= Solution
{c}
The time-translation <Killing vector field> of the <Schwarzschild metric> gives a conserved <Killing energy> per unit <rest mass>,
$$
\mathcal E=-g_{tt}\frac{dt}{d\tau}=c^2f(r)\frac{dt}{d\tau}.
$$
For unit <rest mass>, this is the total <energy> relative to infinity, including <rest energy>. At release $dr/d\tau=0$, so the <four-velocity> normalization gives $dt/d\tau=f(r_0)^{-1/2}$ and
$$
\boxed{\mathcal E=c^2\sqrt{1-\frac{2GM}{r_0c^2}}
=c^2-\frac{GM}{r_0}+O\left(\frac{G^2M^2}{r_0^2c^2}\right).}
$$
Substitution into the radial <four-velocity> normalization yields
$$
\left(\frac{dr}{d\tau}\right)^2=\frac{\mathcal E^2}{c^2}-c^2f(r)
=2GM\left(\frac1r-\frac1{r_0}\right).
$$
Choosing the inward square root and evaluating at the ground proves
$$
\boxed{\frac12\left(\frac{dr}{d\tau}\right)^2_{r=R_E}
=GM\left(\frac1{R_E}-\frac1{R_E+h}\right).}
$$
This identity is exact within the assumed <Schwarzschild metric>, despite its familiar <Newtonian mechanics> form. The quantity $dr/d\tau$ is not exactly the locally measured <speed>. A <Schwarzschild static observer> measures $v_{\rm loc}=(dr/dt)/f=(dr/d\tau)/\sqrt{f(r_0)}$. Thus the distinction disappears in the stated <weak-field approximation>, giving $v_{\rm loc}^2\simeq2gh$ for $h\ll R_E$.
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