= Angular momentum and the escape energy threshold
{title2=$\varepsilon=\tfrac12\dot r^2+\tfrac{h^2}{2r^2}-\tfrac\mu r$}
In the isolated model, <conservation of energy> and <angular momentum conservation> reduce the exterior <test particle>'s motion to
$$
\varepsilon=\frac{E}{m}=\frac{\dot r^2}{2}+\frac{h^2}{2r^2}-\frac{\mu}{r},\qquad h=|\mathbf r\times\mathbf v|,\qquad \mu=GM.
$$
Here $\varepsilon$ is <specific orbital energy> and $h$ is <specific angular momentum>. The <effective potential> is $\Phi_{\mathrm{eff}}=h^2/(2r^2)-\mu/r$. A zero radial <velocity> is a radial <turning point>, and need not mean zero total <speed>: at the <periapsis> or <apoapsis> of a nonradial <Kepler orbit>, tangential <velocity> remains. Thus the vertical gap between $\varepsilon$ and $-\mu/r$ represents total <kinetic energy> per unit <mass>, whereas the gap above $\Phi_{\mathrm{eff}}$ represents its radial part.
For $h>0$, the <orbital eccentricity> obeys $e^2=1+2\varepsilon h^2/\mu^2$. Negative $\varepsilon$ gives an <elliptic orbit>, including a <circular orbit> at its minimum allowed <energy>; zero $\varepsilon$ gives a <parabolic trajectory>; positive $\varepsilon$ gives a <hyperbolic Kepler orbit>. The <effective potential> tends to zero from below at infinity, so its outer region creates no extra positive escape barrier. This explains why <escape velocity> depends on <energy> even though <angular momentum> changes the path. The $h=0$ case is radial motion, rather than a nondegenerate <parabola> or <hyperbola>; inward radial motion can end in collision.
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