Repulsive Coulomb scattering: the unique outer turning point, global non-colliding motion, the asymptotic impact distance and $\mathrm{\Delta }=\pi \mathrm{/}2-\arctan (\kappa \mathrm{/}(2Eb))$

The derivative $V^{\mathrm{\prime }}(r)=-\kappa \mathrm{/}{r}^2-J^2\mathrm{/}(\mu r^3)$ is negative on $(0,\mathrm{\infty })$, so $V$ is strictly decreasing by [L2]; $\kappa \mathrm{/}r\to \mathrm{\infty }$ and $J^2\mathrm{/}(2\mu r^2)\to \mathrm{\infty }$ as $r\downarrow 0$, while both terms tend to $0$ as $r\to \mathrm{\infty }$. Hence $V$ is a continuous strictly decreasing bijection $(0,\mathrm{\infty })\to (0,\mathrm{\infty })$; since $E>0$ lies in the range, $r_{\ast }=V^{-1}(E)$ is the unique solution of $V=E$, it is simple because $V^{\mathrm{\prime }}<0$ everywhere, and $V(r)<E$ for $r>r_{\ast }$. This is clause 1.

Global continuation: energy and angular momentum are conserved by [L1], and $V(\rho )\le E$ forces $\rho \ge r_{\ast }$ since $V$ is strictly decreasing. Also $\mathrm{\mid }{r}^{\mathrm{\prime }}\mathrm{\mid }\le c:=\sqrt{2E\mathrm{/}\mu }$ because $U>0$. At any finite endpoint, $\mathrm{\mid }r\mathrm{\mid }$ is bounded above by its value at a fixed time plus $c$ times the elapsed time, and $r$ stays at least $r_{\ast }$ from the origin. Thus the Cartesian state $(r,r^{\mathrm{\prime }})$ stays in a compact subset of $\{(x,v):\mathrm{\mid }x\mathrm{\mid }>0\}$ near that endpoint. The Cartesian field $(v,\kappa x\mathrm{/}(\mu \mathrm{\mid }x{\mathrm{\mid }}^3))$ is smooth there; compact continuation [L3] extends the solution past every finite right endpoint. For a finite left endpoint apply the same argument to the time-reversed state $(r(-s),-r^{\mathrm{\prime }}(-s))$, which solves the same autonomous Cartesian system; uniqueness [L3] then makes the global extension unique. The maximal extension therefore has domain $\mathbb{R}$.

Branch shape: ${\rho }^{\mathrm{\prime }\mathrm{\prime }}=-V^{\mathrm{\prime }}(\rho )\mathrm{/}\mu >0$ for all times by step 1.1. Hence ${\rho }^{\mathrm{\prime }}$ is strictly increasing, vanishes at $t_{\ast }$ by the energy identity, is negative before $t_{\ast }$ and positive after it. Consequently the apsis is unique and $\rho >r_{\ast }$ elsewhere. Fixing any $t_1>t_{\ast }$ gives ${\rho }^{\mathrm{\prime }}(t)\ge {\rho }^{\mathrm{\prime }}(t_1)>0$ for $t\ge t_1$, so $\rho \to \mathrm{\infty }$ forward; the same argument backward gives $\rho \to \mathrm{\infty }$ there. On either open branch, compact substitution [L4] and then limits give $\mathrm{\mid }t-t_{\ast }\mathrm{\mid }={\int }_{r_{\ast }}^{\rho (t)}ds\mathrm{/}\sqrt{(2\mathrm{/}\mu )(E-V(s))}$. Its lower endpoint is integrable since $E-V(s)$ is comparable to $s-r_{\ast }$ by $V^{\mathrm{\prime }}(r_{\ast })<0$; its upper endpoint diverges since the integrand tends to $1\mathrm{/}c>0$. This proves the branch and quadrature claims without a finite-range hypothesis.

