An orbit held still by the Earth's pear shape

Most of what the subject of station-keeping says is about paying. An orbit has a yearly cost, the gravity field, the Moon and the Sun all pull a satellite away from where its owners want it, and propellant restores it for as long as there is propellant. In a few special cases, however, the perturbations cancel, and a satellite placed in the right spot stays there. A geostationary satellite has longitudes it drifts towards by itself. And one of the most widely used orbits around the Earth is special in this way, for a reason that has nothing to do with height and everything to do with a detail of the planet’s shape.

Almost all satellites that map the sea surface from above fly a near-circular orbit with its eccentricity set to about 0.001 and its perigee over the North Pole. These numbers look like the outcome of a design compromise. They are in fact a fixed point. If the eccentricity is any other value, it will change with time. At precisely this value, and with precisely that orientation, it does not change. The orbit is said to be frozen.

The balance that produces this is a nice illustration of how unequal forces can trade. $J_2$, the flattening of the Earth, is more than four hundred times as large as $J_3$, the coefficient for its slight pear shape. A force so much smaller should be a minor correction. In most respects it is. Here it determines the whole answer, because the large force’s first-order effect on a near-circular orbit has a peculiar null: it turns the eccentricity without changing its size, and it can do that about any centre.

A vector that turns #

The usual elements are a poor description of a nearly circular orbit, as the argument of perigee stops meaning anything when there is no clearly defined perigee to measure. The cure is to combine the eccentricity and the argument of perigee into two components of a vector, $k = e\cos\omega$ and $h = e\sin\omega$. That vector is the natural variable here, and it changes smoothly through zero.

Under $J_2$ alone, the orbit’s perigee rotates at a steady rate, and the size of the eccentricity does not change:

$$\dot\omega = \frac{3}{4}nJ_2\left(\frac{R}{a}\right)^2(5\cos^2 i - 1).$$

So the vector goes round the origin in a circle. The same rate was what gave the first measurements of the Earth’s shape. Now add $J_3$. This term’s force is asymmetric between north and south, so it pushes the orbit a little differently on either side of the equator. Its effect on the vector is a constant drift along the $k$ direction. A rotation combined with a constant drift is a rotation about a different centre, just as a thing turning on a moving platform goes round a displaced point. The centre sits on the $h$ axis, at

$$h^* = -\frac{J_3}{2J_2}\frac{R}{a}\sin i.$$

The point is displaced by a ratio of two small coefficients, and that is why it exists at a usable value. $J_3$ for the Earth is negative at about $-2.5\times10^{-6}$ and $J_2$ is about $1.08\times10^{-3}$, so their ratio is a little over two parts per thousand, and half of that times a factor near one gives about 0.001. This number is a statement about the planet. It is set by the small hemispheric asymmetry in the Earth’s mass, first measured from the orbit of Vanguard 1 in 1958 and 1959, and popularised at the time as the pear shape of the Earth. It has been built into the design of an entire class of spacecraft ever since.

The sign is significant, too. $J_3$ negative puts the centre at positive $h$, with $\omega = 90$ degrees: perigee over the northernmost point of the orbit. On a world with $J_3$ of the other sign, the same balance would put it over the south.

Why the small term decides #

The picture offers an answer to what seems like a paradox. A force four hundred times weaker cannot usually determine a stable state against a stronger one. Here it can, because the stronger force never tries to set the eccentricity’s size. Its secular effect on $e$ is exactly zero to first order, and all it does is turn the vector. So the comparison is not between two forces but between a rotation and a shift, and any shift, no matter how small, moves the rotation’s centre by a proportional amount.

That same reasoning sets a limit on how much the centre matters. The rotation’s radius, as seen in the figure, depends on where the orbit starts. Any eccentricity not at the centre circles round it, and the size of the swing is as large as the separation.

A nearly circular orbit is therefore not naturally circular, and it does not stay still when placed at $e=0$. Launch a satellite into the most perfectly circular orbit available, and within two months its eccentricity will have grown to twice the frozen value, as it swings out past the far side of the circle. It will come back two months later. Perfect circularity is not the undisturbed state, in the same way that the true equilibrium of a tilted moon is not the plane nearest it. The equilibrium for this planet lies at $e = 0.001$.

