Calculating Orbital Velocities in a Binary Star System

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Discussion Overview

The discussion revolves around calculating orbital velocities in a binary star system with two stars of unequal masses. Participants explore concepts related to the orbital period, the center of mass, and the application of Kepler's laws in this context.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether the period of rotation of two unequal masses in a binary system is the same, suggesting that both bodies take the same amount of time to complete one revolution about their center of mass.
  • Another participant affirms this by explaining that the center of mass remains stationary in the absence of external forces, implying that the orbital period must be equal for both bodies.
  • A participant seeks clarification on how to determine the orbital period, considering the use of Kepler's third law and the implications of varying distances between the two bodies during their orbit.
  • There is a suggestion to refer to the concept of reduced mass to aid in calculations related to the binary system.
  • A participant proposes a formula for calculating the radial acceleration experienced by the bodies and attempts to derive the orbital velocities based on gravitational forces and the concept of reduced mass.

Areas of Agreement / Disagreement

Participants generally agree that the two bodies in a binary system have the same orbital period. However, there is no consensus on the specific method for calculating the orbital velocities, as different approaches and formulas are being discussed.

Contextual Notes

Participants express uncertainty about the assumptions involved in applying Kepler's laws, particularly regarding the nature of the orbits and the distances between the bodies. The discussion also highlights the complexity of calculating orbital velocities in systems with unequal masses.

Who May Find This Useful

This discussion may be useful for individuals interested in astrophysics, orbital mechanics, and the dynamics of binary star systems.

theneedtoknow
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I just have a quick question about a binary star system consisting of 2 starts of unequal masses. Is the period of rotation of the 2 masses about their centre of mass equal? Do they both take the same amount of time to complete 1 revolution about the centre of mass?
 
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Yup. The two-body system's center of mass cannot move because there are no external forces, so the line between the two bodies must always pass through the center of mass. Imagine a bar connecting the two bodies; the bar's rotating, expanding, and contracting, but not bending or moving. You can imagine that the two bodies will have the same orbital period.
 
Thanks! :) That makes sense.
So how do I determine this period of orbit of the 2 bodies around the centre of mass? What if, for example, I was considering a planet orbiting around a star. Would I use the general version of Kepler's 3rd law? [ P^2 = (4pi^2*a^3)/ (G(m1 + m2)) ] with a being the average distance between the 2 bodies (since, as you said, this distance will contract and expand as they orbit) (I'll assume circular orbits for simplicity)? Or would I calculate the period in some other way?
 
Have a look at http://en.wikipedia.org/wiki/Reduced_mass"
 
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So , the radial acceleration the 2 bodies would experience towards each other (and therefore toward the centre of mass) is F/mreduced = G*m1*m1/(distance between objects)^2 /mreduced ?
and the radial acceleration (assuming circular orbit) is equal to v^2 / (distance of body to centre of mass), so I can calculate the orbital velocities of the 2 bodies as:

velocity = square root [ (G*m1*m1/(distance between objects)^2 ) * (distance of body to centre of mass)/(mreduced) ]

is this correct?
 

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