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For potential energy $\Phi (r)=-mk/r^2$, the inward central force has magnitude $2mk/r^3$. With conserved specific angular momentum $h=r^2\dot{\theta }$, the radial equation is $\ddot{r}=(h^2-2k)/r^3$. At the critical value $h^2=2k$, this reduces to $\ddot{r}=0$. An inward initial radial velocity $-v$, with $v>0$ and $r(0)=a$, therefore gives $r=a-vt$ until collision at $t_{\ast }=a/v$.

For $h>0$, integration of $\dot{\theta }=h/(a-vt{)}^2$ gives $\theta -\theta (0)=(h/v)(1/r-1/a)$. Thus the orbit is a hyperbolic spiral and the particle completes infinitely many turns before the finite collision time. The tangential speed $h/r$ diverges. The potential energy and kinetic energy individually diverge with opposite signs, while their sum remains the finite conserved mechanical energy $m{v}^2/2$.

The initial transverse velocity gives $h=a(\sqrt{2k}/a)=\sqrt{2k}$. The initial mechanical energy isSince $\Phi (1/u)=-mk{u}^2$ and $m{h}^2/2=mk$, the $u^2$ terms cancel in the conserved energy equation:The initial radial velocity fixes the sign, because $\dot{r}=-h{u}^{\prime }$ gives $u^{\prime }(0)=1/a$. The continuous derivative $u^{\prime }$ cannot change sign while its square remains $1/a^2$. As $u(0)=1/a$, this givesThe future orbit is a hyperbolic spiral: it starts at $(a,0)$, turns counterclockwise and winds indefinitely towards the origin. Its radial distance decreases on every turn. The positive-$r$ continuation has $\theta >-1$, but only $\theta \ge 0$ is needed after release.

To decide whether infinitely many turns require infinite time, use the conserved angular momentum:Integrating from $0$ givesConsequentlyThe particle reaches the origin in finite time, in the sense that $r\to 0$ as $t\uparrow t_{\ast }$. Meanwhile $\theta =t/(t_{\ast }-t)\to \infty$. The central force is undefined at the origin, so these formulas describe motion up to collision. This is finite-time collision in an inverse-cube central force.