Gravity Differential Equation Is it possible to solve this system of differential equations in terms of $G$?
$$r''=\frac{-Gr}{\sqrt{\left(r^2+(1-s)^2\right)^3}},\ s''=\frac{G(1-s)}{\sqrt{\left(r^2+(1-s)^2\right)^3}}$$
with initial conditions $(r(0),s(0))=(0,0),\ (r'(0),s'(0))=(1,0)$. Moreover, $r$ and $s$ are functions of $t$.
I want to find $G$ for which its associated solution satisfies $(r(1),s(1))=(1/2,1)$. I tried a numerical approximation and got that $G\approx 2$ (very roughly). But I need the exact solution. Any ideas?
 A: The standard parametrization to get the Kepler rules is
$$
z=r+i(s-1)=\frac{e^{i\phi(t)}}{u(\phi(t))}.
$$
To incorporate the velocity data we need the first derivative of this parametrization,
$$
\dot z=e^{i\phi}\left(\frac{i}{u}-\frac{u'}{u^2}\right)\dot\phi=z\left(i-\frac{u'}{u}\right)\dot \phi.
$$
Per the naming conventions of the Kepler laws you then get the first law as
$$
u(\phi)=\frac{1+e\cos(\phi-\phi_{peri})}R\text{ with }R=\frac{L^2}{G}\tag{K1}
$$
and the second law as
$$
\dot\phi=Lu(\phi)^2\tag{K2}
$$
From your position data you have

*

*$0+i(0-1)=1e^{-i\pi/2}$, so $u(-\frac\pi2)=1$, and

*$1/2+i(1-1)=\frac12e^{i·0}$, so $u(0)=2$.

The velocity data point gives at $\phi=-\frac\pi2$, $u=1$
$$
1+i·0=(-i)\left(i-u'(\phi)\right)\dot \phi
$$
which allows to identify $\dot\phi=1$, $u'=0$, making this point the apoapsis. As the periapsis is opposite this, $\phi_{peri}=\frac\pi2$. The area velocity now follows as $L=1$.
In view of the first Kepler law, the positions now give the equations
\begin{align}
R&=1-e\\
2R&=1+e·0
\end{align}
giving $R=e=\frac12$ and thus $G=\frac{L^2}{R}=2$ exactly, as you had found numerically.
The period of the orbit is $T=2\pi ab$, where $a$ and $b$ are the long and short half-axis of the orbit $3r^2+\frac94(s-\frac23)^2=1$. Thus $T=2\pi·\frac23·\frac1{\sqrt3}=\frac{4\pi}{3\sqrt3}$.
Details of the derivation of the Kepler laws
Inserting the second derivative of $z$ into the gravity equation gives
\begin{align}
-\frac{Gz}{|z|^3}=\ddot z&=z\left(i-\frac{u'}{u}\right)^2\dot \phi^2
+z\left(-\frac{u''}{u}+\frac{u'^2}{u^2}\right)\dot \phi^2
+z\left(i-\frac{u'}{u}\right)\ddot \phi,
\\
\implies
-Gu^3&=\left(i-\frac{u'}{u}\right)\left(\ddot\phi-2\frac{u'}{u}\dot\phi^2\right)+\left(-1-\frac{u''}{u}\right)\dot\phi^2
\end{align}
As the imaginary part on the left is zero, so it has to be on the right, giving the second Kepler law
$$
\frac{\ddot\phi}{\dot\phi}=2\frac{u'(\phi)\dot\phi}{u(\phi)}\implies\dot\phi=Lu(\phi)^2.
$$
What remains is the first Kepler law,
$$
\frac{G}{L^2}=u''+u
$$
with its solution parametrized as in (K1)
