My question is how can I get to know the velocity of a particle in a elliptical trajectory if a particle is affected by a potential
\begin{equation} U(r) = -\frac{k}{r} \end{equation}
I have a particle of mass $m$ moving under the action of $F(r) = -k/r^2$ for, obviously in correspondence with the gravitational force, $k > 0$. In a point called $P$ distant $a$ from the origin the velocity is $v_0 = \frac{k}{2ma}$ and the velocity in this instant is perpendicular to the position vector in $P$.
I have found the effective potential
\begin{equation} U_{eff} = \frac{ka}{4r^2} - \frac{k}{r} \end{equation}
Using that the angular momentum of this system is a constant of the motion. Then, if you have that the total energy is also a constant of the motion because the force is independent of the time we can find $E$ using
\begin{equation} E = \frac{mv_0^2}{2} + U_{eff}(a) \end{equation}
But then my problem is that I get that the Kinetic Energy is the same both for the $r_{máx} = a$ and for the $r_{mín} = a/3$ the apsidal distances of this orbit. Is the assumption that $U_{eff}(r_{máx}) = U_{eff}(r_{mín})$ wrong ? I'm using this because of the Plot of the equation of the $U_{eff}$. We have two values that satisfy $U_{eff}(r) = -3k/4a = U_{eff}(a)$.
What is the Kinetic Energy for $r_{mín}$?