Asymptotic velocities and their speed: since $V(\rho (t))\to 0$ along each branch (as $\rho (t)\to \mathrm{\infty }$), $\mathrm{\mid }{\rho }^{\mathrm{\prime }}\mathrm{\mid }\to \sqrt{2E\mathrm{/}\mu }$. Put $c:=\sqrt{2E\mathrm{/}\mu }$ and $c_0:=c\mathrm{/}2>0$. The radial-speed limit implies $\mathrm{\mid }{\rho }^{\mathrm{\prime }}(t)\mathrm{\mid }\ge c_0$ for sufficiently large $\mathrm{\mid }t\mathrm{\mid }$; integrating on each monotone branch gives $\rho (t)\ge c_0\mathrm{\mid }t\mathrm{\mid }-C_0$ for a finite constant $C_0$. In particular $\rho (t)\ge (c_0\mathrm{/}2)\mathrm{\mid }t\mathrm{\mid }$ after increasing the tail cutoff. The acceleration of the Cartesian curve is $\mu r^{\mathrm{\prime }\mathrm{\prime }}=-(U^{\mathrm{\prime }}(\rho )\mathrm{/}\rho )r$, so $\parallel r^{\mathrm{\prime }\mathrm{\prime }}{\parallel }_2=\mathrm{\mid }{U}^{\mathrm{\prime }}(\rho )\mathrm{\mid }\mathrm{/}\mu =\kappa \mathrm{/}(\mu {\rho }^2)\le C\mathrm{/}(1+\mathrm{\mid }t\mathrm{\mid }{)}^2$ for large $\mathrm{\mid }t\mathrm{\mid }$, an integrable bound along each half-line; the Cauchy criterion for the vector integral (the integral of $r^{\mathrm{\prime }\mathrm{\prime }}$ over $[s,t]$ is bounded by ${\int }_s^t\parallel r^{\mathrm{\prime }\mathrm{\prime }}{\parallel }_2$) therefore gives existence of $v_{\pm }={\lim }_{t\to \pm \mathrm{\infty }}{r}^{\mathrm{\prime }}(t)$. Passing to the limit in the energy identity $E=\frac{1}{2}\mu \parallel r^{\mathrm{\prime }}{\parallel }_2^2+U(\rho (t))$ with $U(\rho (t))\to 0$ gives $\mathrm{\mid }{v}_{\pm }\mathrm{\mid }=\sqrt{2E\mathrm{/}\mu }$.

Half-angle quadrature: on compact subintervals of either open branch use ${\theta }^{\mathrm{\prime }}=J\mathrm{/}(\mu {\rho }^2)$ and $\mathrm{\mid }{\rho }^{\mathrm{\prime }}\mathrm{\mid }=\sqrt{(2\mathrm{/}\mu )(E-V)}$ in [L4]. Taking limits gives the same positive half-angle on both sides of the apsis, $\mathrm{\Delta }={\int }_{r_{\ast }}^{\mathrm{\infty }}\frac{Jds}{s^2\sqrt{2\mu (E-V(s))}}={\int }_0^{u_{\ast }}\frac{bdu}{\sqrt{1-(\kappa \mathrm{/}E)u-b^2{u}^2}}\mathrm{.}$ Each substitution is first on an interior compact interval. The integral converges at $r_{\ast }$ by the square-root bound of step 3.1 and at infinity by the bound $C\mathrm{/}{s}^2$, so it gives finite limits ${\theta }_{\pm }$ with ${\theta }_{+}-{\theta }_{\ast }={\theta }_{\ast }-{\theta }_{-}=\mathrm{\Delta }$.

Evaluation with scaled roots: put $w=bu$, $A=\sqrt{1+{\alpha }^2\mathrm{/}4}$ and $w_{\ast }=(\sqrt{{\alpha }^2+4}-\alpha )\mathrm{/}2=b{u}_{\ast }$, where $\alpha =\kappa \mathrm{/}(Eb)$. Then $1-\alpha w-w^2=A^2-(w+\alpha \mathrm{/}2{)}^2$. Substitute $w+\alpha \mathrm{/}2=A\sin \psi$ on interior compact intervals, where $\psi \in ({\psi }_0,\pi \mathrm{/}2)$ and ${\psi }_0=\arcsin (\alpha \mathrm{/}(2A))\in (0,\pi \mathrm{/}2)$. The derivative is $A\cos \psi >0$ and the positive denominator is $A\cos \psi$, so the integral equals $\pi \mathrm{/}2-{\psi }_0$ after taking limits. Since $\sin {\psi }_0=\alpha \mathrm{/}\sqrt{{\alpha }^2+4}$ and $\cos {\psi }_0=2\mathrm{/}\sqrt{{\alpha }^2+4}$, $\tan {\psi }_0=\alpha \mathrm{/}2$, giving ${\psi }_0=\arctan (\alpha \mathrm{/}2)$ by [L5]. Hence $\mathrm{\Delta }=\pi \mathrm{/}2-\arctan (\kappa \mathrm{/}(2Eb))\in (0,\pi \mathrm{/}2)$. This proves clause 3.

