An answer from two bodies, flown through a flattened Earth

Two places and a clock decide the path: given a starting point, a destination and a time of flight, Lambert’s theorem says there is a conic — or a small number of them — that connects the two in exactly that time, and a well-chosen variable finds it in three or four iterations. Every flight-dynamics library contains a Lambert solver, and every rendezvous, every interplanetary trajectory and every orbit-transfer plan starts with one.

The theorem is exact for two bodies. A spacecraft in low Earth orbit is not in a two-body problem. The Earth is flattened — its equatorial radius exceeds its polar radius by twenty-one kilometres — and the extra mass around the equator adds a force to the central inverse square, about a thousandth of it at low altitude. A thousandth sounds negligible. Flown for an hour, it moves a spacecraft by tens of kilometres.

A thousandth of the force, an hour of flight #

The size of the effect is easy to estimate. The bulge is described, to first order, by a single coefficient $J_2 = 1.083 \times 10^{-3}$, and the extra acceleration it causes at the Earth’s surface is of order $J_2$ times the surface gravity — about a centimetre per second squared. At 400 kilometres altitude, where the radius is 1.06 times the Earth’s and the bulge’s term falls as the fourth power of the radius, it is about a fifth less, and it varies with latitude by a factor of two.

A constant acceleration $a$ acting for a time $t$ displaces a body by $\tfrac12 a t^2$. For an hour that is about fifty kilometres. The real effect is not a constant acceleration in a fixed direction — its direction rotates with the orbit and its strength varies with latitude — so parts of it average away. But the parts that do not average away accumulate, and over a transfer of an hour the miss is tens of kilometres, which is what the figure shows.

This is not a small correction to a precise answer. It is a precise answer to a different problem. The two-body Lambert solution is the exact velocity for a field the spacecraft is not in, and the error it carries is a hundred thousand times the precision a rendezvous needs.

Where the error goes #

The three components have three different origins.

Along the track is the largest, and it comes from the bulge’s effect on the orbital period. For an orbit inclined less than about 55 degrees, the extra mass around the equator adds to the inward pull on average, which makes the orbit slightly faster than a two-body orbit of the same size; above 55 degrees the average reverses and the orbit runs slightly slow. A spacecraft that runs a fraction of a per cent faster than predicted is ahead of its predicted position by an amount that grows with every minute. This is the error the time-of-flight equation is most sensitive to, since Lambert’s problem is fundamentally a problem about timing.

Across the track comes from the bulge’s tendency to twist the orbit’s plane. The extra mass at the equator pulls an inclined orbit toward the equatorial plane as the spacecraft crosses it, and the result is a slow regression of the line where the orbit crosses the equator — the nodal regression that was first used to measure the Earth’s shape. For an orbit at 51.6°, the node moves westward by about five degrees a day; during a seventy-minute transfer it moves by about a quarter of a degree, which at the orbit’s radius is thirty kilometres of sideways displacement at the point farthest from the node.

Radially the error oscillates rather than grows: the bulge’s pull varies twice per orbit as the spacecraft passes over the equator and the high latitudes, and the orbit’s radius responds with a small oscillation of a few kilometres.

The same transfer, at every tilt #

The decomposition explains why the same transfer misses by such different amounts on differently tilted orbits.

An equatorial orbit stays over the bulge all the time. The extra pull is strongest there, it is always radial, and it changes the orbit’s period more than for any other inclination, so the along-track error is largest. An equatorial orbit’s plane does not regress — it already lies in the plane the bulge pulls toward — so there is no cross-track error at all, but the along-track error alone exceeds a hundred kilometres for a transfer of three-quarters of an orbit.

As the inclination rises, the spacecraft spends less time over the equator and the average extra pull weakens. Near 55 degrees the bulge’s effect on the orbit’s average angular speed passes through zero, and near 63.4 degrees the rotation of the orbit’s long axis does — which is why the Molniya orbits are flown there. The along-track error falls through that range; the cross-track error from the plane’s regression grows with inclination, then falls again as the orbit approaches the pole and the regression stops. The minimum of the total lies where the two compromise, near seventy degrees for this transfer.

The retrograde side mirrors the prograde one. A sun-synchronous orbit at 98 degrees, inclined just past the pole so that its plane regresses at exactly the rate needed to keep pace with the Earth’s motion around the Sun, has a two-body error well below the equatorial one, not far from the minimum.

Lying to the solver #

The obvious repair is to solve the real problem — a boundary-value problem in the flattened field — directly. That requires numerical integration inside an iteration, and before computers were fast enough to do it routinely, a cheaper method was devised that keeps the two-body solver and changes only what it is asked.

