Defining the Initial Conditions for a Planetary Motion to Have a Circular Orbit. The motion of a planet in a central force field has been restricted to a single plane. The equations below describe its motion in the $x$ and $y$ directions.
$x''=−4\pi^2x/((x^2+y^2)^{3/2})$
$y''=−4\pi^2y/((x^2+y^2)^{3/2})$
I'm supposed to show that in order to get a circular orbit, I should choose an initial velocity $v = 2\pi/\sqrt{r}$, perpendicular to $\mathbf{r}$, where $\mathbf{r}$ is the position vector of the planet and $r = (x^2+y^2)^{1/2}$.
I know that I can just start by assuming that the orbit is circular, and use the equations of uniform circular motion to show that the velocity has the magnitude and direction required, as below:
Centripetal force on body  = $F = mv^2/r$ where $m$ is the mass of the planet and $r$ the radius of its orbit.
Also, $F = GMm/r^2 = 4\pi^2m/r^2$ where $G$ is the universal gravitational constant and $M$ the mass of the central body. (Here, units are years and A.U so $GM = 4\pi^2$ (why is this?))
Then, $v^2/r = 4\pi^2/r^2 \implies v = 2\pi/\sqrt{r}$.
I'm sceptical of this solution. Is there a better way to approach this problem?
 A: Problem Statement 
Suppose that a particle has a radial acceleration of the form
$${\bf{a}} =  - {{GM} \over {{r^2}}}{{\bf{r}} \over r}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$ 
where $G$ and $M$ are physical positive constants, $r = \left\| {\bf{r}} \right\|$ is the absolute value of the position vector, and ${\bf{r}}$ is the position vector . Prove that if the initial velocity is perpendicular to the initial position vector and has the absolute value $\sqrt {{{GM} \over {{r_0}}}} $, i.e.
$$\left\{ \matrix{
  {{\bf{r}}_0}.{{\bf{v}}_0} = 0,\,\,\,\,{{\bf{r}}_0} \ne 0,\;\;\;{{\bf{v}}_0} \ne 0 \hfill \cr 
  {v_0} = \left\| {{{\bf{v}}_0}} \right\| = \sqrt {{{GM} \over {{r_0}}}}  \hfill \cr}  \right. \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$$
then the path of the particle will be a circle.

Solution
I tried to write an abstract solution for the problem and consequently the shortest one. Although it is not short! The strategy is to obtain the path in a general form and then apply the assumptions we have. First of all, we will show that this is a planar motion. For this purpose, consider the following
$$\left\{ \matrix{
  {\bf{r}} \times {\bf{a}} = {\bf{r}} \times  - {{GM} \over {{r^2}}}{{\bf{r}} \over r} = {\bf{0}} \hfill \cr 
  {\bf{r}} \times {\bf{a}} = {\bf{r}} \times {{d{\bf{v}}} \over {dt}} = {d \over {dt}}\left( {{\bf{r}} \times {\bf{v}}} \right) \hfill \cr}  \right.\,\,\,\,\, \to \,\,\,\,\,\,\,{d \over {dt}}\left( {{\bf{r}} \times {\bf{v}}} \right)\, = 0\,\,\,\, \to \,\,\,{\bf{r}} \times {\bf{v}} = {\bf{c}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3)$$
where by initial conditions you can find ${\bf{c}}$ as 
