The eccentricity of binary stars' orbits

  • Thread starter shirin
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  • #1
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Hi
As two stars orbit their mutual center of mass in elliptical orbits, why are their eccentricity the same?
And why is it the same as the one of reduced mass around center of mass?

Thanks
 

Answers and Replies

  • #2
984
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Their positions are weighted mirror images of each other:

x1 = - m2/(m1+m2) * x12
x2 = m1/(m1+m2) * x12
where
x12 = x2 - x1
and their center of mass is fixed:
m1*x1 + m2*x2 = 0

Let's now consider mutual interactions:

m1*d2(x1)/dt2 = - f
m2*d2(x2)/dt2 = + f

The forces are reversed in sign from Newton's Third Law, with the consequence that

d2(m1*x1+m2*x2)/dt2 = 0

Plug the formulas for x1 and x2 into these equations of motion, and one gets
mreduced*d2(x12)/dt2 = + f

where mreduced = m1*m2/(m1+m2)

Alternatively,

d2(x1)/dt2 = -(1/m1)*f
d2(x2)/dt2 = (1/m2)*f

with
d2(x12)/dt2 = (1/m1 + 1/m2)*f
 
  • #3
46
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thank you very much!
 
  • #4
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So, why are the eccentricities the same?
 
  • #5
46
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As I understood, the distance of M1 to CM, x1, is linearly proportional to the distance of M2 to CM, and also it is proportional to the distance between these two, x12.
Since x1 is an ellipse with eccentricity e1, its eccentricity remains constant, but its semimajor gets multiplied, when it is multiplied by a constant number. So we have a bigger/smaller ellipse with the same eccentricity.
 
  • #6
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Ok, I couldn't figure it out. Thanks a lot!
 
  • #7
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I am wondering how the eccentricity of a circumbinary (binary star system) planet is defined? To my knowledge Kepler elements, including e, are non-inertial elements. So, how is the planets eccentricity measured?
 
  • #8
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Well, the planet's orbit will have to be stable enough to be an ellipse, which assumes it's quite far away from the apoapsis distance of the stars (AFAIRemember >12*a).
Anyway, a circumbinary orbit should have it's Kepler elements defined w.r.t the binary system's barycenter.
 

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