 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 $\ddot{\mathbf{r}}=-G{M}_T\mathbf{r}/r^3$. 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 isFor $h\ne 0$, the trajectory is a conic:Negative energy gives an ellipse, including the circular limit $e=0$; zero energy gives a parabola; positive energy gives a hyperbola. Bound ellipses obey $\epsilon =-G{M}_T/(2a_o)$ andThus Kepler's laws of planetary motion follow from angular momentum conservation and the inverse-square force. For a circular orbit of radius $r$, $v_c=\sqrt{G{M}_T/r}$, while escape from that radius requires $v_{\mathrm{esc}}=\sqrt{2G{M}_T/r}$. Energy determines whether the orbit is bound; angular momentum also determines its shape. These ideal results neglect extended-body perturbations, drag and relativistic corrections.