Confused about binary star systems?

[21joanna12]

I was thinking about the motion of two stars in a binary star system, but there is something I cannot quite figure out. Suppose you have a binary star system with two stars masses m1 and m2 with m2>m1 so that m2 is closer to the centre of mass of the system. Then when the two stars are as far away from each other as possible, their centre of mass satisfies $\frac{r_1m_1 + r_2m_2}{m_1+m_2}$, so at this position, the velocity of star 1 would be found by $\frac{m_1v_1^2}{r_1}=\frac{Gm_1m_2}{(r_1+r_2)^2}$
But then as the two stars move closer together, both their centre of mass, and thus their distances from the centre of mass r1 and r2, and the gravitational attraction between them, change. So I can't quite figure out what their eventual motion will be...

Thank you in advance for any help :)

 

[Bandersnatch]

First of all, the expression $F=mV^2/r$ is in conflict with this sentence

as the two stars move closer together

can you see why?

As to how to calculate velocities, best to use conservation of energy. You may also want to check the following page:
http://en.wikipedia.org/wiki/Two-body_problem

 

[Ken G]

Strangely, that Wiki article does not define the reduced mass ("mu") that it uses, you have to follow the link to the "center of mass frame" to get that defined. If you put those two articles together, you will see how to do 2-body problems with a central force.

 

[21joanna12]

Oh! I see that I made a wrong assumption that the centre of mass moves, but it cannot because there is not net force ( so we put the centre of mass in the rest frame) and hence the two bodies must always be opposite each other in their orbits!

 

[Bandersnatch]

Yes, but remember that formulation you used only works for the special case of circular orbits.

 

[TEFLing]

Oh! I see that I made a wrong assumption that the centre of mass moves, but it cannot because there is not net force ( so we put the centre of mass in the rest frame) and hence the two bodies must always be opposite each other in their orbits!

isn't this true generally? Even for elliptical, parabolic and hyperbolic orbits?

 

[Ken G]

Yes that much is always true. It was the expression that connects the force to v² that is only true for circular orbits.