= Solution
In the planetocentric <hyperbolic Kepler orbit>, the speed at infinity is $\Delta v_{pc}$ and the <impact parameter> is $b$. Conservation of <specific angular momentum> gives
$$
h=b\Delta v_{pc}=a_mv_{c,t},
\qquad
\boxed{v_{c,t}=\frac{b\Delta v_{pc}}{a_m}}.
$$
Conservation of <specific orbital energy> at the moon's radius gives
$$
v_{c,r}^2+v_{c,t}^2
=\Delta v_{pc}^2+\frac{2GM_p}{a_m}
=\Delta v_{pc}^2+2v_m^2,
$$
where $v_m^2=GM_p/a_m$. Thus
$$
\boxed{v_{c,r}=\pm\left[\Delta v_{pc}^2+2v_m^2
-\left(\frac{b\Delta v_{pc}}{a_m}\right)^2\right]^{1/2}}.
$$
The moon moves tangentially at $v_m$, so
$$
\Delta v_{mc}^2=v_{c,r}^2+(v_{c,t}-v_m)^2
=3v_m^2-2v_m\Delta v_{pc}\frac b{a_m}+\Delta v_{pc}^2.
$$
Therefore the correctly dimensioned reading of the displayed result is
$$
\boxed{\Delta v_{mc}=\left(3v_m^2-2v_m\Delta v_{pc}\frac b{a_m}
+\Delta v_{pc}^2\right)^{1/2}}.
$$
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