Circular solution of Kepler Problem

Question

How can I get a circular solution of the $2-$dimensional Kepler problem of the form

$$q=\exp(Kt)a$$

being

$$\exp(Kt)= \begin{pmatrix} \cos(t) & \sin(t)\\ -\sin(t) & \cos(t)\\ \end{pmatrix} $$

and $a$ a constant vector?

Don't know how to start.

Thank you very much

Answer 1

The solution I got is:

Kepler Problem: (1) $\ddot{q}=\frac{-\mu}{||q||^3}q$

As we want a solution of the form $q=(cos(t)*a_1+sin(t)*a_2, -sin(t)*a_1+cos(t)*a_2)$, we simply look for its derivatives:

$\dot{q}=(-sin(t)*a_1+cos(t)*a_2, -cos(t)*a_1-sin(t)*a_2)$ $\ddot{q}=(-cos(t)*a_1-sin(t)*a_2, sin(t)*a_1-cos(t)*a_2)$

$||q||= ||a||= \sqrt{a_1^2+a_2^2}$

Substituing in (1) we find: $||a||=\sqrt[3]{\mu}$

Taking $a_2=0$ we find a circular solution of the Kepler problem: $$q=(\sqrt[3]{\mu}cos(t), -\sqrt[3]{\mu}sin(t)$$ which are cercles of radius $\sqrt[3]{\mu}$