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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?

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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

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

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  • $\begingroup$ Is it possible to use a similar method to calculate the (time)average orbital speed? I think there's not a closed formula because the ellipse perimeter doesn't have one... $\endgroup$
    – Luca M
    Commented Aug 18, 2022 at 10:19
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    $\begingroup$ @LucaM Energy conservation gives $v(r)$. From this and $r(\theta)$ you get $v(\theta)$. Following the above approach of converting the time integral into an angular integral, you get $\langle v \rangle$ as an integral over $\theta$. I'm not sure whether it can be evaluated analytically, but I suspect that it probably can. $\endgroup$
    – Ghoster
    Commented Aug 21, 2022 at 7:15
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    $\begingroup$ @LucaM I realized from your comment that there is a much simpler approach: the average speed is the circumference of the elliptical orbit divided by the orbital period. This involves a complete elliptic integral of the second kind -- which to a physicist IS a closed formula; it's just another "special function" -- and there isn't a simpler exact expression. However, it could be expanded in a power series in the eccentricity if you want an arbitrarily good approximation that doesn't involve special functions. $\endgroup$
    – Ghoster
    Commented Aug 22, 2022 at 4:39

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