The average depends on what is being averaged

Every table of the solar system gives a planet’s mean distance from the Sun, and for the Earth the figure quoted is one astronomical unit. The phrase reads like a measurement of the same kind as a mass or a period. It is not one, since the orbit is an ellipse with the Sun at a focus and the distance therefore changes continuously, so a mean over what has to be settled before there is a number to quote.

There are four obvious things to average over: the clock; the angle a direction to the body supplies; the geometric angle that makes the algebra tractable; and the reciprocal distance rather than the distance. Two of the four answers are exactly the semi-major axis at every eccentricity, the other two are not, and the two reasons for exactness are different reasons.

Four numbers where one quantity was expected #

With $r$ the distance from the occupied focus, $a$ the semi-major axis, $b$ the semi-minor and $e$ the eccentricity:

$$\langle r\rangle_t = a\left(1 + \tfrac{1}{2}e^2\right), \qquad \langle r\rangle_\nu = a\sqrt{1-e^2} = b,$$

$$\langle r\rangle_E = a, \qquad \frac{1}{\langle 1/r\rangle_t} = a.$$

The subscript names the variable the average runs uniformly over. The true anomaly $\nu$ is the angle at the focus and the observable one — a direction to a body is a statement about $\nu$ and nothing else. The eccentric anomaly $E$ is the angle at the centre of the circle the ellipse is a vertical compression of, a construction with no instrument behind it.

Three claims follow, all checkable. The first is an ordering: $b \le a \le \langle r\rangle_t$ at every eccentricity, with equality only on the circle, so the semi-major axis is the middle of the four. The second is that $a$ is the mean distance in exactly two senses out of four, which matters because $a$ is the length the subject is built on — the size in Kepler’s third law, the energy in the vis-viva equation.

The third is that the average anyone would reach for first, the distance recorded against the clock, is one of the two that is not $a$. Sampling a planet’s distance daily for a year and dividing by 365 gives a number larger than the semi-major axis by $ae^2/2$.

One substitution, and three of the four fall out #

All four are cheap at once because one variable linearises the radius and the time together. In the eccentric anomaly the radius is a single cosine, $r = a(1 - e\cos E)$, and Kepler’s equation $M = E - e\sin E$, differentiated, supplies the time. Since the mean anomaly runs at a constant rate,

$$dt = \frac{T}{2\pi}(1 - e\cos E)\,dE,$$

so the weight that converts an average over $E$ into an average over time is the radius itself, in units of $a$. The rest is that one fact used three ways.

Uniformly in $E$, the mean of a cosine over a turn is zero:

$$\langle r\rangle_E = \frac{a}{2\pi}\int_0^{2\pi}(1 - e\cos E)\,dE = a.$$

Averaging in time squares the bracket, and the mean of $\cos^2$ is one half:

$$\langle r\rangle_t = \frac{a}{2\pi}\int_0^{2\pi}(1 - e\cos E)^2\,dE = a\left(1 + \tfrac{1}{2}e^2\right).$$

The harmonic mean is not an integral at all: the weight is $(1 - e\cos E)$ and the integrand its reciprocal over $a$, so they cancel at every point of the orbit before any summing begins:

$$\langle 1/r\rangle_t = \frac{1}{2\pi a}\int_0^{2\pi} \frac{1 - e\cos E}{1 - e\cos E}\,dE = \frac{1}{a}.$$

The two exact results are exact for different reasons: one cancellation happens after integration, out of a symmetry of the cosine, the other before it, node by node — which is why it survives however coarse the sum.

The average over angle is the semi-minor axis #

The fourth average needs the second law rather than Kepler’s equation. The radius vector sweeps area at a constant rate, which is to say $r^2\,d\nu/dt = h$ for a fixed specific angular momentum $h$. So an average over true anomaly can be turned into one over time:

$$\int_0^{2\pi} r\,d\nu = \int_0^{T} r\,\frac{h}{r^2}\,dt = h\int_0^T \frac{dt}{r} = \frac{hT}{a},$$

using the harmonic-mean result at the last step. Dividing by $2\pi$ and substituting $h = \sqrt{\mu a(1-e^2)}$ and $T = 2\pi\sqrt{a^3/\mu}$ leaves $\langle r\rangle_\nu = a\sqrt{1-e^2} = b$.

The average distance over the angle a telescope measures is therefore the semi-minor axis — the one length of an ellipse astronomy otherwise never uses, arriving as the answer to a question about distance rather than shape. It is also the smallest of the four, because uniform steps in $\nu$ under-weight the far half of the orbit, where a body covers angle slowly.

Invisible for a planet, unavoidable for a comet #

The two moving averages depart from $a$ at second order in the eccentricity, by $\pm ae^2/2$, which makes the distinction a curiosity for the planets and a necessity elsewhere.

