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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.

Taking the dot product of the Laplace-Runge-Lenz vector with $\hat{\mathbf{r}}$ gives $A\cos \theta =L^2/r+mk$. Thus