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