$${\bf{c}} = {{\bf{r}}_0} \times {{\bf{v}}_0} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)$$
and hence by assumptions ${\bf{c}} \ne {\bf{0}}$. Consequently, we have ${\bf{r}}.{\bf{c}} = 0$ which implies that ${\bf{r}}$ lies in the plane whose normal is ${\bf{c}}$. Before continuing, I just remark two identities which I will make use of it
$${\bf{r}}.{{d{\bf{r}}} \over {dt}} = {d \over {dt}}\left( {{1 \over 2}{\bf{r}}.{\bf{r}}} \right) = {d \over {dt}}\left( {{1 \over 2}{r^2}} \right) = r{{dr} \over {dt}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(5)$$  
$${\bf{a}} \times \left( {{\bf{b}} \times {\bf{c}}} \right) = \left( {{\bf{a}}.{\bf{c}}} \right){\bf{b}} - \left( {{\bf{a}}.{\bf{b}}} \right){\bf{c}}$$
Now we go on. Consider the followings
$$\eqalign{
  & \left\{ \matrix{
  {\bf{a}} \times {\bf{c}} = {{d{\bf{v}}} \over {dt}} \times {\bf{c}} = {d \over {dt}}\left( {{\bf{v}} \times {\bf{c}}} \right) \hfill \cr 
  {\bf{a}} \times {\bf{c}} =  - {{GM} \over {{r^2}}}{{\bf{r}} \over r} \times \left( {{\bf{r}} \times {\bf{v}}} \right) =  - {{GM} \over {{r^3}}}{\bf{r}} \times \left( {{\bf{r}} \times {{d{\bf{r}}} \over {dt}}} \right) \hfill \cr 
  \,\,\,\,\,\,\,\,\,\, =  - {{GM} \over {{r^3}}}\left[ {\left( {{\bf{r}}.{{d{\bf{r}}} \over {dt}}} \right){\bf{r}} - \left( {{\bf{r}}.{\bf{r}}} \right){{d{\bf{r}}} \over {dt}}} \right] \hfill \cr 
  \,\,\,\,\,\,\,\,\,\, =  - {{GM} \over {{r^3}}}\left( {r{{dr} \over {dt}}{\bf{r}} - {r^2}{{d{\bf{r}}} \over {dt}}} \right) \hfill \cr 
  \,\,\,\,\,\,\,\,\,\, = GM\left( {{1 \over r}{{d{\bf{r}}} \over {dt}} - {1 \over {{r^2}}}{{dr} \over {dt}}{\bf{r}}} \right) = GM{d \over {dt}}\left( {{{\bf{r}} \over r}} \right) \hfill \cr}  \right.\,\,  \cr 
  &  \to \,\,\,{d \over {dt}}\left( {{\bf{v}} \times {\bf{c}}} \right) = {d \over {dt}}\left( {GM{{\bf{r}} \over r}} \right)\,\,  \cr 
  &  \to \,\,{\bf{v}} \times {\bf{c}} = \,GM{{\bf{r}} \over r} + {\bf{b}} \cr} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(6)$$
Again use the initial conditions to determine ${\bf{b}}$ from the result of $(6)$ as follows
$$\eqalign{
  & {\bf{b}} = {{\bf{v}}_0} \times {\bf{c}} - \,GM{{{{\bf{r}}_0}} \over {{r_0}}} = {{\bf{v}}_0} \times \left( {{{\bf{r}}_0} \times {{\bf{v}}_0}} \right) - \,GM{{{{\bf{r}}_0}} \over {{r_0}}}  \cr 
  & \,\,\,\, = \left( {{{\bf{v}}_0}.{{\bf{v}}_0}} \right){{\bf{r}}_0} - \left( {{{\bf{r}}_0}.{{\bf{v}}_0}} \right){{\bf{v}}_0} - \,GM{{{{\bf{r}}_0}} \over {{r_0}}}  \cr 
  & \,\,\,\, = \left( {v_0^2 - {{GM} \over {{r_0}}}} \right){{\bf{r}}_0} - \left( {{{\bf{r}}_0}.{{\bf{v}}_0}} \right){{\bf{v}}_0} \cr} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(7)$$
Now, dot product the last equation in $(6)$ by ${\bf{r}}$ to eliminate ${\bf{v}}$ and obtain the equation of the path
$$\eqalign{
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,{\bf{r}}.\left( {{\bf{v}} \times {\bf{c}}} \right) = \,GM{{{\bf{r}}.{\bf{r}}} \over r} + {\bf{r}}.{\bf{b}}  \cr 
