Computing the rate of orbital decay We know that the estimated time until a binary system merges is given by: $$\dfrac{5c^5r^4}{256G^3(m_1m_2)(m_1+m_2)}$$ But what is the formula to determine rate of decay, in some sort of distance unit, per unit of time? Also, what are terms are $r$ and $m_i$ written in?
Here's one way to arrive at ProfRob's answer step by step:
$$t = \dfrac{5c^5r^4}{256G^3(m_1m_2)(m_1+m_2)}$$
If we lump all the constant terms together:
$$C = \dfrac{5c^5}{256G^3(m_1m_2)(m_1+m_2)}$$
then the problem suddenly looks a lot simpler:
$$t = r^4 C$$
$$\frac{dt}{dr} = 4r^3 C$$
$$\frac{dr}{dt} = \frac{1}{4r^3} \frac{1}{C}$$
$dt$ is how much longer the system will live if the starting radius is increased by $dr$. We know that if we watch it decay, the rate of decrease per unit time $dr/dt$ will be negative, so we'll add a minus sign.
Now put it back together:
$$\frac{dr}{dt} = -\frac{1}{4r^3} \frac{1}{C}$$
$$\frac{dr}{dt} = -\frac{1}{4r^3} \dfrac{256G^3(m_1m_2)(m_1+m_2)}{5c^5} $$
$$\frac{dr}{dt} = -\frac{1}{r^3} \dfrac{64G^3(m_1m_2)(m_1+m_2)}{5c^5} $$
$$\frac{dr}{dt} = -\dfrac{64G^3(m_1m_2)(m_1+m_2)}{5c^5r^3} $$
With gravitational constant $G = 6.674 \times 10^{-11} \text{m}^{3} \text{kg}^{-1} \text{s}^{-2}$ the speed of light $c=2.997 \times 10^{8} \text{m/s}$ each body with a solar mass $M_{\text{Sol}} = 1.9885 \times 10^{30} \text{kg}$ and an almost-touching separation of a million kilometers $r=1\times 10^{9} \text{m}$, then $C = 1.0102 \times 10^{-20}$ and the rate $dr/dt = -2.4748 \times 10^{-8} \text{s}^{-1}$ or -78 centimeters per year.