= Solution
For two isolated <point masses>, introduce total <mass> $M_T=M+m$, <reduced mass> $\mu=Mm/M_T$ and relative separation $r$. The relative <acceleration> is $\ddot{\mathbf r}=-GM_T\mathbf r/r^3$. Conserved <specific angular momentum> $h=|\mathbf r\times\dot{\mathbf r}|$ fixes a plane and a constant areal <velocity> $h/2$. The <specific orbital energy> is
$$
\varepsilon=\frac{\dot r^2}{2}+\frac{h^2}{2r^2}-\frac{GM_T}{r}.
$$
For $h\ne0$, the trajectory is a conic:
$$
\boxed{r(\varphi)=\frac{h^2/(GM_T)}{1+e\cos(\varphi-\varphi_0)},\qquad
e^2=1+\frac{2\varepsilon h^2}{G^2M_T^2}.}
$$
Negative <energy> gives an <ellipse>, including the circular limit $e=0$; zero <energy> gives a <parabola>; positive <energy> gives a <hyperbola>. Bound <ellipses> obey $\varepsilon=-GM_T/(2a_o)$ and
$$
T^2=\frac{4\pi^2a_o^3}{GM_T}.
$$
Thus <Kepler's laws of planetary motion> follow from <angular momentum conservation> and the <inverse-square force>. For a <circular orbit> of <radius> $r$, $v_c=\sqrt{GM_T/r}$, while escape from that <radius> requires $v_{\rm esc}=\sqrt{2GM_T/r}$. \b[<Energy> determines whether the orbit is bound; <angular momentum> also determines its shape.] These ideal results neglect extended-body perturbations, drag and relativistic corrections.
Back to article page
