Adventures in Parallel Dimension #1729 III

For this, we will use the energy of an object in free fall, which is

$E=mc^{2}\frac{dt}{d \tau} \left( 1-\frac{r_{s}}{R} \right)$

where $m$ is the rest mass, $c$ is the speed of light, $\frac{dt}{d \tau}$ is the ratio of how much time passes on a clock at $r=\infty$ to a clock at $r=R$ according to the Schwarzschild Metric.

Note that this equation takes into account special relativistic effects such as time dilation.

Since this definition of energy is conserved during free fall:

$mc^{2}\frac{dt}{dt_{ship}}\left(1-\frac{r_{s}}{R}\right)=mc^{2}\frac{dt}{d \tau} \left( 1-\frac{r_{s}}{r_{clock}} \right)$

where $d \tau$ is the rate at which the falling clock is ticking and $R$ is the coordinate radius of the parked ship.

We can then use the fact that $\frac{dt_{ship}}{d \tau}$ was given to be $2$ in the problem the moment when $r_{clock}=\frac{R}{2}$ . Thus:

$\frac{dt}{d \tau} = \frac{dt}{dt_{ship}} \times \frac{dt_{ship}}{d \tau} = 2 \frac{dt}{dt_{ship}}$

$\Rightarrow mc^{2}\frac{dt}{dt_{ship}} \left(1-\frac{r_{s}}{R} \right) = 2mc^{2} \frac{dt}{dt_{ship}} \left(1-\frac{r_{s}}{\frac{R}{2}} \right)$

After some nifty cancellation:

$1-\frac{r_{s}}{R} = 2 - \frac{4r_{s}}{R} \Rightarrow R=5r_{s} \Rightarrow \boxed{\frac{R}{r_{s}}=5}$