What is the separation of two masses when they have their average speed in orbit relative to their center of mass?
Present your answer as a ratio with the orbit's semimajor axis, and as a function of the orbit's eccentricity.
Mike, I have the utmost contempt for the kind of evasion that you answer exemplifies. If you don't know the answer, or how to find the answer, then either be silent or else be honest enough to say that you don't know. I am not a cheating student. I'm a 60-year-old retired physicist, and I already know the answer to the question I've asked.
The other answerer, Ronald 7, is merely confused. There is an exact and calculable answer, which I will provide hereafter in an answer of my own.
The distance between two masses when they have their average speed in their orbit around their mutual center of mass is found as follows:
circumference of an ellipse,
C = 4a ∫(0,π/2) √(1−e²sin²θ) dθ
period in an elliptical orbit
P = 2π√[a³/(GM)]
speed in orbit as a function of distance
v = √[GM(2/r − 1/a)]
average speed in orbit
vₐ = C/P
4a ∫(0,π/2) √(1−e²sin²θ) dθ / 2π√[a³/(GM)] = √[GM(2/r−1/a)]
r = 2 { [4/(π²a)] [ ∫(0,π/2) √(1−e²sin²θ) dθ ]² + 1/a }⁻¹
r = 2 { [4/(π²a)] [ ∫(0,π/2) √(1−e²sin²θ) dθ ]² + 1/a }⁻¹
The true anomaly in the orbit at which the object has its average speed in orbit, where r is the value calculated just above, is
θ = ± arccos{ [ (a/r)(1−e²) − 1 ] / e }
Note that
if e=0, then r=a & θ is indefinite
if e→0⁺, then r=a & θ→±π/2
if e>0, then r>a & θ is toward the apoapsis from ±π/2
if e→1⁻, then r→1.423199122a & θ→π