The method works because the error is small and nearly linear. If the bulge displaces the arrival point by a vector $\boldsymbol\delta$ when the spacecraft aims at the target, it will displace it by very nearly the same $\boldsymbol\delta$ when the spacecraft aims at a point slightly different from the target — the perturbation hardly depends on which of two nearby two-body arcs is flown. So aiming at the target minus $\boldsymbol\delta$ puts the arrival at the target, to within the change in $\boldsymbol\delta$ between the two arcs, which is of order $J_2$ times $\boldsymbol\delta$ — a thousandth of a thousandth of the original error, scaled by how sensitive the perturbation is to the arc.

That is the same reasoning as the differential correction that aims an interplanetary trajectory to within kilometres of Mars, with one difference that makes it simpler: the corrected-target method never needs the matrix that says how the arrival depends on the departure. The Lambert solver supplies that relationship implicitly, every time it is asked for a new aim point. The iteration is Newton’s method with the two-body problem standing in for the derivative of the real one.

The convergence is not guaranteed in general. It works because the perturbation is small compared with the central force and smooth across the family of nearby arcs. For a trajectory with no good first guess — near a Lagrange point, where the dependence of arrival on departure is exponential — the two-body arc is not a nearby approximation to anything, and the corrected target diverges. For a transfer in low Earth orbit, where the perturbation is a thousandth of the force and the dynamics are nearly linear, it converges by two orders of magnitude a pass.

What the iteration converges to is the model, not the sky. Each pass measures the miss by integrating a model of the field, and the iteration drives that modelled miss to zero. If the model omits a force — drag on a day when the Sun is active, a harmonic of the gravity field that matters for that orbit — the corrected burn hits the modelled target perfectly and the real one by whatever the omitted force adds. The iteration’s convergence to centimetres in the figure is a statement about how well the two-body answer can be corrected toward a $J_2$ field; the real accuracy of a burn is set by how well the field and the spacecraft’s state are known, which is why rendezvous approaches are flown in stages, with tracking and a fresh correction before each burn.

What it costs in velocity #

The correction moves the aim point, and moving the aim point changes the departure velocity. That change is the bulge’s cost.

For most transfers the correction is a few metres to a few tens of metres a second. Against the 7.7 kilometres a second of the orbit it is under one per cent, and the two-body answer is a fine first estimate. Against the precision of a rendezvous burn — which must be good to a few centimetres a second, because every centimetre a second of error becomes metres of miss at docking — it is a thousand times too large, and the two-body answer is useless as a command.

The rise near 180 degrees is the most instructive feature. The variable that makes Lambert’s problem easy removes every singularity but one: at exactly 180 degrees the two position vectors and the centre are collinear, and the plane of the transfer is not determined by them. Near 180 degrees the plane is determined, but weakly — a small sideways displacement of the target swings the transfer plane through a large angle. The bulge supplies exactly such a displacement, the cross-track drift from nodal regression, and correcting for it near 180 degrees means swinging the plane hard. The corrected target inherits the two-body problem’s geometry, including its one remaining weakness.

Why the two-body solver is kept #

It might seem perverse to keep a solver for the wrong problem and correct its answers, when the right problem can be integrated directly. There are three reasons, and they explain why the corrected target is still how most targeting is done.

The two-body solution is the right starting point. Any method for the real problem needs an initial guess, and a guess within a fraction of a per cent of the answer — which is what the two-body solution is — converges in a few steps by any method. Without it, a solver for the real boundary-value problem has to search.

The two-body solver is robust. It is a single root-find in one variable, always well posed away from 180 degrees, with no integration inside it. Wrapping it in an iteration that only moves the target keeps all of that robustness and adds a correction whose size is known in advance to be small.

The structure of the problem is preserved. A two-body Lambert problem has branches — short way and long way, zero and several revolutions — and the multiple-revolution solutions are a fold structure that a mission designer chooses among deliberately. A direct solver of the perturbed problem finds whichever branch its first guess lies near; the corrected-target method keeps the branch the two-body solver was asked for, and moves it only by the perturbation.

A rendezvous is a relative problem #

A rendezvous does not, in the end, need either spacecraft to arrive at a point fixed in space. It needs the chaser to arrive where the target is, and the target is in orbit too, perturbed by the same bulge. Much of the bulge’s effect is common to both: two spacecraft on nearby orbits have their planes turned by nearly the same amount and their periods changed by nearly the same fraction, and in the frame that moves with the target those common effects cancel.

