<p>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.</p>
<p><strong>Can we know the distance between the two binary stars at any time t?</strong></p>
<p>From ProfRob's answer,</p>
<p><span class="math-container">$$\frac{da}{dt} = -\dfrac{64G^3(m_1m_2)(m_1+m_2)}{5c^5a^3} $$</span></p>
<p>If you want a closed form solution, we can rewrite the above as the differential equation</p>
<p><span class="math-container">$$a'=C_1a^{-3}$$</span></p>
<p>where the constant <span class="math-container">$C_1= -\dfrac{64G^3(m_1m_2)(m_1+m_2)}{5c^5}$</span>.  If the distance between the binary stars at time <span class="math-container">$t=0$</span> is <span class="math-container">$a_0$</span>, then <span class="math-container">$a(0)=a_0$</span> is the initial value.  The solution to the differential equation with initial value is:</p>
<p><span class="math-container">$$a(t) = (a_0^4+4C_1t)^{1/4}$$</span></p>
<p>So the closed form solution or the distance between the stars at time t is given by:</p>
<p><span class="math-container">$$a(t) = \left(a_0^4-\dfrac{256G^3(m_1m_2)(m_1+m_2)}{5c^5}t\right)^{1/4}$$</span></p>
<p><strong>edit:</strong></p>
<p>We can also get to a closed form solution to <span class="math-container">$a(t)$</span> by recognizing that <span class="math-container">$\tau a^4$</span>, where <span class="math-container">$\tau = \dfrac{5c^5}{256G^3(m_1m_2)(m_1+m_2)}$</span> is a countdown time to merge, that is, the countdown time goes down as time goes up.  This means for a binary system at <span class="math-container">$a_0$</span> distance at some initial time <span class="math-container">$t=0$</span>, it will take <span class="math-container">$\tau a_0^4$</span> time to merge. So time <span class="math-container">$t=\tau a_0^4 - \tau a^4$</span> properly describes an increasing time and decreasing distance for an initial time and distance. Solve this algebraic equation for <span class="math-container">$a$</span> to get:</p>
<p><span class="math-container">$$a(t) = \left( a_0^4 - \frac{t}{\tau}\right)^{1/4}$$</span></p>
<p>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.</p>

CC BY-SA 4.0
2021-03-18T23:44:49.147Z
+1 Thanks so much! That saved me from many hours of calculation ;)
WarpPrime
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2021-03-18T17:55:14.827Z
2021-03-23T17:55:12.507Z
2021-03-23T17:55:12.507Z
Connor Garcia
<p>Here's one way to arrive at <a href="https://astronomy.stackexchange.com/a/42000/7982">ProfRob's answer</a> step by step:</p>
<p><span class="math-container">$$t = \dfrac{5c^5r^4}{256G^3(m_1m_2)(m_1+m_2)}$$</span></p>
<p>If we lump all the constant terms together:</p>
<p><span class="math-container">$$C = \dfrac{5c^5}{256G^3(m_1m_2)(m_1+m_2)}$$</span></p>
<p>then the problem suddenly looks a lot simpler:</p>
<p><span class="math-container">$$t = r^4 C$$</span></p>
<p><span class="math-container">$$\frac{dt}{dr} = 4r^3 C$$</span></p>
<p><span class="math-container">$$\frac{dr}{dt} = \frac{1}{4r^3} \frac{1}{C}$$</span></p>
<p><span class="math-container">$dt$</span> is how much <em>longer</em> the system will live if the starting radius is <em>increased</em> by <span class="math-container">$dr$</span>. We know that if we watch it decay, the rate of decrease per unit time <span class="math-container">$dr/dt$</span> will be negative, <strong>so we'll add a minus sign.</strong></p>
<p>Now put it back together:</p>
<p><span class="math-container">$$\frac{dr}{dt} = -\frac{1}{4r^3} \frac{1}{C}$$</span></p>
<p><span class="math-container">$$\frac{dr}{dt} = -\frac{1}{4r^3} \dfrac{256G^3(m_1m_2)(m_1+m_2)}{5c^5} $$</span></p>
<p><span class="math-container">$$\frac{dr}{dt} = -\frac{1}{r^3} \dfrac{64G^3(m_1m_2)(m_1+m_2)}{5c^5} $$</span></p>
<p><span class="math-container">$$\frac{dr}{dt} = -\dfrac{64G^3(m_1m_2)(m_1+m_2)}{5c^5r^3} $$</span></p>
<p>With <a href="https://en.wikipedia.org/wiki/Gravitational_constant" rel="nofollow noreferrer">gravitational constant</a> <span class="math-container">$G = 6.674 \times 10^{-11} \text{m}^{3} \text{kg}^{-1} \text{s}^{-2}$</span> the speed of light <span class="math-container">$c=2.997 \times 10^{8} \text{m/s}$</span> each body with a solar mass <span class="math-container">$M_{\text{Sol}} = 1.9885 \times 10^{30} \text{kg}$</span> and an almost-touching separation of a million kilometers <span class="math-container">$r=1\times 10^{9} \text{m}$</span>, then <span class="math-container">$C = 1.0102 \times 10^{-20}$</span> and the rate <span class="math-container">$dr/dt = -2.4748 \times 10^{-8} \text{s}^{-1}$</span> or -78 centimeters per year.</p>

