M_p^3 = \frac{V_s^3}{2\pi G} PM_s^2Solve Binary Star Mass Homework Equation

AI Thread Summary
In a binary star system, the mass of one star, denoted as M_p, can be expressed using the equation M_p^3 = (V_s^3 / (2πG)) P M_s^2. This relationship is derived from gravitational dynamics and Kepler's laws, where the gravitational force and orbital mechanics are considered. The discussion highlights that the mass of the second star (M_p) is often negligible compared to the mass of the first star (M_s), simplifying calculations. The user is seeking clarification on this equation as part of their introductory astrophysics studies. Understanding this relationship is essential for analyzing binary star systems in astrophysics.
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Homework Statement



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How do I show that for a binary star system, if one star has mass ##M_s##, speed ##V_s##, period ##P##, the mass of the other star is given by: ##M_P^3 \approx \frac{V_s^3}{2\pi G} PM_s^2##?

Homework Equations


The Attempt at a Solution



\frac{GM_pM_s}{(a_p+a_s)^2} = \frac{M_s v_s^2}{a_s}
Substituting in ##PV_s=2\pi a_s##:
M_p = \frac{2\pi(a_p+a_s)^2V_s}{PG}
Using kepler's second law: ## P^2 = \frac{(a_p+a_s)^3(2\pi)^2}{G(M_p+M_s)} ##:
M_p^3 = \frac{V_s^3}{2\pi G} P (M_p+M_s)^2
 
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Seems correct to me. The mass of the planet is more often than not negligible compared to the mass of a star, so M_{p} << M_{S}
 
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Fightfish said:
Seems correct to me. The mass of the planet is more often than not negligible compared to the mass of a star, so M_{p} << M_{S}

Thanks alot! I'm doing an introductory course to my first ever astrophysics module, so I'm not quite familiar with these things.
 
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