- #1

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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

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- Thread starter shirin
- Start date

- #1

- 46

- 0

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

- #2

- 984

- 174

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*d

m2*d

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

d

Plug the formulas for x1 and x2 into these equations of motion, and one gets

m

where m

Alternatively,

d

d

with

d

- #3

- 46

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thank you very much!

- #4

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So, why are the eccentricities the same?

- #5

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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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- #8

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Anyway, a circumbinary orbit should have it's Kepler elements defined w.r.t the binary system's barycenter.

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