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Put $u=1/r$. Since $h=r^2\dot{\theta }$, the radial equation becomes the Binet equationwhere primes denote derivatives with respect to the polar angle. Choosing the angular origin at periapsis givesWriting the semi-latus rectum as $h^2/\mu =a(1-e^2)$ yields the Kepler orbitHence the specific angular momentum and specific orbital energy are

In axes directed toward periapsis and along the motion there, the position is $\mathbf{r}=r(\cos f,\sin f)$. Differentiating it and using $\dot{f}=h/r^2$ givesDifferentiating the orbit equation gives $\dot{r}=(\mu e/h)\sin f$, and substitution simplifies the components to

An orbit is unbound precisely when its specific orbital energy is nonnegative, equivalently $a/a^{\prime }\le 0$. At periapsis, $f=0$, and for $e=1/2$ the condition from part (d) isThe directions are sampled from the uniform distribution on a circle. The fraction satisfying $\cos \varphi \ge x$ is $\arccos x/\pi$; setting it equal to $1/6$ gives $x=\sqrt{3}/2$. ThereforeThe positive root is

The specific angular momentum of an elliptic Kepler orbit is $h=\sqrt{\mu a(1-e^2)}$. At periapsis, $r=a(1-e)$, soThe mean motion understates instantaneous angular speed near periapsis. A particle in an exterior mean-motion resonance can briefly overtake the planet in angle despite completing full orbits more slowly.

A passing star perturbs a planetary or cometary system through its differential Newtonian gravitational field. A distant encounter is predominantly tidal; a sufficiently close one can substantially change specific orbital energy and angular momentum, eject a comet, or shift its periapsis into or out of the planetary region.