Where is the time-average position of a planet on an elliptical orbit? Assume that a Planet has an elliptical orbit around the Sun and follows the Kepler's laws; no relativistic shenanigans. Let the Sun be in the focal point $f_1$ of the elliptical orbit. Given that the Planet moves much faster when it's closer to $f_1$, the time average of the planet position cannot be $f_1$ or the geometric center of the ellipse. What is the Planet's time-average position?
By symmetry it's obviously on the line connecting the foci, and it would be aesthetically pleasing if the answer was the other focus $f_2$, but I have no idea what the answer actually is.
 A: I feel like Newton would have come up with a clever geometric proof of this for the Principia, but I can't think of one, so I will use calculus.
Put the major axis at $(-M,0)$ to $(M,0)$, the minor axis at $(0,-m)$ to $(0,m)$, and the Sun at $(f_1,0)$. By Kepler's law you can do a triangle-area-weighted integral:
$$x_\text{avg} = \frac{\int_{-M}^{M} x A(x)\,dx}{\int_{-M}^{M} A(x)\,dx}$$
where $A(x)\,dx$ is the area of the triangle with vertices $(x,y(x))$, $(x{+}dx,y(x{+}dx))$, and $(f_1,0)$, and $y(x)=m\sqrt{1-x^2/M^2}$.
The bottom integral is half the ellipse area, or $\frac12 π M m$.
The triangle area is half the wedge product of two of the sides:
$$A(x)\,dx = \frac12 (f_1{-}x,-y(x)) \wedge (dx,y'(x)dx) = \frac12 (f_1 y'(x) - x y'(x) + y(x))\,dx$$
so
$$\begin{align} x_\text{avg} &= \frac{1}{πMm} \int_{-M}^{M} (f_1 x y'(x) - x^2 y'(x) + x y(x)) \, dx \\
&= \frac{1}{πMm} \int_{-M}^{M} f_1 x y'(x) \, dx \qquad \text{(other terms are odd)} \\
&= -\frac{1}{πMm} \frac{m}{M^2} f_1 \int_{-M}^{M} \frac{x^2}{\sqrt{1-(x/M)^2}} dx \\
&= -f_1/2
 \end{align}$$
i.e., the average position is halfway between the center and the other focus.
