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At the encounter radius $r=a_p$, the planet's circular Kepler orbit has speedThe comet's speed follows from the vis-viva equation:Its component parallel to the planet's prograde tangential velocity is fixed by its specific angular momentum,The relative velocity therefore satisfiesand hence

Let $h=r^2\dot{f}$ be the specific angular momentum. The radial Kepler orbit equation has semi-latus rectum $p=h^2/\mu$, so comparison withgives $h^2=\mu a(1-e^2)$ for an elliptic orbit and $h^2=\mu a(e^2-1)$ for a hyperbolic Kepler orbit. At either apsis, $\dot{r}=0$ and $v=h/r$. Substituting the apsidal radius and angular momentum into part (i), or equivalently using the vis-viva equation, givesThus the signs are $C=\mp \mu /(2a)$ in the convention of the question.

The limiting trajectory grazes the planet at periapsis $R_p$. Equating the asymptotic and periapsis values of specific angular momentum and using the vis-viva equation givesHence the comet avoids the planet when

For a Kepler orbit with semi-major axis $a$, the vis-viva equation isFor the convention that a hyperbolic Kepler orbit has positive semi-major-axis magnitude $a$, its final term instead has a plus sign.