighting a torque that never stops.

A spacecraft at the Sun–Earth $L_2$ point, a million and a half kilometres beyond the Earth, is in the opposite situation. The collinear Lagrange points are unstable: a body placed at one with the slightest error drifts off exponentially along a direction the manifolds of the point organise into tubes. And yet a large observatory there holds its orbit for two to four metres per second a year — a fifteenth to a twenty-fifth of the geostationary budget, spent on a point that actively throws it away.

That comparison reverses the usual intuition about what stability costs, and the reason can be written as a single expression.

How fast the point throws a spacecraft away #

Near a collinear point the motion in the rotating frame can be linearised. With the secondary’s share of the mass written μ and the point’s distance from the secondary γ, one coefficient does most of the work,

$$c_2 = \frac{\mu}{\gamma^3} + \frac{1-\mu}{(1 \pm \gamma)^3}$$

and the in-plane motion splits into an oscillation and an exponential. The exponential grows at a rate λ times the orbital mean motion $n$, with

$$\lambda^2 = \tfrac{1}{2}\left[c_2 - 2 + \sqrt{9c_2^2 - 8c_2}\right]$$

When the secondary is small, both $L_1$ and $L_2$ sit at the edge of its Hill sphere, $c_2$ tends to 4, and λ tends to $\sqrt{1 + 2\sqrt{7}} = 2.508$. That number belongs to no particular system. It is the rate at which a body at the boundary of any small secondary’s gravitational domain is pulled one way or the other, in units of how fast the secondary goes round.

The e-folding time is the orbital period divided by 2πλ. For the Sun–Earth system that is 365.25 days divided by 2π × 2.484, or 23.4 days: an error doubles every 16 days and grows a hundredfold in 15 weeks. For the Earth–Moon system the growth rate in natural units is almost the same, 2.16, and the period is 27.3 days, so the e-folding time is two days. The Earth–Moon $L_2$ point is thirteen times less forgiving than the Sun–Earth one for no reason except that the Moon goes round faster than the Earth does.

What an uncorrected error does #

The shape of the curve contains the whole strategy. For the first part of an e-folding time nothing dramatic happens: an error of a centimetre a second becomes a kilometre in a day and tens of kilometres in a month, and at those distances the linear picture is excellent. Then the exponential dominates, and a month later the same error is hundreds of kilometres and growing by a factor of e every three weeks.

An instability that grows exponentially is cheap to cancel early and dear to cancel late. The corrective velocity needed at any moment is about the velocity error along the unstable direction at that moment, and that grows with the drift. Catch it within an e-folding time and the correction is of the same size as the original error; wait four e-folding times and it is fifty-five times larger.

The same error at a slower system shows how completely the timescale is inherited from the orbit.

A spacecraft at Jupiter’s $L_2$ point could go most of a year between corrections. The point is no more stable in any natural unit; the whole difference is that the instability is measured in fractions of an orbit, and Jupiter’s orbit takes twelve years. The same reasoning makes the Earth–Moon points the most demanding in the solar system that any mission has used: fast orbit, same instability, and a correction schedule measured in days.

The optimal interval, and its price #

That argument has an optimum, and its value is the expression in the first figure.

Suppose each correction can null the unstable component only to an accuracy σ — the combined error of knowing where the spacecraft is and how fast it is moving, and of executing the burn. Over the interval T until the next correction, that residual error grows by $e^{T/\tau}$, and the next correction must remove it. So each correction costs about $\sigma e^{T/\tau}$ and a year of corrections costs

$$\Delta v_{\rm year} = \frac{365.25}{T}\,\sigma\,e^{T/\tau}$$

The first factor penalises frequent corrections, because each one adds its own σ; the second penalises infrequent ones. Differentiating, the minimum is at $T = \tau$ exactly, and the cost there is

$$\Delta v_{\rm year,min} = \frac{365.25\, e\, \sigma}{\tau}$$

— e times the error, once per e-folding time.

Three things about that result are worth drawing out. The optimal interval does not depend on σ at all: it is a property of the point, not of the spacecraft. The cost is proportional to σ, so it is a statement about navigation and execution, not about the strength of any force. And the minimum is broad: correcting at half the e-folding time costs about a fifth more and at twice it about a third more, which is why operators schedule manoeuvres at convenient intervals of about three weeks and lose little by it.

