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On gravitational wave radiation changing orbital parameters

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Two celestial objects of mass $m_{1}$ and $m_{2}$ are orbiting around each other on a very elongated ellipse($\epsilon\approx1$). The system is isolated, there is no other proximate celestial body. How would the parameters of the elliptical orbit change as a consequence of radiation of gravitational waves. How can the time of the process be estimated when the orbit changes from an ellipse to a circle?

Here is my work:
The power emitted by gravitational waves is given by:
$$P_{GW}= \frac{c^5}{G}\left(\frac{GM}{c^5l}\right)^5$$

Very compact binaries will lose energy rapidly by GW radiation.
If we assume that the two bodies making up the binary lie in the $x-y$ plane and their **orbits are circular** ($\epsilon=0$), then only non-vanishing components of quadrupole tensors are:
$$\boxed{Q_{xx}=-Q_{yy}=\frac{1}{2}(\mu)a^{2}\cos2\Omega t}$$ and 
$$\boxed{Q_{xy}=Q_{ya}=\frac{1}{2}(\mu)a^{2}\sin2\Omega t}$$
Where $\Omega$ is the orbital velocity, $\mu=\dfrac{m_{1}m_{2}}{m}$ is the reduced mass and where $m=m_{1}+m_{2}$

The luminosity of the system can be deduced as:
$$L^{GM}=\frac{32}{5}\frac{G}{c^5}\mu^2 a^4\Omega^6=\frac{32}{5}\frac{G^4}{c^5}\frac{M^3 \mu^2}{a^5}$$

The latter part is obtained from Kepler third law: $\Omega^2=\frac{GM}{a^3}$
As the gravitating system loses energy by emitting radiation, the distance between the 2 bodies shrinks at a rate:
$$\boxed{\frac{da}{dt}=\frac{64}{5}\frac{G^{3}M\mu}{c^5 a^3}}$$
The binary would hence colase at a time :
$$\tau=\frac{5}{256}\frac{c^5}{G^3}\frac{a_{0}^4}{\mu M^4}$$

I am having problems in converting this concept to that mentioned in the question, I.e., for a system of binaries of highly elliptical orbit(in my calculations, I have assumed a circle) and how would the elliptical parameters change with time. I also require help in calculating the time of the process when the orbit changes from an ellipse to a circle.

I have obtained the following integral for finding the decay time of the orbits( the orbits would eventually decay into a circular one):
$$T(a_{0},e_{0})=\frac{12(c_{0}^4)}{19\gamma}\int_{0}^{e_0}{\frac{e^{29/19}[1+(121/304)e^2]^{1181/2299}}{(1-e^2)^{3/2}}}de\tag1$$
Where $$\gamma=\frac{64G^3}{5c^5}m_{1}m_{2}(m_{1}+m_{2})$$
For $e_{0}$ close to $1$ the equation becomes:
$$T(a_{0},e_{0})\approx\frac{768}{425}T_{f}a_{0}(1-e_{0}^2)^{7/2}\tag2$$
Where $$T_{f}=\frac{a_{0}^4}{4\gamma}$$
Here $T$ depends on both $a_{0}$ an de $e_{0}$ as the former determines the latter.
And $\gamma$ has the same value as define do above.
I am stuck with $(1)$ and $(2)$; any help would be appreciated.
**My approach:**
I have tried to use Appell hypergeometric integration and obtained the following result:(let $x=e$)
>$$\boxed{\int\frac{x^(\frac{3}{2}+y)[1+Ax^2]^{\frac{1}{2}+z}}{(1-x^{2})^\frac{3}{2}}dx=2x^{y+\frac{5}{2}}\frac{F_{1}(\frac{y}{2}+\frac{5}{4};\frac{3}{2},-z-\frac{y}{2}+\frac{9}{4};x^{2},-Ax^2)}{2y+5}}$$
Where $y=\frac{55}{38},z=\frac{63}{4598},A=\frac{121}{304}$
Here $F_{1}(a,\beta,\beta^{'},c;x,y)$ is appells function and I read that it can be expressed as the following integral:
>$$\boxed{F_{1}(a,\beta,\beta^{'},c;x,y)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_{0}^{1}u^{a-1}(1-u)^{b-a-1}(1-ux)^{-\beta}(1-uy)^{-\beta^{'}}du}$$
Is this right? How am I to solve this integral? And also how am I to solve for $e_{0}$ in equation (2)?

asked Oct 29, 2016 in Astronomy by Naveen (45 points) [ revision history ]
edited Oct 31, 2016 by Naveen

I have to say it is not clear to me at all how does equation (1) come about. For instance, do you take relativistic precession into account, or do you assume Newtonian trajectories at every point of the decay? (I.e., what is your approximation, Newtonian+linearized gravity waves, post-Newtonian...) The loss of eccentricity can be essentially seen from the slightly different behavior of energy and angular momentum - how would that look in your case?

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