For the Earth $e^2/2$ is $1.39\times10^{-4}$. The time average of the Sun’s distance exceeds the semi-major axis by 20,900 km and the semi-minor axis falls short of it by the same 20,900 km, out of 149.6 million — both dwarfed by the 2.5 million kilometres by which the Sun sits off the centre, which is why an orbit can look exactly like a circle and still have four distinct mean radii. For Mercury the spread is 2.14% and the extremes lie 4.25% of $a$ apart, 0.3788 AU against 0.3953 AU, which is beginning to be a difference a table ought to name.

For a comet it stops being a subtlety. Halley’s orbit has $e = 0.96714$ and $a = 17.83$ AU, and the four averages of its distance are 26.17 AU, 17.83 AU, 17.83 AU and 4.53 AU. A quoted mean distance for that orbit conveys nothing unless it says which mean, since the largest is 5.8 times the smallest — one inside Jupiter’s orbit, the other beyond Uranus’s.

The year’s sunlight is the area law and nothing else #

One average has a physical consequence rather than a bookkeeping one, and it is not an average of $r$ at all: what a climate responds to is the flux, which goes as $1/r^2$. That average takes two lines, neither containing an integral that has to be done. Since $d\nu/dt = h/r^2$,

$$\int_0^T \frac{dt}{r^2} = \frac{1}{h}\int_0^{2\pi} d\nu = \frac{2\pi}{h},$$

and since the area swept in one period is $A = \pi ab$ at the constant rate $h/2$, giving $h = 2A/T$,

$$\left\langle \frac{1}{r^2}\right\rangle_t = \frac{2\pi}{hT} = \frac{\pi}{A} = \frac{1}{ab}.$$

The annual mean insolation is the total angle swept divided by the total area swept, and therefore nothing but the second law. Against a circular orbit of the same semi-major axis it is $1/\sqrt{1-e^2}$: it rises with eccentricity, and at second order.

A fifth mean hides in that result, and it is the one with a physical meaning attached: since the annual total is $1/(ab)$, a circular orbit of radius $\sqrt{ab}$ — the geometric mean of the semi-axes — delivers exactly the sunlight the ellipse does. That, not $a$, is the radius an eccentric planet should be compared against when asking whether it lies inside the band where liquid water is possible.

Why the ice ages do not run on eccentricity #

A factor of 240 between the seasonal and the annual consequence of one number is large enough to decide a question in another subject.

The Earth’s eccentricity is not fixed: it varies between about 0.005 and 0.058 with a period near 100,000 years, and the annual sunlight total varies with it, from +0.001% to +0.169% above the circular case. That is the whole available range of the effect — under two parts in a thousand of the year’s energy, across the entire eccentricity cycle. No plausible climate sensitivity turns it into an ice age.

The obliquity does far better without changing the annual total at all: it moves sunlight between latitudes and seasons, and the leverage is enormous where the ice sheets are. Both effects were worked out from celestial mechanics alone in the 1920s and confirmed from ocean sediment cores half a century later, and the cores found the obliquity and precession periods rather than a straightforward eccentricity signal. Tilting the axis and turning it slowly redistributes the year’s sunlight; changing the orbit’s shape barely changes how much of it there is.

Eccentricity is not absent from the theory, but its role is indirect: it sets the amplitude of the precession term, since the seasonal contrast that precession moves around the orbit is itself proportional to $e$. A modulation is not a forcing.

The same identity, read as an energy #

Of the four, the harmonic mean in time looked like an oddity: an average of a reciprocal, chosen because it comes out exactly. It is the deepest of them. The specific potential energy on a Kepler orbit is $-\mu/r$, so its time average is $-\mu\langle 1/r\rangle_t = -\mu/a$ exactly, while the total specific energy is $\varepsilon = -\mu/2a$. The averaged potential is therefore exactly $2\varepsilon$, the averaged kinetic energy is $\varepsilon - 2\varepsilon = -\varepsilon$, and

$$2\langle T\rangle_t = -\langle U\rangle_t.$$

That is the virial theorem, holding exactly for one body on one closed orbit rather than statistically for a crowd. Averaging the vis-viva relation the same way gives its useful form, $\langle v^2\rangle_t = \mu(2\langle 1/r\rangle_t - 1/a) = \mu/a$. The root-mean-square orbital speed is exactly the circular speed at $r = a$, at every eccentricity — 29.78 km s⁻¹ for the Earth, and the same for a comet of equal semi-major axis on a wildly different path. That step from a time-averaged reciprocal distance to a mass is how an elliptical galaxy is weighed, where the orbits are unknown, unresolved and never repeat and the statistical version of the identity is all there is. An average nobody would think to take of one planetary orbit turns out to supply the masses of galaxies, which is what happens when the averaging variable is chosen to match the physics instead of the instrument.

The average with no closed form #

Every result above came out in elementary functions, and one obvious member of the family does not. It is worth following, because the reason is the same reason an ellipse has no simple perimeter.