For the Sun–Earth $L_2$ point with σ of 2 cm/s the model gives under a metre per second a year. Real missions spend more — a few metres per second — because the model is one-dimensional and because solar radiation pressure on a large, flat sunshield is a steady error source the navigation has to keep absorbing. But the order of magnitude, and the schedule, come straight out of the expression.

The expression’s linearity in σ is its most practical consequence, and the reason the budget of an $L_2$ mission is set in the tracking station rather than in the propulsion system.

Doppler tracking measures a spacecraft’s line-of-sight velocity to a fraction of a millimetre per second, and ranging fixes its distance to metres. The limiting error at $L_2$ is therefore rarely the measurement itself. It is the unmodelled accelerations between measurements — radiation pressure on a sunshield whose reflectivity is known to a per cent or two, outgassing, the small velocity changes that come with desaturating reaction wheels — and the execution error of burns a few centimetres per second in size. Every one of those enters the budget as σ, linearly.

Why the stable orbit costs more #

Put the two budgets side by side. The geostationary satellite pays about fifty metres per second a year for its north–south control. The $L_2$ observatory pays a few. The difference is not that one is stable and the other unstable. It is what each is paying to cancel.

The geostationary orbit’s plane is pushed by a torque that is always there. The lunisolar pull tilts the plane by about 0.85° a year whether the satellite has been corrected recently or not, and the velocity change needed to undo a tilt of that size at geostationary speed is fixed by the geometry: about fifty metres per second per year of accumulated tilt. Better navigation does not reduce it. Correcting more often does not reduce it. It is a force budget.

The $L_2$ spacecraft is on an equilibrium where the net force, to first order, is zero. Nothing pushes it; it drifts away only because it was never exactly on the equilibrium in the first place. The budget is therefore set entirely by how far from exact it is placed each time, which is a navigation quantity, and the instability’s growth rate sets only how often that knowledge has to be refreshed. An unstable equilibrium held carefully is a navigation budget; a stable orbit pushed steadily is a force budget, and force budgets are the expensive kind.

There is a mirror of the argument on the geostationary side. The east–west budget of a geostationary satellite — the part that holds its longitude against the pull of the Earth’s elliptical equator — is only one or two metres per second a year. That drift, too, is slow and caught early. The expensive part of geostationary station-keeping is the one term that is a steady torque.

A point that cannot be held without manifolds #

The station-keeping picture also explains why getting to $L_2$ is cheap. The same instability that makes a spacecraft drift off the point makes trajectories drift onto it: the stable manifold of the halo orbit is a family of paths that arrive with no braking at all. A spacecraft launched onto one coasts into orbit about $L_2$ and needs only small corrections to arrive, exactly as a spacecraft on the orbit needs only small corrections to stay.

That symmetry is the reason missions to Sun–Earth $L_2$ plan their mid-course corrections the way they do. A trajectory that would overshoot onto the unstable manifold on the far side must be slowed, and slowing down means thrusting towards the Sun. A spacecraft whose sunshield must always face the Sun cannot do that, so it is launched on a trajectory aimed deliberately slightly short, with every correction in the direction it can thrust. The same one-sidedness applies on station: corrections are biased so that the drift is always towards the side the spacecraft can push against. It raises the budget a little and removes the failure mode entirely.

Why nobody sits on the point itself #

No mission has ever parked at a collinear point. They fly halo or Lissajous orbits round it, hundreds of thousands of kilometres across, and the reasons are about light and radio rather than about dynamics.

The Sun–Earth $L_1$ point lies directly between the Earth and the Sun. A spacecraft there would be seen from the Earth against the Sun’s disc, and its radio signal would be drowned in the Sun’s own radio emission; an orbit round the point keeps the spacecraft several degrees from the Sun as seen from the ground. The $L_2$ point lies directly behind the Earth, 1.5 million kilometres out, just beyond the tip of the Earth’s umbra, which is about 1.4 million kilometres long — a shadow’s length is set by the ratio of the two radii and the distance between them. A spacecraft at the point would sit in the Earth’s penumbra, partially eclipsed, with its power and its thermal balance disturbed; a halo orbit of a few hundred thousand kilometres keeps it in full sunlight.

