3. Alpha Centauri A and B are Sun-like stars, and together they form the binary star Alpha Centauri AB. Alpha Centauri A has 1.1 times the mass of the Sun while Alpha Centauri B has 0.907 times the Sun’s mass. (Sun’s mass is 1.989 x 1030 kg.) The pair orbit about a common center with an orbital period of 79.91 years. Their elliptical orbit is eccentric, so that the distance between A and B varies, but if we assume it is circular, we can calculate the average velocities. Find the average velocities of the two stars.

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°

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.