Answer:
8722.8 m/s, average speed of B relative to A (approximate)
Explanation:
The total mass of the two stars,
M = M₁ + M₂ = 2.007 M☉ = 3.99101985e+30 kg
G = 6.6743e-11 m³ kg⁻¹ sec⁻²
GM = 2.663726378e+20 m³ sec⁻²
The orbital period,
P = 79.91 years = 2.521767816e+9 sec
The semimajor axis of the orbit,
a = ∛[P²GM/(4π²)]
a = 3.50090e+12 meters
Assuming that the orbit is circular, with radius equal to the calculated semimajor axis, the average speed in orbit,
V = C/P
where
C = 2πa = 2.19968e+13 meters
V ≈ 8722.8 m/s, average speed of B relative to A (approximate)
However, we can look up the eccentricity of the orbit of α Cen A relative to α Cen B, and then we can calculate the average speed in orbit more accurately.
e = 0.5179
C = 4a ∫(0,π/2) √(1−e²sin²θ) dθ
C = 2.04379e+13 meters
V = 8104.6 m/s, average speed of B relative to A (more accurate)
As you can see, the more nearly correct value is considerably different than the approximation obtained when the orbit is assumed circular.
Incidentally, we can find that
The periapsis distance is 1.68779e+12 meters.
The apoapsis distance is 5.31402e+12 meters.
The semilatus rectum is 2.56189e+12 meters.
The semiminor axis is 2.99482e+12 meters.
The focal parameter is 4.94669.
We can also calculate the separation of A and B when the orbital speed has this average value:
r = 2a { 1 + [ (2/π) ∫(0,π/2) √(1−e²sin²θ) dθ ]² }⁻¹
r = 3.75778349e+12 meters
Moreover, that separation occurs when the stars reach the part of their orbit having true anomalies of
θ = ± arccos{ [ 1 − (a/r) (1+e²) ] / e }
θ = 110.518° and 249.482°
