Average distance between an object and the body it's orbiting over time

Question

I became interested in finding the average distance of an orbiting object over time from its parent, expressed mathematically as $\frac{\int_{0}^{T}r\left(t\right)dt}{T}$, where $T$ is the orbital period and $r(t)$ is the orbital distance at a given point in time. After googling didn't work, I wrote some code to find this number at different orbital apoapsides, and plotted it with the orbital apoapsis (the periapsis was always 1), hoping the equation would reveal itself that way. Alas, it did not. Does anyone here know the answer?

Answer 1

The time-averaged distance over the orbital period $T$ is

$$\langle r \rangle \equiv \frac1T \int_0^T r(t)\,dt = a \left( 1+\frac12 e^2 \right),\tag1$$

where $a$ is the semi-major axis of the orbit and $e$ the eccentricity.

Evaluating this integral is non-trivial. First, there is no simple expression for $r(t)$ to use! But there is a simple expression for $r(\theta)$, where $\theta$ is the angle around the orbit, namely

$$r = \frac{a(1-e^2)}{1+e\cos\theta}.\tag2$$

(This is Kepler's First Law).

So the trick for evaluating the integral for the time-average is to use

$$dt=\frac{d\theta}{\dot\theta}\tag3$$

to convert it from an integral over $t$ to an integral over $\theta$:

$$\langle r \rangle = \frac1T \int_0^{2\pi} \frac{r(\theta)}{\dot\theta}\,d\theta.\tag4$$

Clearly for this to work, we need to be able to express $\dot\theta$ in terms of $\theta$.

We can do this using Kepler's Second Law, which says that the rate at which the orbital body sweeps out area is constant:

$$\frac{dA}{dt} = \text{const} = \frac{A}{T}.\tag5$$

Geometrically, the area of the infinitesimal triangle formed by the orbital segment $r\,d\theta$ at distance $r$ is

$$dA=\frac12 r^2 d\theta,\tag6$$

so

$$\dot\theta = \frac{2}{r^2}\frac{dA}{dt} = \frac{2}{r^2}\frac{A}{T}.\tag7$$

Furthermore, the geometrical area of the elliptical orbit is

$$A = \pi ab = \pi a^2(1-e^2)^{1/2}\tag8$$

where

$$b = a(1-e^2)^{1/2}\tag9$$

is the semi-minor axis.

Thus we have

$$\dot\theta = \frac{2}{r^2}\frac{\pi a^2(1-e^2)^{1/2}}{T}.\tag{10}$$

Since we know $r$ as a function of $\theta$, we now also know $\dot\theta$ as a function of $\theta$.

Putting this together, we have

$$\frac{r}{\dot\theta} = \frac{Tr^3}{2\pi a^2(1-e^2)^{1/2}}\tag{11}$$

and

$$\langle r \rangle = \frac{1}{2\pi a^2(1-e^2)^{1/2}}\int_0^{2\pi}r(\theta)^3\,d\theta = \frac{a(1-e^2)^{5/2}}{2\pi}\int_0^{2\pi}\frac{d\theta}{(1+e\cos\theta)^3}.\tag{12}$$

The integral can be done by contour integration (or, more simply, by a computer algebra system) and evaluates to

$$\int_0^{2\pi}\frac{d\theta}{(1+e\cos\theta)^3} = \frac{\pi(2+e^2)}{(1-e^2)^{5/2}},\tag{13}$$

giving

$$\langle r \rangle =a \left( 1+\frac12 e^2 \right).\tag{14}$$