  &  \to \,\,\,\,\,\,\,{\bf{c}}.\left( {{\bf{r}} \times {\bf{v}}} \right) = GM{{{r^2}} \over r} + {\bf{r}}.{\bf{b}}  \cr 
  &  \to \,\,\,\,\,\,\,{\bf{c}}.{\bf{c}} = GMr + {\bf{r}}.{\bf{b}}  \cr 
  &  \to \,\,\,\,\,\,\,\,{c^2} = GMr + {\bf{r}}.{\bf{b}} \cr} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(8)$$
and finally the equation of path is
$${c^2} = GMr + {\bf{r}}.{\bf{b}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(9)$$
where by $(4)$ and $(7)$ we have $c$ and ${\bf{b}}$ in terms of initial position vector and velocity, and hence we have the general form of the path in terms of initial conditions. When does equation $(9)$ correspond to a circle? Consider the case ${\bf{r}}.{\bf{b}}=0$. But according to ${\bf{b}}$ being a constant vector and ${\bf{r}} \ne {\bf{0}}$ which varies with time, this can happen only when ${\bf{b}} = {\bf{0}}$ which is equivalent to
$$\left( {v_0^2 - {{GM} \over {{r_0}}}} \right){{\bf{r}}_0} - \left( {{{\bf{r}}_0}.{{\bf{v}}_0}} \right){{\bf{v}}_0} = {\bf{0}}\,\,\,\,\, \to \,\,\,\,\,\,\left\{ \matrix{
  v_0^2 = {{GM} \over {{r_0}}} \hfill \cr 
  {{\bf{r}}_0}.{{\bf{v}}_0} = 0 \hfill \cr}  \right.\,\,\,\,\, \to \,\,\,\,\,\,\,\,\left\{ \matrix{
  {v_0} = \sqrt {{{GM} \over {{r_0}}}}  \hfill \cr 
  {{\bf{r}}_0}.{{\bf{v}}_0} = 0 \hfill \cr}  \right. \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(10)$$
where I used the linear in-dependency of ${{{\bf{r}}_0}}$ and ${{\bf{v}}_0}$. In this case, equation $(9)$ becomes 
$${c^2} = GMr + 0\,\,\,\,\, \to \,\,\,\,\,\,r = {{{c^2}} \over {GM}} = {{r_0^2v_0^2} \over {GM}} = {{r_0^2{{GM} \over {{r_0}}}} \over {GM}} = {r_0} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(11)$$
which indeed is the equation of the circle. Finally, we are done! :) 
A: Problem Statement: 
$${\ddot {\bf{r}}} =  - {{k} \over {{{\left\| {\bf{r}} \right\|}^3}}}{\bf r}\qquad k\gt 0$$ 
$$\left\{ \matrix{
  {{\bf{r}}(0)}.{\dot{\bf{r}}(0)} = 0\hfill \cr \left\| \dot{\bf{r}}(0) \right\|^2 = {k \over {\bf{r}}(0)}  \hfill \cr}  \right.$$
we want to prove:
$${\left\| {\bf{r}} \right\|}=constant$$
Lemma: $${\left\| {\bf{r}} \right\|}=constant \quad \iff \quad {{\bf{r}}}.{\dot{\bf{r}}} \equiv 0 $$
So I prove in this case we have:
$$u:={{\bf{r}}}.{\dot{\bf{r}}} \equiv 0$$
Suppose
$${\bf A}:={\dot u}\,{\bf r}-u\,{\dot{\bf{r}}}$$
It is easy to prove that ${\dot{\bf{A}}}\equiv 0$ and so we have ${\bf{A}}=constant$.
But $u(0)={\dot u}(0)=0$ (why?) So: 
$${\bf{A}}\equiv {\bf{A}}(0)=0 $$
So:
$${\bf{A}}.{\dot{\bf{r}}}\equiv 0$$ 
Or:
$${\dot u}u-u\,{\dot{\bf{r}}}.{\dot{\bf{r}}}\equiv 0$$
Or:
$$u({\dot u}-{\dot{\bf{r}}}.{\dot{\bf{r}}})\equiv 0$$
But ${\dot u}-{\dot{\bf{r}}}.{\dot{\bf{r}}}$ is nowhere zero because:
$${\dot u}-{\dot{\bf{r}}}.{\dot{\bf{r}}}=\bf{r}.{\ddot{\bf{r}}}=- {{k} \over {{{\left\| {\bf{r}} \right\|}}}}$$
So:
$$u\equiv 0$$
That's it :) I love also Solution of my friend H. R.