Direction limits: differentiating $r=\rho n(\theta )$ gives $r^{\mathrm{\prime }}={\rho }^{\mathrm{\prime }}n(\theta )+(J\mathrm{/}(\mu \rho ))n_{\perp }(\theta )$, where $n_{\perp }(\theta )=-\sin \theta e_1+\cos \theta e_2$. By step 4.1, ${\rho }^{\mathrm{\prime }}\to c$ forward and ${\rho }^{\mathrm{\prime }}\to -c$ backward, while the tangential term tends to zero. Step 4.2 therefore gives $v_{+}=cn({\theta }_{+})$ and $v_{-}=-cn({\theta }_{-})$. The addition formulas [L5] and ${\theta }_{+}-{\theta }_{-}=2\mathrm{\Delta }$ show the velocity rotation $2\mathrm{\Delta }-\pi$ and the displayed cosine. There is no hidden extra winding: the radial coefficient ${\rho }^{\mathrm{\prime }}$ changes sign only once, the tangential coefficient is always positive, so the velocity direction relative to $n(\theta )$ varies continuously from $\pi$ to $0$ through the upper half-plane; its net change is $2\mathrm{\Delta }-\pi$.

Impact limit: put $q(s)=\sqrt{(2\mathrm{/}\mu )(E-V(s))}$ for $s>r_{\ast }$. On the outgoing branch the remaining angle is $\mathrm{\Psi }(t)={\int }_{\rho (t)}^{\mathrm{\infty }}J\mathrm{/}(\mu s^{2}q(s))ds$, so $\rho (t)\mathrm{\Psi }(t)=\frac{J}{\mu }{\int }_1^{\mathrm{\infty }}\frac{du}{u^{2}q(\rho (t)u)}\mathrm{.}$ As $q(s)\to c>0$, for any $\epsilon >0$ one has $\mathrm{\mid }1\mathrm{/}q(s)-1\mathrm{/}c\mathrm{\mid }<\epsilon$ for every $s\ge S$ for some $S$. Once $\rho (t)\ge S$, the difference between the integral above and $1\mathrm{/}c$ is at most $\epsilon {\int }_1^{\mathrm{\infty }}{u}^{-2}du=\epsilon$. Thus $\rho \mathrm{\Psi }\to J\mathrm{/}(\mu c)$ and $\mathrm{\Psi }\to 0$; the elementary limit $\sin \mathrm{\Psi }\mathrm{/}\mathrm{\Psi }\to 1$ gives $\rho \sin \mathrm{\Psi }\to J\mathrm{/}(\mu c)=b$. By step 6.1 this is exactly $\mathrm{\mid }r(t)\times v_{+}\mathrm{\mid }\mathrm{/}\mathrm{\mid }{v}_{+}\mathrm{\mid }$. The incoming computation uses $q=\mathrm{\mid }{\rho }^{\mathrm{\prime }}\mathrm{\mid }$ and the remaining angle $\theta (t)-{\theta }_{-}$ and gives $\mathrm{\mid }r(t)\times v_{-}\mathrm{\mid }\mathrm{/}\mathrm{\mid }{v}_{-}\mathrm{\mid }\to b$ as well.

For every $b>0$ set $J(b)=\sqrt{2\mu E}b$ and take the turning radius of clause 1. The initial data stated in clause 4 have angular momentum $J(b)e_3$ and energy $V(r_{\ast })=E$, radial derivative zero and positive radial second derivative. Local existence [L3] supplies a trajectory and conservation [L1] forces $\rho \ge r_{\ast }$; the preceding continuation, angle and impact-limit steps apply to it. Uniqueness [L3] determines it from its apsis state. Thus this actual family realises the deflection formula for all $b>0$, and all clauses are proved.