On what timescale does gravitational wave emission circularise an orbit? Gravitational waves remove both energy and angular momentum from a binary orbit. Both rates are enhanced in non-circular (eccentric) orbits and I presume that (like tidal friction) the net effect will be to circularise the orbits over time.
But is there a handy formula for the circularisation timescale and is it always (or never?) shorter than the timescale for merger?
 A: I have not seen any nice formula in the literature yet. There are some with baroque rational powers, though. At least for point masses circularisation may not happen fast enough to matter, surprisingly enough.
Peters, P. C., & Mathews, J. (1963). Gravitational radiation from point masses in a Keplerian orbit. Physical Review, 131(1), 435. gives a formula for the average power loss over one period of a Keplerian orbit as $$\langle P \rangle = \frac{32}{5}\frac{G^4}{c^5}\frac{m_1^2m_2^2(m_1+m_2)}{a^5(1-e^2)^{7/2}}\left(1+\frac{73}{24}e^2+\frac{37}{96}e^4\right ),$$ noting that it is equivalent to the standard circular orbit formula multiplied by an enhancement factor of $$f(e)=\frac{1+(73/24)e^2+(37/96)e^4}{(1-e^2)^{7/2}}.$$ where $f(0.6)\sim 10, f(0.8)\sim 100, f(0.9)\sim 1000$. So we should expect circularisation to happen on a timescale less than $1/f(e)$ of the normal energy loss timescale.
Peters then went on analysing the decay rate of eccentricity over time in Peters, P. C. (1964). Gravitational radiation and the motion of two point masses. Physical Review, 136(4B), B1224 as $$\left\langle \frac{de}{dt}\right\rangle = -\frac{304}{15}\frac{G^2}{c^5}\frac{m_1m_2(m_1+m_2)}{a^4(1-c^2)^{5/2}}e\left(1+\frac{121}{304}e^2\right)$$ which can be combined with his expression for $\langle da/dt\rangle$ to get an equation for $\langle da/de\rangle$ and $$a(e)=\frac{c_0 e^{12/19}}{1-e^2}\left (1+\frac{121}{304}e^2\right )^{870/2299}$$ where $c_0$ is determined by the initial condition. 

An eccentric system loses a lot of angular momentum until the eccentricity is $<0.5$ and then things level out - but to get rid of the last few percent eccentricity the semi-major axis has to shrink a lot more. 
Peters then calculates the lifetime for a system starting at $a_0,e_0$ as $$T(a_0,e_0)=\frac{12}{19}\frac{c_0^4}{\beta}\int_0^{e_0}\frac{e^{29/19}\left(1+\frac{121}{304}e^2\right )^{1181/2299}}{(1-e^2)^{3/2}} de$$ where $\beta=(64/5)(G^4/c^5)m_1m_2(m_1+m_2)$. 

The result, compared to a circular system, is that initially eccentric systems have lifetimes that actually are shorter roughly like the $f(e)$ factor: they radiate away so much energy by being in eccentric orbits that the reduction in eccentricity doesn't have the time to "take" before final infall. 
However, this is all for point masses. Tidal effects will also allow spin-up and dissipation in actual objects, like in this paper on white dwarfs near black holes.  This numerical paper found that black hole pairs in-spiralling over just 9 orbits with $e\leq 0.8$ circularised by merger time, so the process looks very fast in the more extreme cases. 