What a repeating height is worth #

The practical value of this lies in what the eccentricity does to the height of the satellite. An orbit with $e = 0.001$ at 800 kilometres has a range of about fourteen kilometres between perigee and apogee. That is small, but an altimeter measures the height of the sea surface by radar to a few centimetres, and the satellite’s height enters that measurement directly.

The profile shows why oceanographers want this. Their signal is the change in the sea surface over time, a few centimetres or less, measured by comparing passes over the same ground track. If each pass over a given latitude comes at a different height, then the corrections for drag, for the signal’s path and for the gravity that a satellite feels all change from one pass to the next. A gravity field felt more strongly at lower altitude would appear as an apparent change in sea level unless modelled exactly. With a frozen orbit the height repeats. At any given latitude the satellite always passes at about the same altitude, and repeat-track comparison becomes differencing a signal with itself. It removes a whole category of error that no later correction could remove as cleanly.

There’s also a saving that follows directly: an orbit near its centre needs very few manoeuvres to stay there. Since the frozen point is a stable centre in this plane, a small error produces a small circle, and it does not grow unless driven. Sun-synchronous imaging satellites, for which a consistent height means consistent scale and illumination, generally use it for the same reason. So do most recent altimetry and weather platforms. It adds nothing to the launch cost and nothing to the mission’s station-keeping budget beyond a brief period spent settling onto it.

From an oddity in Vanguard to a design rule #

The chain of events behind this is short and rather pleasing. Vanguard 1, launched in March 1958, was a six-inch sphere intended as a test of a launch vehicle. Its orbit had an unusually long life for its time, and John O’Keefe, Ann Eckels and Kenneth Squires noticed that its perigee height oscillated with the period of its perigee’s rotation, by about five kilometres. They interpreted this as the effect of an odd harmonic, in 1959, and the result was widely reported as the discovery that the Earth is shaped like a pear, the difference between the hemispheres being a few tens of metres. The announcement was a curiosity for the newspapers. It was also, in the language of this account, the first measurement of a circle’s displaced centre, though nobody needed to call it that.

The oscillation they saw is what the second figure shows, at a larger eccentricity. Any orbit that happens not to be at the centre rotates around it, and the size of its perigee swing tells of a term that the flattening cannot supply. Once that swing had been explained, the idea of removing it was straightforward. Writing the centre down as a condition for an orbit followed in the 1960s, in the analytical work of Geoffrey Cook and others on the long-term behaviour of satellite orbits. And Seasat, the first dedicated ocean-observing altimeter satellite, launched in 1978, flew it for exactly the reason given here. Since then it has been the default.

It is uncommon for a discovery in geodesy to turn into an engineering specification so directly. The value of the coefficient has been improved by a few parts in ten thousand since 1959, and the frozen eccentricity has changed by the same small fraction. The only real change has been in what can be measured with the result. A radar altimeter on a frozen orbit now records a global rise in sea level of a few millimetres a year, and it does so by repeating the same tracks at the same heights for three decades, across a sequence of satellites deliberately placed on one another’s paths. That record of the ocean rests in part on a six-inch test sphere’s unexpected wobble.

A manoeuvre to reach the frozen point is not expensive, but it does have to be done correctly. A newly launched satellite is measured for a few days, its mean elements are separated from the oscillating ones, and then two small burns half an orbit apart move the vector to the centre, at a cost usually below a metre per second. Compared with the several metres per second a year required by drag, this is a negligible price.

How it depends on the orbit #

The position of the centre depends on the orbit in a simple way.

The $\sin i$ factor reflects the geometry of an asymmetry between north and south. An orbit near the equator never samples the difference between hemispheres, so it feels no such drive and its centre is at zero. A polar orbit samples it fully. Radius changes the centre only modestly, through the ratio $R/a$, which is one reason so many missions end up at nearly the same value. Higher zonal harmonics, $J_5$ and above, shift it by small amounts in detail, and in practice the frozen point is found by numerically integrating an orbit with the full gravity model, not from this formula. Even so, its first-order version comes within about ten per cent.

The rate at which the vector circles about the centre has a different behaviour, which is the more interesting for what it implies.

