How do you find the angular velocity required to keep a constant distance $r_0$ between two unequal spherical masses?

Question

I am having trouble coming up with this solution. Here is where I am at. I defined two variables, $r_1$ and $r_2$, where the former is the distance between $m_1$ and the center of mass, and the latter is the distance between $m_2$ and the center of mass.

C.o.m = $m_1*r_1 + m_2*r_2$/$m_1 + m_2$.

Then I said, $F_c$ = $(m_1*(r_0^2*w_0^2))$/c.o.m = $F_g$ = $Gm_1m_2/r_0^2$. Where I replaced the $v$ in centripetal force using $v=rw$. Is there a simplification I am missing here? It gets sort of ugly.

Thank you.

Answer 1

If you want to find the angular velocity, let me show you how,

Let two uniform spheres [cosidering as point masses] of masses $m_{1}$ & $m_{2}$. And their initial separation is $d_{0}$. Recaling the motion of a Binary star system,

The center of mass of the system relative to $m_{1}$ is at $$r_{1}=\frac{m_{2}d_{0}}{m_{1}+ m_{2}}$$ And, relative to $m_{2}$ is at, $r_{2}=d_{0}\left(1-\frac{m_{2}}{m_{1}+m_{2}}\right)=\frac{m_{1}d_{0}}{m_{1}+m_{2}}$.

And here, $d_{0}=r_{1}+r_{2}$.

The necessary centripetal force [Inertial frame of ref.] is supplied by the Gravitational attractive force, $$F_{c}=F_{g}$$ For $m_{1}$ $$m_{1}\omega^{2}r_{1}=\frac{Gm_{2}m_{1}}{d_{0}^{2}}$$

Substituting $r_{1}$,

$$\omega = \sqrt{\frac{G(m_{1}+m_{2})}{d_{0}^{3}}}$$

You will get the same result for $r_{2}$