 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 $\mathbf{L}\ne 0$. 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 $\mathbf{K}=\mathbf{L}\times \dot{\mathbf{r}}-\varphi (r)\mathbf{r}$. Since $\ddot{\mathbf{r}}=-{\varphi }^{\prime }(r)\hat{\mathbf{r}}/m$ and $\mathbf{L}$ is constant, the vector triple product givesThereforeIt vanishes for general motion precisely when $(r\varphi {)}^{\prime }=0$, givingThis 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 $\varphi$ 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.