The factor $5\cos^2 i - 1$ is the critical inclination’s signature. Both rotation and drive contain that factor, so the location of the centre, their ratio, stays where it was while the rate falls to zero. At exactly the critical inclination nothing circulates at all, and any eccentricity is frozen to first order. This has become the most familiar special orbit of all: the Soviet Molniya communications satellites, with an eccentricity of about 0.7 and an inclination of 63.4 degrees, keep their apogee fixed over the high northern latitudes they serve. These two orbits have seemed very different things, a highly elliptical communications orbit and a nearly circular oceanographic orbit, and both are the same frozen condition: one with the rotation term switched off, and one with its centre moved by the next harmonic.

The same balance elsewhere #

Each planet has its own coefficients, and the same formula gives very different answers.

Mars has a substantial pear shape, thanks to the large difference in crustal thickness between its two hemispheres, and its $J_3$ is more than ten times the Earth’s in relative terms. Mars mapping orbiters have consistently used frozen orbits, with periapsis near the south pole, where the elevation and the gravity would otherwise bring an orbiter closest to the surface. Mars Global Surveyor and Mars Reconnaissance Orbiter both flew in that way. The balance sets where the periapsis must be, and that happens to be where it does most good.

The Moon is a very different case. Its field has a strong sequence of higher harmonics and mass concentrations, and the first low lunar orbiters discovered how fast perigee can fall into the surface. A two-term balance tells little there, since the harmonics beyond $J_3$ matter at least as much, and low polar orbits around the Moon decay in months without active maintenance. There are stable inclinations for low orbits, but they are found numerically and do not follow from this reasoning. A related frozen solution at high altitude, in which the Earth’s tide replaces $J_2$, is what is used to keep a navigation constellation over the lunar south pole. Venus has so nearly spherical a shape that a formal solution of this kind gives an unrealistically large eccentricity. The formula then says little about the true behaviour, and it reminds a reader that this is a first-order result for near-circular orbits.

A balance between unlike things #

The most surprising thing here is how little has to be known. An orbit’s frozen state can be reached with a value of $J_3$ that nobody could have measured before satellites. A hemispheric difference of a few tens of metres in the Earth’s mass distribution, too small to be seen from the surface, has come to determine the eccentricity of most of the Earth-observing fleet.

It is also a useful correction to a natural way of thinking about perturbation theory. A hierarchy of forces ordered by size suggests a hierarchy of effects ordered in the same way. That is often true and often very misleading. When the largest force has a null in the quantity in question, the next largest sets the answer, and the theory that expands in the strongest force has no business being surprised. A similar logic shows up in the graveyard orbit, where sunlight sets a term on an eccentricity vector in a way that the much stronger gravity leaves untouched. The vector is the object in which these effects add, and a circle about a displaced centre is the shape they take.

What the picture cannot show #

The equations used here are averaged over one orbit, and they include only two harmonics. Over each revolution there are short-period oscillations in the eccentricity of comparable size to the frozen value itself, so a real satellite’s osculating elements, computed at any moment, can look nothing like the mean values plotted here. Getting a satellite exactly onto the frozen point requires that the mean elements be separated from the osculating ones with care, because the two can differ by a factor of two or more in eccentricity.

The Moon and Sun are omitted, as are drag and solar radiation pressure. For most low orbits those are small effects on the eccentricity over a year, but at the end of a solar maximum drag can matter a good deal, and over many years lunisolar forces matter. The height profile uses a very simple ellipsoid and an assumed 800-kilometre mean, and it ignores everything else that makes the height above the actual sea vary. The body comparison uses single choices of altitude and inclination for each world and the coefficients available in round numbers. So its numbers do not describe particular missions.

Finally the circles themselves conceal a limitation. The real centre has a small drift of its own under higher-order terms, and a mission holds it there with occasional burns. The orbit is frozen, in other words, only in an approximate sense, and it is not costless, merely cheap.

Still open: how far the method reaches with a changing constellation #

The frozen condition was worked out for a single satellite and a static gravity field. Now large constellations in low orbit want to keep thousands of satellites on predictable tracks, and the same conditions apply, with a further requirement that satellites in neighbouring planes must not come close to each other at the same latitude. Whether a whole shell of orbits can be arranged so that the frozen points of all its members keep them apart, with mutual safety built into their geometry, has been studied with encouraging results. It has not been shown to hold in practice over a decade with storm-time drag changes and the slow change in the Earth’s gravity field. The most delicate question is what happens when a frozen satellite fails and stops being maintained: its mean elements will continue around a small circle for a long time, and whether it keeps its protected height or wanders into its neighbours’ is a matter that the same simple diagram helps to address and doesn’t settle.