Solving $\frac{\mathrm d^2\mathbf{q}}{\mathrm dt^2} = -\frac{\mathbf{q}}{|\mathbf{q}|^3}$ I am reading a set of course notes and it has this example of a system of differential equations given by
$$\frac{\mathrm d^2\bf{q}}{\mathrm dt^2} = -\frac{\bf{q}}{|\bf{q}|^3}$$
All it says is that Newton showed that this equation is solved by the conic sections but I cannot find any references or do any work toward showing this. How do you solve this equation?
I believe $\mathbf{q} = \left[\begin{array}{c} x(t) \\ y(t) \end{array}\right]$ but I could be wrong. 
 A: It is called the Kepler Problem. You can find a detailed analysis in its Wikipedia page.
A: Yeah, that is the Two-Body problem. I know that there are closed solutions up to $\mathbf{q}\in \mathbb{R}^3$. A complete and detailed solution to this problem can be found in the chapter 2 of H.D. Curts, Orbital Mechanics for Engineer Student. There the complete solution is given in polar form. The principal step for the solution of that problem was found by Kepler. You can easily proove that the quantity
$
h = q \times \dot q
$
where $\times$ is the cross product, is a constant along the time history of the system, i.e.
$
\frac{dh}{dt} = 0
$
In fact $h$ is the angular momentum of the two-masses system. That means also the the motion of $q$ lies always in the same plane, so the current form of the solar system makes sense ;). Also you can write $q$ in polar form using just 2 coordinates.
From that you can derive the Kepler's second law: "equal areas are swept out in equal times", and the complete solution of the problem. I recommend to you reading that chapter mentioned above. Have a nice reading. 
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Note that ( this appears in any ${\tt\mbox{Classical Mechanics}}$ book ):
$$
\totald[2]{\bf q}{t}=-\,{{\bf q} \over \verts{\bf q}^{3}}
=\nabla_{\bf q}{1 \over \verts{\bf q}}
$$
which is derived from the Lagrangian
$\ds{L = \half\,\pars{\totald{\bf q}{t}}^{2} - {1 \over q}\ \mbox{where}\
q \equiv \verts{\bf q}}$. Now, write the Lagrangian in Cylindrical Coordinates ( we can choose $\ds{z = 0}$ ):
$$
L = \half\pars{\dot{q}^2 + q^{2}\dot{\theta}^{2}} - {1 \over q}
$$

From this equation we see that $\ds{p_{\theta} = \partiald{L}{\dot{\theta}} = q^{2}\dot{\theta}}$ is time independent such that $\ds{L}$ becomes:
  $$
L=\half\,\dot{q}^{2} + \half\,q^{2}\pars{p_{\theta} \over q^{2}}^{2} - {1 \over q}
=\half\,\dot{q}^{2} - \pars{{1 \over q} - {p_{\theta}^{2} \over 2q^{2}}}
$$

With Euler-Lagrange equations:
$$
\totald{}{t}\pars{\partiald{L}{\dot{q}}}
=\partiald{}{q}\pars{{1 \over q} - {p_{\theta}^{2} \over 2q^{2}}}
\ \imp\ \ddot{q}=-\,{1 \over q^{2}} + {p_{\theta}^{2} \over q^{3}}
$$ 

Also
  \begin{align}
\dot{q}&=\dot{\theta}\,\totald{q}{\theta}
={p_{\theta} \over q^{2}}\,\totald{q}{\theta}
=-p_{\theta}\,\totald{\pars{1/q}}{\theta}=
-p_{\theta}\,\totald{\xi}{\theta}\,,\qquad\xi \equiv {1 \over q}
\\[3mm]
\ddot{q}&=-p_{\theta}\dot{\theta}\,\totald[2]{\xi}{\theta}
=-\,{p_{\theta}^{2} \over q^{2}}\,\totald[2]{\xi}{\theta}
=-p_{\theta}^{2}\xi^{2}\,\totald[2]{\xi}{\theta}\ \imp\
-p_{\theta}^{2}\xi^{2}\,\totald[2]{\xi}{\theta}
=-\xi^{2} + p_{\theta}^{2}\xi^{3}
\end{align}

$$
\totald[2]{\xi}{\theta} + \xi={1 \over p_{\theta}^{2}}
$$
