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16 March, 18:06

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?

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  1. 16 March, 18:57
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    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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