# Double Star Mass

1. Aug 20, 2009

### mkphysics

1. The two components of a double star are observed to move in circles of radii r_1 and r_2. What is the ratio of their masses? (Hint: Write down their accelerations in terms of the angular velocity of rotation, w)

The answer is m1/m2 = r2/r1.

How does one cancel the velocity from the problem to get this result?

2. Relevant equations

For an isolated two body system

m1.a1=-m2.a2
where m are the body masses and a are the accelerations

w=v/r
angular velocity equals velocity divided by radius

a=dw/dt
angular acceleration

3. The attempt at a solution
m1.dw1/dt = m2.dw2/dt

therefore m1.w1=-m2.w2

m1/m2 = -w2/w1

m1/m2 = -(v2/r2)/(v1/r1)

m1/m2 = -v2r1/v1r2

Last edited: Aug 21, 2009
2. Aug 20, 2009

### kuruman

Do the stars rotate about any which point or about a specific point? What are this point's properties?

3. Aug 21, 2009

### mkphysics

Unfortunately all information is contained in the question. This question is taken directly from chapter 1 problem 2 of Classical Mechanics by Kibble and Berkshire. I'm not sure if this is a good book yet or not. From your reply it seems the question does not contain enough information to find the solution given. Can one infer the conditions of the system from the result given (m1/m2 = r2/r1)?

4. Aug 21, 2009

### kuruman

I am not familiar with the textbook, but it does not matter. Does "center of mass" ring a bell?

5. Apr 4, 2010

### iva

I know this is an old post but for anyone looking up this problem, here's how I got to that answer:

Using Newtons law for 2 isolated bodies

m1a1 = -m2a2

The bodies are rotating ( here i guessed at the same velocity).
The motion is rotational so you can use tangential acceleration

aT = w2r and substitute this into above to get

m1 w2r1 = -m2w2r2

cancelling out w2 on both sides, move m's to left and r's to right to get

m1/m2 = -r2/r1