Prove this Relation of Chirp Mass In my astrophysics homework, I was asked to prove this equation of chirp mass
$$\frac{(m_1m_2)^{3/5}}{(m_1+m_2)^{1/5}} = \left[\frac{-5}{192\pi}\frac{\mathrm{d}T}{\mathrm{d}t}\right]^{3/5}\frac{c^3T}{2\pi G}$$
The hint reads

Use the radiation power of the binary star gravatational wave
$$L_\text{gw} = \frac{32}{5}\frac{G^4m_1^2m_2^2(m_1+m_2)}{c^5a^5}$$

Any idea on how to solve this problem?
 A: The general idea of this exercise is to approximate the orbit by a succession of (almost) Keplerian orbits.
We can start with expression with the time of a period $T$ as a function of the masses and their separation $a$. It is given by
$$ T = \left[\frac{4\pi^2}{GM}\right]^{1/2}a^{3/2} $$
where $M=m_1+m_2$ is the total mass.
The time derivative is then
$$ \dot T = \left[\frac{4\pi^2}{GM}\right]^{1/2}\frac{3}{2}a^{1/2}\dot a $$
This equation can be used to express $\dot a$ as a sole function of $T$ and $\dot T$.
$$\dot a = \frac{2}{3}T^{-1/3}\dot T \left(\frac{GM}{4\pi^2}\right)^3$$
The second ingredient is the orbital energy of the binary system, which we approximate here to be Newtonian.
The energy is given by
$$E_{\rm grav} = -\frac{G m_1 m_2}{a}$$
The change of orbital energy in time (and that is the very critical assumption) is then given energy radiated in gravitational wave which you were given in the exercise. You thus have to equate
$$ \dot E_{\rm grav} = -L_{\rm gr} $$
(The minus sign comes from the fact that any energy lost, will  decrease $E_{\rm grav}$ - remember it is negative - and thus decrease also the orbital separation $a$. The bodies approach each other over time).
We can then calculate $ \dot E_{\rm grav}$. It is given by
$$ \dot E_{\rm grav} =  \frac{G m_1 m_2}{a^2}\dot a$$
With these equations you can then replace all $a$'s and $\dot a$'s, and you should be able to derive the expression for the chirp mass given above.