CC BY-SA 4.0
2021-03-18T12:57:33.403Z
Is there such a graph relating an initial $m_1$, $m_2$, $r$, with time, and rate?
WarpPrime
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2021-03-18T13:13:01.133Z
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?
WarpPrime
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2021-03-18T14:07:58.987Z
Asked a question regarding the above comment: https://math.stackexchange.com/questions/4066853/deriving-closed-form-of-function
WarpPrime
CC BY-SA 4.0
2021-03-18T18:15:03.053Z
@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!
Connor Garcia
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2021-03-18T23:19:20.493Z
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.
ProfRob
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2021-03-18T23:35:18.293Z
@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.
uhoh
CC BY-SA 4.0
2021-03-18T01:38:31.063Z
2021-03-18T18:43:48.837Z
2021-03-18T18:43:48.837Z
uhoh
<p>For a circular orbit you can write the rate of change of the orbit semi-major axis <span class="math-container">$a$</span> as
<span class="math-container">$$ \frac{da}{dt} = -\frac{64 G^3 (m_1+m_2)m_1m_2}{5a^3c^5} \ .$$</span></p>
<p>The components in the formula can be written in whatever units you like; bit of course that will determine the units of <span class="math-container">$da/dt$</span>.</p>
<p>If you want the time dependence of <span class="math-container">$a(t)$</span>, <span class="math-container">$\omega(t)$</span> or even the orbital period <span class="math-container">$T(t)$</span>, they are (by simple integration)
<span class="math-container">$$
a(t) = a_0\left( 1 - \frac{t}{\tau}\right)^{1/4}\, ,
$$</span>
<span class="math-container">$$
\omega(t) = \omega_{0}\left( 1 - \frac{t}{\tau} \right)^{-3/8}
$$</span>
<span class="math-container">$$
T = \frac{2\pi}{\omega} = T_0\left(1 - \frac{t}{\tau} \right)^{3/8}\, ,
$$</span>
where <span class="math-container">$\tau$</span> is the original merger timescale that you started with, and where the subscript refers to the values at <span class="math-container">$t=0$</span>.</p>

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2021-03-17T23:12:52.757Z
2021-03-18T23:28:07.853Z
2021-03-18T23:28:07.853Z
ProfRob
<p>We know that the estimated time until a binary system merges is given by: <span class="math-container">$$\dfrac{5c^5r^4}{256G^3(m_1m_2)(m_1+m_2)}$$</span> 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 <span class="math-container">$r$</span> and <span class="math-container">$m_i$</span> written in?</p>

CC BY-SA 4.0
2021-03-17T21:41:42.443Z
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. https://en.wikipedia.org/wiki/Gravitational_constant gives a $G$ value using parsecs & solar masses.
PM 2Ring
CC BY-SA 4.0
2021-03-17T21:50:42.957Z
There are some rate equations here. https://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity#Gravitational_radiation
PM 2Ring
CC BY-SA 4.0
2021-03-17T22:22:15.127Z
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!
Connor Garcia
<orbit><gravity><orbital-mechanics><gravitational-waves><orbital-decay>
Computing the rate of orbital decay
WarpPrime
# Computing the rate of orbital decay

(Asked by: username_0 on 2021-03-17)

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?

**Tags:** orbit, gravity, orbital-mechanics, gravitational-waves, orbital-decay
> **Comments:**
> **username_1**: 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. https://en.wikipedia.org/wiki/Gravitational_constant gives a $G$ value using parsecs & solar masses.  
> **username_1**: There are some rate equations here. https://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity#Gravitational_radiation  
> **username_2**: 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!  

---

## 3 answers

### Answer by username_2 on 2021-03-18. Score: 2 | Accepted answer

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

$$a'=C\_1a^{-3}$$

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:

$$a(t) = \left(a\_0^4-\dfrac{256G^3(m\_1m\_2)(m\_1+m\_2)}{5c^5}t\right)^{1/4}$$

**edit:**

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.

> **Comments:**
> **username_0**: +1 Thanks so much! That saved me from many hours of calculation ;)  

---

### Answer by username_3 on 2021-03-18. Score: 6

Here's one way to arrive at [ProfRob's answer](https://astronomy.stackexchange.com/a/42000/7982) 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](https://en.wikipedia.org/wiki/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.

> **Comments:**
> **username_0**: Is there such a graph relating an initial $m_1$, $m_2$, $r$, with time, and rate?  
> **username_0**: 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?  
> **username_0**: Asked a question regarding the above comment: https://math.stackexchange.com/questions/4066853/deriving-closed-form-of-function  
> **username_2**: @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!  
> **username_4**: 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.  
> **username_3**: @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.  

---

### Answer by username_4 on 2021-03-17. Score: 4

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 $t=0$.


---

