= Solution
Let $h=J/m=bv_0$ be the <specific angular momentum>, and measure the orbital angle $\vartheta$ from <periapsis>. Set $w=1/r$. From $\dot\vartheta=h/r^2$, differentiating gives $\dot r=-h\,dw/d\vartheta$. The radial <equation of motion> $\ddot r-r\dot\vartheta^2=-\mu_J/r^2$ becomes the <Binet equation>
$$
\frac{d^2w}{d\vartheta^2}+w=\frac{\mu_J}{h^2}.
$$
Its solution symmetric about <periapsis> is
$$
w=\frac{\mu_J}{h^2}(1+e\cos\vartheta),\qquad
e=\sqrt{1+\frac{v_0^2h^2}{\mu_J^2}}
=\sqrt{1+\frac{b^2v_0^4}{\mu_J^2}}.
$$
For example, the expression for the <orbital eccentricity> follows by substituting $w$ and $\dot r=-hw'$ into the <energy> equation: $E/m=\mu_J^2(e^2-1)/(2h^2)=v_0^2/2$. Since $e>1$, this is a <hyperbolic Kepler orbit>.
The outgoing asymptote occurs at $\vartheta_\infty$ with $\cos\vartheta_\infty=-1/e$, where $\pi/2<\vartheta_\infty<\pi$. The incoming <velocity> points inward along the incoming radial asymptote, whereas the outgoing <velocity> points outward. Therefore the change in <velocity> direction is $\phi=2\vartheta_\infty-\pi$, not $2\vartheta_\infty$. It follows that
$$
\sin\frac\phi2=\frac1e,\qquad
\boxed{\cot\frac\phi2=\sqrt{e^2-1}=\frac{bv_0^2}{GM_J}.}
$$
For $b>0$, this gives $0<\phi<\pi$: a large <impact parameter> produces a small deflection, and the ideal point-mass limit $b\to0^+$ approaches a reversal.
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