= Solution
The <Lagrangian> has no explicit <time> dependence, so its <energy function> is constant. With total <speed> $v^2=\dot r^2+r^2\dot\phi^2$, division by $M_p$ gives
$$
\epsilon=\frac12v^2-\frac\mu r=\text{constant}.
$$
Consequently, between the initial point $(r_i,v_i)$ and any other point,
$$
\boxed{v^2(r)=v_i^2+2GM_*\left(\frac1r-\frac1{r_i}\right).}
$$
This energy relation holds for nonradial motion too. Radially outward motion is the stipulated approximation for the subsequent probe calculation; <angular momentum> is then zero.
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