Indeed. The *Molniya* orbit is a highly eccentric orbit around the Earth which is used for high-latitude communications satellites. It is "stationary" (at least in terms of its angles from the ground) for most of its orbit, and hence can be used with a fixed dish on the ground very much like a dish for a geostationary orbit. By putting several satellites in the same orbit, but in different phases, continuous communications can be assured. (To generalize, a geostationary orbit might be considered a Molniya orbit for latitude zero, with the other orbits for higher latitudes having increasing eccentricities.) Needless to say, the whole point of the Molniya orbit is that the *average* position (in terms of time) is near the focal line of the satellite dish. https://en.wikipedia.org/wiki/Molniya_orbit At 03:49 PM 7/14/2016, Brent Meeker wrote:

You need to think about whether you're averaging with respect to time or position. In an elliptical orbit the planet travels faster when it is nearer the center-of-mass so it spends more time near the farther focus.

Brent

On 7/14/2016 3:31 PM, John Aspinall wrote:

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On 07/14/2016 04:35 PM, James Propp wrote:

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Ignoring the effects of bodies other than the Earth and the Sun, and treating the Earth as a point mass, what is the average position of the Earth over the course of a year as it travels an elliptical orbit with the Sun at one focus?

If you ask me "Are you interested in the average position of Earth relative to the the Sun, or relative to the position of the center of mass of the two-body system?", my answer is "I'm interested in both". The equivalence between the two frames of reference is fairly simple, and expressed by the reduced mass: https://en.wikipedia.org/wiki/Reduced_mass

As to the average position, I would expect it to be the center of mass. The equation of motion is decomposable into two one-dimensional equations in orthogonal coordinates, reducing the question to "what is the average of a sinusoid".

In this view ("scaffolding", to borrow Henry's term from another recent email) the ellipse is not the fundamental solution. The ellipse is assembled from two more basic bits: two one-dimensional sinusoids. Various ellipses are possible by varying amplitudes and relative phase, but neither amplitude nor phase affects the average value.

- John

Apologies to people whose email doesn't recognize nonascii characters.  With the Sun at the origin, the equation for the orbital ellipse, of eccentricity e, is r = p / (1 + e cosθ), from which it follows that p = a(1 - e^2), a = semi-major axis.  Taking the orbiting body to have unit mass, its angular momentum, a conserved constant, is L = r^2 (dθ/dt), so that (dθ/dt) = L / r^2. The energy, also a conserved constant, is E = Kinetic + Potential = (1/2)((dr/dt)^2 + r^2 (dθ/dt)^2) - GM / r. At perihelion r = a(1 - e); at aphelion r = a(1 + e); and at both extrema (dr/dt) = 0, so that E = L^2 / (2 r^2) - GM/ r. Setting the energies at perihelion and aphelion equal yields L^2 = GM a (1 - e^2). We now have equations that allow p and L to be expressed in terms of a and e.  It follows, though not needed for the main proof, that E = - GM / (2 a). The orbital period T is T = int( dt, t = 0...T) = int( dθ / (dθ/dt), θ = -π...π)  = int ( dθ r^2 / L, θ = -π...π) = (p^2 / L) I2, where I2 = int ( dθ / (1 + e cosθ)^2, θ = -π...π). Using the usual calculus bag of tricks (I used z = tan(θ/2) followed by partial fractions), I2 = 2π (1 - e^2)^(-3/2), and (p^2 / L)  = (GM)^(-1/2) a^(3/2) (1 - e^2)^(3/2), so that T = (2π) (GM)^(-1/2) a^(3/2), which is Kepler's third law.  Now we get to the main question.  By symmetry, the mean position of the orbiting body lies on the major axis.  So we need to calculate the x0, the mean x position. x0 = (1/T) int( dt x(t), t = 0...T)   = (1/T) int( dθ x(θ) / (dθ/dt), θ = -π...π)  = (1/T) int( dθ (r cosθ) (r^2 / L), θ = -π...π)  = (p^3 / (T L)) I3, where I3 = int( dθ cosθ /  (1 + e cosθ)^3, θ = -π...π). I3 = (-1/2) (d/de) I2 = - 3π e (1 - e^2)^(-5/2), and (p^3 / (T L)) = (a / (2π)) (1 - e^2)^(5/2), so finally we have x0 = (-3/2) e a. Thus x0 is half-way between the ellipse center at x = - e a and the other focus at x = - 2 e a.   --  Gene From: Tom Rokicki <rokicki@gmail.com> To: Eugene Salamin <gene_salamin@yahoo.com>; math-fun <math-fun@mailman.xmission.com> Sent: Friday, July 15, 2016 1:16 PM Subject: Re: [math-fun] Average position of Earth Can you cite a reference for this, or explain a derivation?  This sounds fascinating.Perhaps the math is straightforward, but I'm leery of trying to go from Kepler'sequal-area-scan result to this one . . . Thanks!   -tom On Fri, Jul 15, 2016 at 1:11 PM, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote: The average position lies on the major axis of the orbital ellipse, on the aphelion side, and half-way between the center of the ellipse and the other focus.   --  Gene       From: James Propp <jamespropp@gmail.com>  To: math-fun <math-fun@mailman.xmission.com>  Sent: Thursday, July 14, 2016 1:35 PM  Subject: [math-fun] Average position of Earth Ignoring the effects of bodies other than the Earth and the Sun, and treating the Earth as a point mass, what is the average position of the Earth over the course of a year as it travels an elliptical orbit with the Sun at one focus? If you ask me "Are you interested in the average position of Earth relative to the the Sun, or relative to the position of the center of mass of the two-body system?", my answer is "I'm interested in both". I'm hoping that there's an embarrassingly easy way to see what the answer is using principles from physics that I must've been taught but haven't fully absorbed. I'm also interested in the limiting case where the ratio of the two masses in a two-body system goes to 0. Even if the answer to my original questions is "It's messy in either coordinate system", this limiting case (in which the two original questions coincide) might behave nicely. Jim Propp