= Solution
Treat Earth as a stationary spherical source, with $\mu=GM_E$, 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, $mv_c^2/r_0=\mu m/r_0^2$, so $K=\mu m/(2r_0)=-U/2$. Hence
$$
\boxed{E_0=K+U=-\frac{\mu m}{2r_0}.}
$$
At 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 are
$$
\boxed{E=\frac{\mu m}{2r_0}(k^2-2)=E_0(2-k^2),\qquad
h=r_0v_0=k\sqrt{\mu r_0}.}
$$
The <pericentre-speed qualification of Kepler orbit classes> requires $k\ge1$ when $r_0$ is genuinely the closest approach. For an ellipse, conservation of h gives $v_m=v_0r_0/r_m$. Equating energies at the two turning points and writing $x=r_m/r_0$ gives
$$
k^2-2=\frac{k^2}{x^2}-\frac2x,\qquad
(x-1)[(k^2-2)x+k^2]=0.
$$
The distinct apocentre root is therefore
$$
\boxed{\frac{r_m}{r_0}=\frac{k^2}{2-k^2},\qquad 1<k<\sqrt2.}
$$
For an unbound orbit, $v^2(r)=2E/m+2\mu/r$ decreases toward infinity. The <asymptotic minimum speed of an unbound Kepler orbit> is
$$
\boxed{v_\infty=\sqrt{\frac{\mu}{r_0}(k^2-2)}.}
$$
It 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. <University Physics Volume 1 by William Moebs and others>[Chapter13, gravitational energy, escape and Kepler orbits]{ref}
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