The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an orbital speed of 36 km/s. The second star, Beta, has an orbital speed of 12 km/s. The orbital period is 137 d. a) What is the mass of the star alpha? b) What is the mass of the star beta?

Given:

Va = 36 km/s = 3.6*10⁴ m/s

Vb = 12 km/s = 1.2*10⁴ m/s

T = 137 d = 1.18*10⁷ s

For each star, circumference = velocity * time:

2π R = V T

R = V T / (2π)

So Ra = Va T / (2π), and Rb = Vb T / (2π).

Sum of the forces on Alpha:

Ma Va² / Ra = G Ma Mb / (Ra + Rb) ²

Va² / Ra = G Mb / (Ra + Rb) ²

Mb = Va² (Ra + Rb) ² / (G Ra)

Similarly, sum of the forces on Beta:

Mb Vb² / Rb = G Ma Mb / (Ra + Rb) ²

Vb² / Rb = G Ma / (Ra + Rb) ²

Ma = Vb² (Ra + Rb) ² / (G Rb)

First, calculate Ra and Rb:

Ra = (3.6*10⁴) (1.18*10⁷) / (2π)

Ra = 6.78*10¹⁰

Rb = (1.2*10⁴) (1.18*10⁷) / (2π)

Rb = 2.26*10¹⁰

Therefore, the mass of Alpha is:

Ma = (1.2*10⁴) ² (6.78*10¹⁰ + 2.26*10¹⁰) ² / (6.67*10⁻¹¹ * 2.26*10¹⁰)

Ma = 7.81*10²⁹ kg

And the mass of Beta is:

Mb = (3.6*10⁴) ² (6.78*10¹⁰ + 2.26*10¹⁰) ² / (6.67*10⁻¹¹ * 6.78*10¹⁰)

Mb = 2.34*10³⁰ kg