 Solution

The time-translation Killing vector field of the Schwarzschild metric gives a conserved Killing energy per unit rest mass,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 $dt/d\tau =f(r_0{)}^{-1/2}$ andSubstitution into the radial four-velocity normalization yieldsChoosing the inward square root and evaluating at the ground provesThis 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 $v_{\mathrm{loc}}=(dr/dt)/f=(dr/d\tau )/\sqrt{f(r_0)}$. Thus the distinction disappears in the stated weak-field approximation, giving $v_{\mathrm{loc}}^2\simeq 2gh$ for $h\ll R_E$.