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?

Explanation:

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

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