The dynamics permit this at no extra cost because the halo orbits are themselves families of periodic solutions near the point, with the same instability and the same e-folding time. The first was flown in 1978 by ISEE-3, which spent four years in a halo orbit round the Sun–Earth $L_1$ point watching the solar wind arrive — and was then sent, by a sequence of lunar flybys that cost almost nothing in propellant, to fly through the tail of a comet. The same property that makes the point cheap to hold made it cheap to leave.

Choosing a gentler instability #

The expression has a lever the point itself does not offer: τ. At a given collinear point the e-folding time of the point is fixed, but the orbits round it are not all equally unstable, and a mission can choose among them.

The halo orbits about a collinear point form a continuous family, from small loops close to the point to large orbits that swing far out of the plane. The three-body problem has no general solution, but its periodic orbits can be found numerically and their stability computed, and along the halo family about the Earth–Moon $L_2$ point the instability varies by a large factor. The members that pass closest to the Moon — near-rectilinear halo orbits, elongated loops that dive to a few thousand kilometres over one lunar pole and climb to tens of thousands over the other — are far more weakly unstable than the small halo orbits near the point: an error grows by a factor of two or three per week-long revolution rather than by e every two days.

That is why the lunar orbit chosen for a long-duration crewed outpost, and flown first by a small navigation-demonstration spacecraft in 2022, is a near-rectilinear halo orbit rather than a halo close to $L_2$. On the expression here, lengthening τ by a factor of several cuts the annual cost by the same factor for the same navigation quality — which turns a budget of tens of metres per second a year into a few, and relaxes the correction schedule from every two days to roughly once per revolution. The orbit family was chosen for its stability number before anything else about it was decided.

The same selection is available at every collinear point and is used at the Sun–Earth ones too: large Lissajous orbits are chosen partly because their correction schedule is relaxed. An unstable equilibrium is not one number; it is a family of orbits with a spectrum of instabilities, and the station-keeping budget is a choice from that spectrum.

There is a price on the other side of the choice, and it is the reason the weakest instability is not always taken. A gentler orbit is also one that responds more sluggishly to deliberate changes, so reaching it and leaving it can take longer or cost more; and the near-rectilinear orbits pass so close to the Moon that their perilune passage is fast, lit and shadowed in quick succession, and sensitive to the Moon’s gravity field. The station-keeping expression prices only one of the costs a trajectory designer weighs. It prices that one exactly, and in the unit that most often decides a mission’s length: metres per second a year, set by how well the spacecraft’s own motion is known.

What the model leaves out #

The cost model is a caricature of one mode. A real halo or Lissajous orbit has an unstable direction, a stable direction and two oscillatory ones, and the navigation error projects onto all of them; a correction chosen to cancel only the unstable component is cheaper than one that nulls everything, and the best strategies do exactly that. The e-folding time is also not quite constant around a large halo orbit, because the linearisation is about the point rather than about the orbit.

The model treats σ as a fixed per-correction error, when in practice it depends on tracking geometry, the time since the last orbit determination and the reliability of the thrusters at very small impulses. And it omits the steady forces that do act at $L_2$ — solar radiation pressure most of all — which convert part of the budget back into a force budget, and which for a spacecraft with a large sunshield are the dominant source of the errors that the navigation has to catch.

An instability sets a rate, and the cost is set by how early it is caught #

An instability sets a rate, not a cost. What it costs to live with depends on how early it is caught, and if it can be caught while the error is still at the level of the measurement, the price is proportional to the measurement’s precision. A stable system under a steady perturbation has no such escape: the perturbation must be paid for in full however well it is measured.

The same distinction runs through a prediction with an expiry date: chaos sets how fast knowledge of a system decays, and the only response is to renew the knowledge, which buys time additively rather than multiplicatively. It applies to balancing a pole on a hand, where the correction is cheap if it is made quickly and impossible if it is made late, and to every control system that holds an unstable plant. The point $L_2$ is useful not despite its instability but because the instability can be bought off with information.

Still open: what the end of the propellant means #

Every station-keeping budget ends. A geostationary satellite is raised a few hundred kilometres into a graveyard orbit with its last propellant, and an $L_2$ spacecraft is steered onto an unstable manifold that carries it away into a heliocentric orbit. How much propellant remains when neither kind of tank has a gauge — how the remaining quantity is estimated from the history of every burn, how that estimate’s error grows with the number of manoeuvres, and what margin a disposal has to keep against it — is the question that decides when a working spacecraft has to be switched off.