= Solution
With $\psi=0$, the <angular momentum> becomes $\mathbf L=m\mathbf r\times\dot{\mathbf r}$ and is constant. The identity $\mathbf L\cdot\mathbf r=0$ confines the trajectory to the plane through the origin perpendicular to $\mathbf L$ when $\mathbf L\ne0$. If $\mathbf L=0$, the <velocity> is radial and a <central force> preserves the radial line; this degenerate motion is also planar.
For the proposed <Laplace-Runge-Lenz vector> convention, differentiate $\mathbf K=\mathbf L\times\dot{\mathbf r}-\phi(r)\mathbf r$. Since $\ddot{\mathbf r}=-\phi'(r)\widehat{\mathbf r}/m$ and $\mathbf L$ is constant, the <vector triple product> gives
$$
\mathbf L\times\ddot{\mathbf r}
=-\phi'(r)(r\dot{\mathbf r}-\dot r\,\mathbf r).
$$
Therefore
$$
\dot{\mathbf K}
=-\phi'(r)(r\dot{\mathbf r}-\dot r\,\mathbf r)
-\phi'(r)\dot r\,\mathbf r-\phi(r)\dot{\mathbf r}
=-[r\phi'(r)+\phi(r)]\dot{\mathbf r}.
$$
It vanishes for general motion precisely when $(r\phi)'=0$, giving
$$
\boxed{\phi(r)=\frac Cr\quad\text{for a constant }C,\qquad\dot{\mathbf K}=0.}
$$
This is an <inverse-square potential>, attractive when $C<0$ and repulsive when $C>0$; $C=0$ is the free-particle case. An additive constant in $\phi$ leaves the force unchanged but would change this particular $\mathbf K$ by a constant multiple of $\mathbf r$, so its definition requires the displayed choice of energy zero.
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