 Solution

Let $h=J/m=b{v}_0$ 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 $\dot{r}=-hdw/d\vartheta$. The radial equation of motion $\ddot{r}-r{\dot{\vartheta }}^2=-{\mu }_J/r^2$ becomes the Binet equationIts solution symmetric about periapsis isFor example, the expression for the orbital eccentricity follows by substituting $w$ and $\dot{r}=-h{w}^{\prime }$ 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 $\varphi =2{\vartheta }_{\infty }-\pi$, not $2{\vartheta }_{\infty }$. It follows thatFor $b>0$, this gives $0<\varphi <\pi$: a large impact parameter produces a small deflection, and the ideal point-mass limit $b\to 0^{+}$ approaches a reversal.