Gravitational wave radiation power from dimensional analysis Let us try to find a formula for the power emitted through gravitational waves (GW) from a binary system in quasi circular orbit. The relevant quantities are the Newton's constant $G_N$, speed of light $c$, a mass scale $M$, orbital frequency $\omega$. So I write $P=G_N^a c^b M^c \omega^d$. Demanding both sides have the same dimension gives
\begin{align}
P_{GW}=\frac{c^5}{G_N}\left(\frac{G_NM\omega}{c^3}\right)^d
\end{align}
with arbitrary exponent $d$. It turns out that $d=10/3$ from the quadrupole approximation. My question is the following: Is there a way to get $d=10/3$ from some sort of argument without having to get into the details of the computation? One possible answer is that we know from post-Newtonian theory that radiation effects start at $1/c^5$, and this implies $d=10/3$. But I thought  maybe there is a better argument.
 A: The short answer is no.  The problem is that a simple dimensional analysis argument tells you that you need the power to be energy per unit time, given the mass $M$, frequency $\omega$, and $G$ and $c$. Well, the energy part is "easy" in the sense that $Mc^2$ has the right dimensions. However, the problem is that there are two quantities with dimensions of time
\begin{equation}
\frac{G M}{c^3}, \ \ \frac{1}{\omega}
\end{equation}
So dimensional analysis is only powerful enough to tell you that the answer must look like
\begin{equation}
P_{\rm GW} = \left(M c^2\right) \times \omega^d \times \left(\frac{c^3}{GM}\right)^{1-d} = \frac{c^5}{G} \left(\frac{GM\omega}{c^3}\right)^d
\end{equation}
for some $d$, as you said. Another way to express the same point is that dimensional analysis only fixes the dependence on dimensionful quantities, but there is a dimensionless ratio $GM \omega/c^3$ which dimensional analysis cannot help you with.
In fact, in general the situation is even worse than this. If you allowed for the two objects in the binary to have different masses, there would even be another dimensionless quantity, which we can take to be the mass ratio $q=m_1/m_2$. The dependence on $q$ is also not fixed by dimensional analysis; doing a more careful calculation ends up telling you that the power depends on the chirp mass. And if you allow the black holes to have spins, you need even more associated dimensionless ratios to describe the system.
So, the bottom line is that you do need information about the leading order scaling of the post-Newtonian corrections to fix $d$ (as well as mass ratio dependence, and spin dependence).
The longer answer, though, is that it isn't too hard to get the leading order PN scaling to fix the frequency dependence, if you are willing to accept that gravity is a spin-2 field and so must couple to the quadrupole moment $Q$. (As @ProfRob pointed out in the comments, there are other ways you can justify the coupling to $Q$. For example: $Q$ is the lowest order multipole that gravitational waves can couple to, given that the monopole and dipole moments correspond to the mass and linear momentum of the system, which are conserved and so cannot contribute to radiation). In order to estimate the frequency dependence, because of Kepler's laws, we also need to know the dependence on the size of the system, $R$. Given that $h \sim \ddot{Q}$ where $Q \sim R^2$, we have $h \sim \omega^2 R^2$. Using Kepler's laws, we use that $R \sim \omega^{-2/3}$, so $h \sim \omega^{2/3}$. Therefore, $E \sim \int dt \dot{h}^2 \sim \omega^2 h^2/\omega \sim \omega^{7/3}$ and $P \sim \dot{E} \sim \omega E \sim \omega^{10/3}$.
A: You can't without further physical arguments, as described by Andrew.
An approach is to assume that the power radiated depends on $G$, $c$ and some time derivative of a mass multipole. It is straightforward to show that the first time-derivative of the monopole and dipole moments is zero because of conservation of mass, linear momentum and angular momentum. One is left then with using the quadrupole moment as the lowest viable order.
Since power is a scalar and must be $\geq 0$, then the time derivative of the quadrupole moment must be squared and thus we can say
$$ P \sim  G^a c^b \left(\frac{\partial^c Q}{\partial t^c}\right)^2\ .$$
Letting $Q$ have dimension of $ML^2$, then dimensional analysis yields $a=1$, $b=-5$ and $c=3$.
In a binary, we have $Q \sim Ma^2 \sin 2\omega t$ (twice the frequency of the binary because the quadrupole moment flips twice per orbital period). Differentiate this three times, square it:
$$ \dddot{Q}^2 \sim M^2 a^4 \omega^3$$
and then note that $a^4 \propto (GM)^{4/3}\omega^{-8/3}$ from Kepler's third law and you get
$$ P \sim \frac{G}{c^5}G^{4/3} M^{10/3}\omega^{10/3} = \frac{c^5}{G}\left(\frac{GM\omega}{c^3}\right)^{10/3}\ .$$
