 Solution

Treat Earth as a stationary spherical source, with $\mu =G{M}_E$, and neglect drag and other bodies. The conserved quantities are total mechanical energy and the angular momentum vector, implying a fixed orbital plane and constant areal velocity. The Kepler problem additionally conserves the Laplace-Runge-Lenz vector. The orbiting object's linear momentum is not constant because gravity exerts a force.

For a circular orbit, $m{v}_c^2/r_0=\mu m/r_0^2$, so $K=\mu m/(2r_0)=-U/2$. HenceAt the closest approach of a noncircular orbit, the velocity is tangential and $v_0=k\sqrt{\mu /r_0}$. Its energy and specific angular momentum areThe pericentre-speed qualification of Kepler orbit classes requires $k\ge 1$ when $r_0$ is genuinely the closest approach. For an ellipse, conservation of h gives $v_m=v_0{r}_0/r_m$. Equating energies at the two turning points and writing $x=r_m/r_0$ givesThe distinct apocentre root is thereforeFor an unbound orbit, $v^2(r)=2E/m+2\mu /r$ decreases toward infinity. The asymptotic minimum speed of an unbound Kepler orbit isIt is the limiting infimum, reached only asymptotically: positive for a hyperbola and zero for a parabola. The turning-point interpretation of $r_0$ is essential; if $0<k<1$, the same instantaneous energy describes a bound orbit whose $r_0$ is the apocentre instead. ^[ref]