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For $k=\kappa >0$, the same asymptotic data give $E=v^2/2$ and $|h|=bv$. At the central-force radial turning point,The positive root and the tangential speed, again with $q=\kappa /v^2$, areFor $b>0$, $\tilde{p}>b$ and $\tilde{u}<v$. The repulsive inverse-square force bends the trajectory away from the origin, as in the right-hand sketch. Its polar equation of a repulsive inverse-square orbit is $r=h^2/[\kappa (e\cos \varphi -1)]$, with $e=\sqrt{1+b^2{v}^4/{\kappa }^2}>1$ and $\varphi$ measured from closest approach.

For $b=0$, the regular radial turning point is $\tilde{p}=2\kappa /v^2$ and its speed is zero; the particle reverses direction there. The formula $bv/\tilde{p}$ still gives zero, while its rationalized form containing $1/b$ should be interpreted only by a limit.

The repulsive force is $\mathbf{F}=k\mathbf{r}/r^3=-\nabla (k/r)$, so the Hamiltonian isThe Hamilton's equations give $\dot{\mathbf{r}}=\mathbf{p}/m$ and $\dot{\mathbf{p}}=k\mathbf{r}/r^3$. Consequentlyso the angular momentum is conserved. Moreover,It follows that the Laplace-Runge-Lenz vector $\mathbf{A}=\mathbf{p}\times \mathbf{L}+mk\hat{\mathbf{r}}$ is conserved. By the Hamiltonian conservation law, these time derivatives are exactly $\{\mathbf{L},H\}$ and $\{\mathbf{A},H\}$, so both Poisson brackets vanish.

The integrals of motion are the energy $H$, the components of $\mathbf{L}$, and the components of $\mathbf{A}$, subject toTaking the dot product of $\mathbf{A}$ with $\hat{\mathbf{r}}$ givesTherefore the polar equation of a repulsive inverse-square orbit iswhere $\lambda =L^2/(mk)\ge 0$ and $e=A/(mk)\ge 0$. For a nonradial physical scattering orbit $H>0$, the identity above gives $e>1$, as expected for a hyperbolic Kepler orbit.