Kepler problem in cartesian coordinates I'm trying to solve the Kepler problem in Cartesian coordinates, that is, I want to show that the trajectory is an ellipse using Cartesian coordinates instead of using polar coordinates, as is usually done. For simplicity, I will solve the one-body problem in which one of the two masses, with mass $M$, is assumed to be static, while only the other mass, $m$ is moving, and $M\gg m$. Assuming gravity is the only force acting on the particle of mass $m$, the equation of motion is
$$m\ddot{\mathbf{r}}=-\frac{GMm}{|\mathbf{r}|^3}\mathbf{r}=-\frac{GMm}{|\mathbf{r}|^2}\hat{\mathbf{r}},$$
which is equivalent to the following equations for the $x$ and $y$ coordinates
$$m\ddot{x}=-\frac{GMmx}{(x^2+y^2)^{\frac{3}{2}}}, \qquad m\ddot{y}=-\frac{GMmy} {(x^2+y^2)^{\frac{3}{2}}},$$
since the motion is confined to a plane. Now that I have these equations I don't know how to proceed and show that the trajectory is an ellipse. How does one show from these equations that the trajectory is a conic section of eccentricity $\epsilon$? Also, how does one derive Kepler's third law from this?
 A: Absorb the dimensional GM into the units, to remove excuses for not recognizing the plane-geometry structure. Note the rotational and translational invariance to be used in fixing your coordinate system below. All vectors are then 2-vectors on that plane,
$$\ddot{ x }=-{ { x}\over r^3},\qquad \ddot{ y }=-{ { y}\over r^3} ~. $$
It is then self-evident that
$$
L=x\dot y -y\dot x
$$
is a constant of the motion (in the z-direction, the only non vanishing one, of course), $\dot L=0$.
Moreover, the mass/normalized LRL 2-vector on that plane,
$$
\vec e= L \begin{pmatrix} \dot y\\-\dot x\end{pmatrix}-{1\over r} \begin{pmatrix} x\\  y\end{pmatrix}
$$
is also conserved, $\dot{\vec e }=0$. Its magnitude will turn out to be the eccentricity.
Dotting by $\vec r$ you have
$$
\vec r\cdot \vec e =L^2 -r, \leadsto \\
r= L^2-\vec r\cdot \vec e , \leadsto \\
x^2+y^2= (L^2-\vec r\cdot \vec e )^2.
$$
You may use rotational invariance to take $\vec e$ along the x-axis, $\vec e=-\epsilon \hat x$
to prettify your ellipse orientation, and work out the r.h.s. square, as a quadratic polynomial in x, with the obvious constants. Elementary algebra leads you to the ellipse your teacher taught you in terms of the constants ε and L. That is, shifting the origin of xs to $L^2 \epsilon/(1-\epsilon^2)$ and taking $\epsilon ^2= 1-b^2/a^2$, you get
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}    = L^4 a^2/b^4.
$$
A: I can't comment directly on posts, so I'll post a note here (for NinjaDarth). The other comment is right: the solution is too over-wrought and (even as noted in the reply) it can be simplified. So, I'll actually do it here, following the suggestions made in the reply. The result is much simpler and more direct.
We'll keep the notation: the Cartesian coordinates are wrapped up in 3D-vector form. The Kepler problem is framed as:
$$
'(t) = (t), \hspace 1em (0) = _0, \\
'(t) = -\frac{μ(t)}{r(t)^3}, \hspace 1em (0) = _0,
$$
where $r = ||$, with a similar convention to denote magnitudes of vectors by using the light-face version of the bold-face letter. We'll change the clock to $G$ by writing
$$\frac{dt}{dG} = r.$$
When this is applied to $$ and $r$, the results - after using some vector algebra - are:
$$
\frac{d}{dG} = r, \hspace 1em \frac{d(r)}{dG} = z + , \\
\frac{dr}{dG} = ·, \hspace 1em \frac{d(·)}{dG} = zr + μ,
$$
where
$$z = v^2 - \frac{2μ}{r}, \hspace 1em  = \frac{μ}{r} - ×(×),$$
both of which prove to be constant, and to be connected to the better-known constants of the Kepler problem:
$$ = ×, \hspace 1em  = \frac{×}{μ} - \frac{}{r}, \hspace 1em H = \frac{v^2}{2} - \frac{μ}{r},
$$
by $ = -μ$ and $z = 2H$.
The Solution:
We will use the functions $(C,S,D,T)$ of $G$ and $z$ given by
$$\frac{d}{dG}(C,S,D,T) = (zS,C,S,D), \hspace 1em (C,S,D,T) = (1,0,0,0) @ G = 0.$$
Noting that the general solution to the problem
$$a'(G) = b, \hspace 1em b'(G) = z a(G) + c, \hspace 1em a(0) = a_0, \hspace 1em b(0) = b_0,$$
is
$$a = a_0 C + b_0 S + c D, \hspace 1em b = b_0 C + (c + a_0 z) S,$$
then it follows that:
$$
 = _0 C + r_0 _0 S +  D, \\
r = r_0 C + _0·_0 S + μ D, \\
 = \frac{r_0 _0 C + ( + z _0) S}{r},
$$
and, upon integration of $dt/dG = r$ with $t(0) = 0$:
$$t = r_0 S + _0·_0 D + μ T.$$
To recover $t_0$ as the sixth Kepler constant (along with the five independent parameters that come out of $(, , H)$), we could reset $G$ so that $G = 0$ when $r$ is at its minimum, at which time, we would have $_0·_0 = 0$; but it's not really necessary, since $_0$ and $_0$ already provide us with six independent parameters, in place of the six Kepler parameters.
Conic Section Equation:
Actual equations can written for $$ in vector form, which will show that it is a conic or a line, by using the identities:
$$C = 1 + z D, \hspace 1em S^2 = 2 D + z D^2.$$
First: for the following, define
$$ =  + z _0, \hspace 1em  =  × _0 =  × _0 + z .$$
Then, second: we have
$$ - _0 = r_0 _0 S +  D,$$
$$\left( - _0\right) × _0 = \left(r_0 _0 S +  D\right) × _0 =  × _0 D =  D,$$
$$ × \left( - _0\right) =  × \left(r_0 _0 S +  D\right) = r_0  × _0 S = r_0  S.$$
Then, from $S^2 = 2D + z D^2$, upon cancellation of the factor of $c^2$, follows:
$$\left| × \left( - _0\right)\right|^2 = {r_0}^2 \left(2·\left( - _0\right) × _0 + \left|\left( - _0\right) × _0\right|^2 \right).$$
Otherwise, we have:
$$ = .$$
When expressed in terms of the constants $$, $$ and $H$, noting that:
$$
\left( - _0\right)×_0 = ×_0 - , \\
 =  + z _0 = -μ  + 2 H _0, \\
 =  × _0 =  × _0 + z  = -μ  × _0 + 2 H ,
$$
we can write the following:
$$
\left|(-μ  + 2H _0) × \left( - _0\right)\right|^2 = {r_0}^2 \left(2 \left(-μ  × _0 + 2 H \right)·\left( × _0 - \right)  + \left| × _0 - \right|^2 \right).
$$
The case $ = $ has to be handled separately. The condition entails that
$$\left( - _0\right) × _0 =  D = ,$$
and may only occur if either $_0 = $ - which gives rise to linear motion $ = _0 C +  D = _0 +  D$ - or $_0 ≠ 0$ - which also also gives rise to linear motion, with $ = _0 + (r_0 S + ·_0/{v_0}^2 D) _0$.
