# What is the separation at which an orbiting object has its average speed, relative to the primary object?

Find a general expression for the separation between two gravitationally bound masses in a elliptical orbits around a common center of mass have their average speed relative to each other. (Hint: it isn't the length of the semimajor axis.)

### 2 Answers

- az_lenderLv 71 year agoFavourite answer
I'll consider a case where a small body is orbiting a large one, so I won't deal with the motion of the large body. My "ANSWER" is noted below as "ANSWER".

The orbital period is 2*pi*a^(3/2)*(GM)^(-1/2), where "a" is the semi-major axis of the orbit. The speed at any particular separation "r" is sqrt[GM(2/r - 1/a)]. These facts are given in various websites, including Wikipedia.

A decent approximation for the arc-length of an ellipse is

L = pi*{3(a + b) - sqrt[(3a + b)(a + 3b)]}. This approximation is due to Ramanujan. So the average speed of the orbiting body would be

{3(a+b) - sqrt[(3a+b) (a+3b)]} * (GM)^(1/2) / [2a^(3/2)].

So we're looking for "r" such that

sqrt(2/r - 1/a) = {3(a+b) - sqrt[(3a+b)(a+3b)] / [2a^(3/2)].

From here on, one could solve for "r" in terms of a and b, by methods of high-school algebra! Whether Ramanujan's approximation is robust enough to support such an answer, I cannot say.

Let's see if I can do it without frying my brain:

(2/r - 1/a)*(4a^3)

= 9(a+b)^2 + (3a^2+10ab+3b^2) - 6(a+b)*sqrt(3a^2+10ab+3b^2) =>

8a^3/r - 4a^2 = 9a^2 + 18ab + 9b^2 + 3a^2 + 10ab + 3b^2 - 6(a+b)*sqrt(3a^2 + 10ab + 3b^2) =>

8a^3/r = 16a^2+28ab+12b^2 - 6(a+b)*sqrt(3a^2+10ab+3b^2) =>

4a^3/r = 8a^2+14ab+6b^2 - 3(a+b)*sqrt(3a^2+10ab+3b^2)

=>

ANSWER: r = (4a^3) /

[8a^2+14ab+6b^2 - 3(a+b)*sqrt(3a^2+10ab+3b^2)].

As a check on reasonableness, let's see what happens when we have a circular orbit (a=b). You'd get

r = 4a^3/(28a^2 - 6a*sqrt(16a^2))

= 4a^3/(28a^2 - 24a^2) = a,

which of course is the right answer for a circular orbit.

Well, at least THAT case tends to support my analysis !!!

- MorningfoxLv 71 year ago
The bunch of words is unclear. Do you want to know where in the orbit, the object's instantaneous speed equals its average orbital speed?