= Solution
Put $\mu=GM$. At the turning point of a radial launch the <speed> vanishes, so <conservation of energy> gives
$$
\frac12v_0^2-\frac\mu R=-\frac\mu{r_0},\qquad
\boxed{r_0=\left(\frac1R-\frac{v_0^2}{2GM}\right)^{-1}=\frac{R}{1-v_0^2/v_{\rm esc}^2},\quad v_{\rm esc}=\sqrt{\frac{2GM}{R}}.}
$$
This result is exact within the exterior Newtonian <gravitational potential>. The condition $r_0\gg R$ means $v_0$ is just below the <escape velocity>: $1-v_0^2/v_{\rm esc}^2\ll1$. Replacing gravity by a constant would be inappropriate in this regime.
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