What survives in the relative frame is the difference — the bulge’s effect on one orbit minus its effect on a slightly different one. For orbits twenty kilometres apart in altitude it is a few per cent of the absolute effect, and the relative-motion equations that rendezvous planning uses, in which the relative orbit breathes once per revolution around the target, are usually written for a spherical Earth with the bulge added as a small correction. The counter-intuitive manoeuvres of rendezvous — catching up by slowing down, dropping to a lower orbit to gain on the target — are two-body results that survive the bulge almost intact, because the bulge acts on both orbits nearly alike.

The absolute miss matters where the two orbits are not nearby: the climb of a freshly launched vehicle from its injection orbit to the station’s altitude, the transfer of a servicing spacecraft between satellites in different planes, the first manoeuvre after separation from a launcher. Those transfers span tens to hundreds of kilometres and many minutes, the two orbits do not share their perturbations, and the corrected target is how the burns are computed.

How large a thousandth is #

The coefficient $J_2$ is one of the best-measured numbers in geophysics, and it was first measured from exactly the effect that spoils the two-body answer. When the first artificial satellites were tracked in 1958, their orbital planes were seen to regress at rates the spherical-Earth model did not predict, and the rates gave the flattening directly — more precisely than a century of surveying had. The Earth’s shape was read off a satellite’s node, and the flattening it gave, one part in 298.3, has hardly changed since.

So the force that makes a two-body answer wrong is also the force that made satellites a geodetic instrument. The cross-track error in the figures is the same nodal regression that measured the Earth; the along-track error is the same change of period that the measurement had to separate from it. A spacecraft that flew the two-body answer and missed its target by thirty kilometres would, if tracked precisely, be measuring the Earth’s flattening along the way.

Beyond the bulge #

The flattening is the largest departure from a two-body field in low Earth orbit, but not the only one, and the same method absorbs the others.

Higher harmonics of the Earth’s gravity field — the pear shape, the ellipticity of the equator, the mountain ranges and ocean trenches mapped by gravity missions — add forces a thousand times smaller again, and which of them a satellite feels depends on its orbit. Atmospheric drag at 400 kilometres decelerates a spacecraft by a few micrometres per second squared, varying with solar activity by a factor of ten, and over an hour it moves the arrival point by tens of metres. The Moon and Sun add tidal accelerations of order a micrometre per second squared.

All of these enter the corrected-target iteration in the same way: the integration that measures the miss includes them, the miss shrinks, and the two-body solver never knows they were there. The method is indifferent to what the perturbation is, provided it is small and smooth — which is the assumption that makes it fail at the edges of its domain, where a perturbation stops being small.

What was actually computed #

The figures use a three-dimensional universal-variable Lambert solver and a fourth-order Runge–Kutta integration of the Earth’s field with its $J_2$ term and nothing else — no higher harmonics, no drag, no Moon or Sun. The departure is a point on a circular orbit at 400 kilometres altitude and the target is a point on a circular orbit at 420 kilometres in the same plane; the time of flight is the transfer angle divided by the mean motion of an orbit halfway between, so that each case is a gentle climb, not a fast transfer. The 180-degree case is left out, where the solver has no unique answer.

Real rendezvous targeting differs in detail: the target is a moving spacecraft whose own orbit is perturbed by the same bulge, so the relative motion matters more than either orbit separately, and much of the bulge’s effect cancels between two spacecraft on nearby orbits. The miss in the relative frame is smaller than the absolute misses drawn here, by an amount that depends on how close the two orbits are. The absolute miss is the right measure for transfers between orbits that differ substantially — the climb from an injection orbit to a station, the transfer of a tug between two satellites — and an upper bound for the final approach.

Still open: when the corrected target does not converge #

The corrected target converges here because every ingredient favours it: a perturbation a thousandth of the central force, an hour of flight, dynamics close to linear. Each ingredient can fail. Long transfers of many revolutions let the perturbation accumulate until it is no longer small compared with the transfer’s own geometry; very low orbits add drag that is large, uncertain and changes from orbit to orbit; and transfers near the 180-degree geometry inherit the two-body problem’s indeterminate plane.

For multi-revolution transfers in particular the bulge changes not only the arrival point but which branch of the two-body solution is cheapest, since the branches differ in how many times they cross the equator and how much nodal regression they accumulate. A cost ranking computed from the two-body problem can be wrong after correction, and a mission planner choosing among branches by their two-body costs can choose the wrong one. Ranking branches by their corrected costs requires correcting every branch, which multiplies the work by the number of branches — and is the point at which solving the perturbed problem directly, with a two-body solution only as the first guess, starts to win.