I am currently studying Classical Mechanics, fifth edition, by Kibble and Berkshire. Problem 2 of chapter 1 is as follows:
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, $\omega$.)
The only relevant information provided by the chapter is as follows:
If we isolate the two bodies from all other matter, and compare their mutually induced accelerations, then according to (1.1) and (1.3),
$$m_1 \mathbf{a}_1 = - m_2 \mathbf{a}_2 \tag{1.7}$$
Since the textbook chapter does not provide enough information to complete this problem, I referred to the Wikipedia article for angular velocity. Writing linear velocity as $v = \omega r$, we get
$$m_1 \mathbf{a}_1 = -m_2 \mathbf{a}_2$$
$$\therefore m_1 \left( r_1 \dfrac{d \omega_1}{dt} \right) = -m_2 \left( r_2 \dfrac{d \omega_2}{dt} \right)$$
$$\Rightarrow \dfrac{m_1}{m_2} = \dfrac{\left( -r_2 \dfrac{d \omega_2}{dt} \right)}{\left( r_1 \dfrac{d \omega_1}{dt} \right)}$$
The only way that I can see to proceed would be to assume that the angular velocities $\omega_1$ and $\omega_2$ are the same (I have no idea if this is implied by the physics of a "double star"):
$$\therefore \dfrac{m_1}{m_2} = - \dfrac{r_2}{r_1}$$
The answer is said to be $\dfrac{m_1}{m_2} = \dfrac{r_2}{r_1}$.
Why are the angular velocities equal? And what happened to the negative sign? I would greatly appreciate it if people would please take the time to clarify this.