# Binary Star System

1. Dec 27, 2005

### G01

A binary star system has a period of T=90 days. Each star is 2X the mass of the sun. They rotate like a dumbell around the center of mass. what is the distance d between them?
OK first, Mass of sun = $$1.99 X 10^{30} kg$$
Lets concentrate on one of the stars rotating around the center of mass. The force of gravity on this star is the centripetal force keeping the star circling the center. So:
$$F_{g} = F_{c}// \frac{\G(2M)(4M)}{r^2} = (2M)r\omega^2// \frac{G(4M)}{\omega^2} = r^3// r = \sqrt [3] {\frac{\G(4M)}{\omega^2}}$$
If $$\omega = 2 \pi f$$ converting T to f and converting to the right units we get
$$\omega = 8.08 X 10^{-7} rad/s$$
Now the answer in the back of my book gives the answer as $$9.33 X 10^{10} m$$ (Knight, Chapt 12 #61) I'm getting double that value somehow. I can't see what I did wrong. I'm beginning to think that the answer in the back is giving the radius of rotation when it should be giving the diameter. Chances are I'm doing something stupid. Hopefully someone here may see what I'm doing wrong. Thanks for you help. Also if someone can tell me how to make something go to a new line in latex my formulas will be much more legible

Last edited: Dec 27, 2005
2. Dec 27, 2005

### maverick6664

You seem to make one or two mistakes in the distance of two stars or the mass of each star in $$F_g$$

My calculation is

Let $$r$$ be the radius (half the distance).

$$F_g = \frac {G(2M)^2}{(2r)^2}, F_c = (2M)r\omega^2$$

so

$$r = \sqrt[3]{\frac {GM}{2 \omega^2} } = 4.66*10^{10} m$$ where $$G= 6.672*10^{-11}$$

$$distance = 2r = 9.33*10^{10} m$$

Last edited: Dec 27, 2005
3. Dec 27, 2005

### G01

ahh i see thank you.

4. Mar 13, 2011

### salu azmat

how did u do the second step? i.e.,r=(GM/2w square)*1/3