= Solution
For a tangential launch, conserved specific <angular momentum> is $j=Rv$. At the outer turning point the radial <velocity> vanishes but the tangential <speed> is $j/r_0$. Thus
$$
\frac12v^2-\frac\mu R=\frac{R^2v^2}{2r_0^2}-\frac\mu{r_0}.
$$
Using <10/a> eliminates the <gravitational potential> difference and gives
$$
\boxed{v=\frac{v_0}{\sqrt{1-R^2/r_0^2}}=v_0\left[1+\frac{R^2}{2r_0^2}+O\left(\frac{R^4}{r_0^4}\right)\right].}
$$
The expression printed in the source is the leading correction for $r_0\gg R$, rather than an exact identity. Equivalently, $v^2=2\mu r_0/[R(r_0+R)]$, so the launch <speed> exceeds the circular value and the surface is the <periapsis> of the nonintersecting ideal orbit.
Back to article page
