Past exam of the general relativity and cosmology course of the University of Oxford / 2018 / 1 / e / Solution

For radial motion outside the Schwarzschild event horizon, the equations needed to determine the acceleration components areThey are respectively the four-velocity norm, velocity-acceleration orthogonality, and the maximum proper acceleration norm. Together with part d they give the stated thrust equation; no integration of the trajectory is required. The positive radial sign corresponds to outward thrust.

Let $r_I=\sqrt{GM/g}$. A first transition from inward to outward motion requires $u^r=0$ at a minimum radius and $d{u}^r/d\tau \ge 0$ there. The largest possible radial derivative at such an event isAt $r=r_I>r_s$, this isThus the ship cannot reverse its inward motion there. It cannot reverse at any smaller exterior radius either: $GM/r^2$ grows while $\sqrt{1-r_s/r}$ decreases. If it first reaches $r_I$ while falling inward, it cannot return home using the stipulated bounded radial thrust. If $r_I\le r_s$, it would instead have reached the black-hole horizon before that radius, which already prevents escape.

The minimum radius for radial turnaround with bounded proper acceleration lies strictly outside the proposed radius. Define $r_c>r_s$ byThe right side is the proper acceleration required to hover, decreases monotonically from infinity to zero, and exceeds $GM/r^2$. A genuine turnaround needs $r_{\mathrm{turn}}>r_c>r_I$; equality gives no positive initial outward acceleration at rest. A sufficiently early burn is also needed to remove the pre-existing inward velocity. The failure at $r_I$ follows from the turning-point condition, not from treating the moving ship as static throughout its fall.