Kinetic Energy in central force motion

Question

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 $r_{mín}$?

Answer 1

You must be mistaken in your calculations.

Let R = max distance from O(according to your diagram), and $v_0$be the velocity at r = R(r is the general radial distance of any point P from O).

Since energy is conserved, we have : $\frac{-k}{R} + \frac{1}{2}m{v_0}^2$ = $\frac{-k}{r} + K(r)$ [K(r) is the kinetic energy for radial distance r].

From this we obtain $K(r) = k(\frac{1}{r} - \frac{1}{R}) + \frac{1}{2}m{v_0^2}$.

Clearly, $K(R) = \frac{1}{2}m{v_0^2} \ne K(R/3) = \frac{2k}{R} + \frac{1}{2}m{v_0^2}$