Velocity to Eccentricity

$v = \sqrt{\mu (\frac{1 + e}{r})}$

$u = \sqrt{\frac{\mu}{r}}$

Thus,

$v : u = \sqrt{1 + e} = \sqrt{1.44} = \boxed{1.2}$

Consider the effective potential to which the satellite is subject: $U_{\rm eff} = \frac{L^2}{2mx^2} - \frac{GMm}{x}\;,$ where $m$ is the satellite mass, $M$ is the Earth mass, $G$ is Newton's gravitational constant, $x$ is the distance from Earth and $L$ is the satellite angular momentum, which is a constant of the motion.

The minimum of this potential corresponds to the circular motion. Therefore, for a circular motion of radius $r$ we need an angular momentum: $L_{\rm circ}^2 = GMm^2r$

Negative values of $U_{\rm eff}$ correspond to an elliptic motion, with perigee and apogee given by the solutions of the following equation: $x^2 - \frac{GMm}{|E|}x + \frac{L^2}{2m|E|}=0\;,$ where $E<0$ is the total mechanical energy of the satellite. The solutions of the above equation are: $x_{\rm A,P} = \frac{GMm}{2|E|}\left(1 \pm \sqrt{1 - \frac{2L^2|E|}{G^2M^2m^3}}\right)\;.$ Now, since the apogee and eccentricity are given: $x_{\rm A} = r\;,$ and $e = \frac{x_{\rm A} - x_{\rm P}}{x_{\rm A} + x_{\rm P}} = 0.44\;,$ we can solve for the energy and the angular momentum, finding $|E| = \frac{0.72GMm}{r}\;,$ and $L_{\rm ell}^2=0.56GMm^2r\;.$ Since the angular momentum is conserved, then $\frac{v}{u} = \frac{L_{\rm ell}}{L_{\rm circ}} = \sqrt{0.56} \approx \boxed{0.748}$