= Solution
The repulsive force is $\mathbf F=k\mathbf r/r^3=-\nabla(k/r)$, so the <Hamiltonian> is
$$
H=\frac{\mathbf p^2}{2m}+\frac kr.
$$
The <Hamilton's equations> give $\dot{\mathbf r}=\mathbf p/m$ and $\dot{\mathbf p}=k\mathbf r/r^3$. Consequently
$$
\dot{\mathbf L}
=\dot{\mathbf r}\times\mathbf p+\mathbf r\times\dot{\mathbf p}=\mathbf0,
$$
so the <angular momentum> is conserved. Moreover,
$$
\dot{\mathbf p}\times\mathbf L
=k\left(\frac{\mathbf r(\mathbf r\cdot\mathbf p)}{r^3}-\frac{\mathbf p}{r}\right)
=-mk\frac{d\widehat{\mathbf r}}{dt}.
$$
It follows that the <Laplace-Runge-Lenz vector> $\mathbf A=\mathbf p\times\mathbf L+mk\widehat{\mathbf r}$ is conserved. By the <Hamiltonian conservation law>, these time derivatives are exactly $\{\mathbf L,H\}$ and $\{\mathbf A,H\}$, so both <Poisson bracket>[Poisson brackets] vanish.
The <integrals of motion> are the energy $H$, the components of $\mathbf L$, and the components of $\mathbf A$, subject to
$$
\mathbf A\cdot\mathbf L=0,
\qquad
A^2=m^2k^2+2mHL^2.
$$
Taking the <dot product> of $\mathbf A$ with $\widehat{\mathbf r}$ gives
$$
A\cos\theta=\frac{L^2}{r}+mk.
$$
Therefore the <polar equation of a repulsive inverse-square orbit> is
$$
r=\frac{L^2/(mk)}{(A/(mk))\cos\theta-1}
=\frac{\lambda}{e\cos\theta-1},
$$
where $\lambda=L^2/(mk)\geq0$ and $e=A/(mk)\geq0$. For a nonradial physical scattering orbit $H>0$, the identity above gives $e>1$, as expected for a <hyperbolic Kepler orbit>.
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