Schwarzschild static observer

An observer at fixed exterior Schwarzschild spatial coordinates. Its proper time is $dT=\sqrt{f}dt$, with $f=1-2GM/(r{c}^2)>0$. It is accelerated, not freely falling; no such static timelike worldline exists on or inside the event horizon.

For radial motion in Schwarzschild spacetime, $a^r=d{u}^r/d\tau +GM/r^2$. At a prospective radial turning point $u^r=0$, bounded outward proper acceleration $g$ permits $d{u}^r/d\tau \le g\sqrt{1-2GM/(c^{2}r)}-GM/r^2$. The equality $g=g_{\mathrm{hover}}$ defines a unique critical radius outside the horizon. A ship already falling inward cannot turn at or below it. The Newtonian proposed radius $\sqrt{GM/g}$ is strictly smaller and is therefore too late for reversal. A radius beyond the critical one is necessary, but braking must also remove the pre-existing inward velocity.

For a radial timelike geodesic of specific Killing energy $k=fdt/d\tau$, the static observer measures $v_{\mathrm{loc}}=(dr/dt)/f=\dot{r}/k$. Its relative Lorentz factor is $k/\sqrt{f}$, giving $v_{\mathrm{loc}}^2=c^2(1-f/k^2)$.

For asymptotic speed $0\le u<c$, $k=(1-u^2/c^2{)}^{-1/2}$ and the local outward speed obeys $v_{\mathrm{loc}}^2=u^2+(2GM/r)(1-u^2/c^2)$. It tends to $c$ as the horizon is approached from outside, though a static observer cannot actually remain on the horizon.

The outgoing radial timelike geodesic with zero speed at infinity has $k=1$ and $dr/d\tau =\sqrt{2GM/r}$. Thus $r(\tau )=[r_0^{3/2}+(3/2)\sqrt{2GM}\tau {]}^{2/3}$ and the local static-observer speed is $\sqrt{2GM/r}$.