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?

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    $\begingroup$ The SI value of $G$ is $$6.67430(15)×10^{-11}\, \mathrm{m^3kg^{-1}s^{-2}}$$ so if you're using that value, then you need to measure $r$ in metres and the masses in kg. And the result will be in seconds. en.wikipedia.org/wiki/Gravitational_constant gives a $G$ value using parsecs & solar masses. $\endgroup$
    – PM 2Ring
    Mar 17, 2021 at 21:41
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    $\begingroup$ There are some rate equations here. en.wikipedia.org/wiki/… $\endgroup$
    – PM 2Ring
    Mar 17, 2021 at 21:50
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    $\begingroup$ I prefer the fff system, in which $G$ is $$4.8966753 \times 10^{-4} fu^3fi^{-1}fo^{-2}$$ where $fu$ is length in furlongs, $fi$ is mass in firkins of water, and $fo$ is fortnights. The point is, as long as the result is in time units, you can use whatever consistent system you like! $\endgroup$
    – Connor Garcia
    Mar 17, 2021 at 22:22

3 Answers 3


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.

  • $\begingroup$ Is there such a graph relating an initial $m_1$, $m_2$, $r$, with time, and rate? $\endgroup$
    – WarpPrime
    Mar 18, 2021 at 12:57
  • $\begingroup$ So now I have a recursive function $t(1)=-\frac{64G^{3}(M_{1}M_{2})(M_{1}+M_{2})}{5c^{5}r^{3}}\cdot\left(86400\cdot365\right)$, $t(2)=-\frac{64G^{3}(M_{1}M_{2})(M_{1}+M_{2})}{5c^{5}\left(r-t(1)\right)^{3}}\cdot\left(86400\cdot365\right)$. Do you have any idea on how to convert this into closed form? $\endgroup$
    – WarpPrime
    Mar 18, 2021 at 13:13
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    $\begingroup$ @fasterthanlight If you want to graph distance between binary stars as a function of time, I've posted an answer with an equation that should allow you to do so, given $m_1$, $m_2$, and an initial distance $r$ (or $a$, equivalently). The x-axis should be fortnights, and the y-axis should be furlongs! $\endgroup$
    – Connor Garcia
    Mar 18, 2021 at 18:15
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    $\begingroup$ Well the way to arrive at it step by step is not to go backwards, but to derive the formula from Kepler's third law and the power emitted in gravitational waves. $\endgroup$
    – ProfRob
    Mar 18, 2021 at 23:19
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    $\begingroup$ @fasterthanlight I think you should un-accept my answer for now and let the answers and comments continue to evolve. I usually wait at least a week after there's no activity before accepting answers. $\endgroup$
    – uhoh
    Mar 18, 2021 at 23:35

For a circular orbit you can write the rate of change of the orbit semi-major axis $a$ as $$ \frac{da}{dt} = -\frac{64 G^3 (m_1+m_2)m_1m_2}{5a^3c^5} \ .$$

The components in the formula can be written in whatever units you like; bit of course that will determine the units of $da/dt$.

If you want the time dependence of $a(t)$, $\omega(t)$ or even the orbital period $T(t)$, they are (by simple integration) $$ a(t) = a_0\left( 1 - \frac{t}{\tau}\right)^{1/4}\, , $$ $$ \omega(t) = \omega_{0}\left( 1 - \frac{t}{\tau} \right)^{-3/8} $$ $$ T = \frac{2\pi}{\omega} = T_0\left(1 - \frac{t}{\tau} \right)^{3/8}\, , $$ where $\tau$ is the original merger timescale that you started with, and where the subscript refers to the values at $t=0$.


This is a response to FasterThanLight's request for a closed form solution. I tried to reply in a comment, but it turned out to be too long.

Can we know the distance between the two binary stars at any time t?

From ProfRob's answer,

$$\frac{da}{dt} = -\dfrac{64G^3(m_1m_2)(m_1+m_2)}{5c^5a^3} $$

If you want a closed form solution, we can rewrite the above as the differential equation


where the constant $C_1= -\dfrac{64G^3(m_1m_2)(m_1+m_2)}{5c^5}$. If the distance between the binary stars at time $t=0$ is $a_0$, then $a(0)=a_0$ is the initial value. The solution to the differential equation with initial value is:

$$a(t) = (a_0^4+4C_1t)^{1/4}$$

So the closed form solution or the distance between the stars at time t is given by:

$$a(t) = \left(a_0^4-\dfrac{256G^3(m_1m_2)(m_1+m_2)}{5c^5}t\right)^{1/4}$$


We can also get to a closed form solution to $a(t)$ by recognizing that $\tau a^4$, where $\tau = \dfrac{5c^5}{256G^3(m_1m_2)(m_1+m_2)}$ is a countdown time to merge, that is, the countdown time goes down as time goes up. This means for a binary system at $a_0$ distance at some initial time $t=0$, it will take $\tau a_0^4$ time to merge. So time $t=\tau a_0^4 - \tau a^4$ properly describes an increasing time and decreasing distance for an initial time and distance. Solve this algebraic equation for $a$ to get:

$$a(t) = \left( a_0^4 - \frac{t}{\tau}\right)^{1/4}$$

This is equivalent to my other answer, but uses a much simpler method. I purposely used the same notation ProfRob uses in his answer, but I can't explain why our answers are different.

  • $\begingroup$ +1 Thanks so much! That saved me from many hours of calculation ;) $\endgroup$
    – WarpPrime
    Mar 18, 2021 at 23:44

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