The mean speed in time is the distance travelled divided by the period — the perimeter of the ellipse over $T$. The perimeter is

$$P = 4a\int_0^{\pi/2}\sqrt{1 - e^2\sin^2\theta}\;d\theta,$$

which is the complete elliptic integral of the second kind and has no expression in elementary functions at any eccentricity but zero. The mean speed therefore has none either, and it is the one average of an orbital quantity in this essay that has to be evaluated numerically or as a series.

That is a curious asymmetry. The mean of the square of the speed is exactly $\mu/a$, in one line; the mean of the speed itself is a transcendental function of the eccentricity. Squaring made the problem easy because $v^2$ is a linear combination of $1/r$ and a constant, and both of those have exact time averages, while $v$ itself is a square root and square roots do not average.

The same asymmetry appears everywhere in the subject and is usually invisible because nobody asks for the harder quantity. Energy is quadratic in velocity and is therefore tractable; momentum is linear and is a vector, so its average round a closed orbit is zero and carries no information. The quantity that is both scalar and linear in speed — the path length — is precisely the one that resists.

There is a practical consequence for anyone integrating an orbit. A step size chosen to be uniform in path length would be the natural choice for controlling truncation error, since the error per step scales with how far the body moves; and it cannot be computed in closed form, so integrators step uniformly in time or in one of the anomalies instead and accept a step size that varies along the orbit.

Stepping uniformly in the eccentric anomaly is the usual compromise, and the reason is visible in the substitution that made three of the four averages easy: uniform steps in $E$ are steps whose time weight is the radius, so they crowd where the body is close and moving fast. That is not uniform in path length, and it is much closer to it than uniform time is.

The variable that makes an integral tractable and the variable that makes it accurate are not always the same one, and where they differ the subject has generally chosen tractability and then measured what it cost.

What is actually measured #

Nothing here is measured as a distance, and the averages are not measured at all.

What instruments record are angles and times. Optical astrometry gives directions against background stars, and the eccentricity and semi-major axis are parameters of a dynamical model fitted to a long run of them, with spacecraft radio tracking added in a modern solution and round-trip light times good to a few metres. Earth’s $e = 0.0167$ is a fitted element of that solution, not the outcome of averaging measured distances, and no instrument’s output is the mean distance of a planet from the Sun.

The semi-major axis in kilometres arrived long after the shape. Kepler’s third law fixes $a$ in astronomical units from the period alone; converting to metres needed a radar echo, first obtained from Venus in 1961, and the astronomical unit is now a defined constant — exactly 149 597 870 700 m, by the resolution of 2012 — so the conversion is a convention and the orbit carries the uncertainty.

The insolation is the one quantity here with a direct instrument. Radiometers above the atmosphere measure a total solar irradiance of 1361 W m⁻², and the ±3.3% annual swing is plainly present in the raw record. It is divided out before publication, by scaling every reading to 1 AU, precisely because it is the largest and best-understood variation in the signal; what remains is the 0.1% swing over the solar cycle, which is the science. The first-order consequence of eccentricity is therefore a calibration step in solar physics and the second-order one is a term in ice-age theory — one number, two fields, two orders.

Mean solar time is the oldest instance of the same move: a varying quantity replaced by a fictitious uniform one, the definition stated before any clock can be called right.

What the picture cannot show #

No orbit that fails to close. Every average here divides by a period, and at $e \ge 1$ there is none. The generator refuses such a request rather than drawing a limit: for an orbit that never comes back the time average of the radius is infinite, and a harmonic mean describes an interval somebody chose.

A fixed ellipse. The figures sweep $e$ as a parameter, not as something that changes. The elements drift under the pull of the other planets, so a planet’s mean distance is also an average over an epoch, and the 100,000-year cycle above is the slow part of that drift.

A flux at the top of the atmosphere, on a surface facing the Sun. The insolation panels compute $1/r^2$ and nothing else. Every step from there to a climate — the obliquity factor, the day’s length, the albedo of ice, an ocean’s heat capacity — is absent, and each is larger than the effect drawn.

Convergence rather than exactness. The two flat curves are flat because a sum of 2,048 terms comes out within $10^{-12}$ of unity, checked at sixty-one eccentricities before anything is drawn. That is a demonstration and not a proof, and it is deliberately the weaker claim: the closed forms are withheld from the drawing and used only to test it, so the figure tests an identity instead of illustrating one.

Where the ladder goes next #

Later rungs: the averages of quantities other than the radius, since $\langle 1/r^3\rangle_t$ is what every secular perturbation calculation needs. Averaging as a method rather than a curiosity — perturbation theory’s standard move is to average the equations of motion over the fast angle and keep the remainder, which turns a wobbling orbit into a drifting one and is where mean elements come from. The difference between mean and osculating elements, which is why two catalogues can disagree about a satellite’s semi-major axis and both be right. And the mean anomaly, the one average here given a name, a symbol and a place among the orbital elements before anybody asked what it averaged.