Schwarzschild metric, speed of ball as measured by observer who catches the ball, just before ball is caught? The Schwarzschild metric, describing the exterior gravitational field of a planet of mass $M$ and radius $R$, is given by$$ds^2 = -(1 - 2M/r)\,dt^2 + (1 - 2M/r)^{-1}\,dr^2 + r^2(d\theta^2 + \sin^2\theta \,d\phi^2).$$A tower has its base on the surface of this planet ($r = R$) and its top at radial coordinate $r = R_1$. A ball is held at rest by an observer at the top of the tower. It is then dropped and caught by an observer at the bottom of the tower.
What is the speed, $v$, of the ball as measured by the observer who catches the ball, just before the ball is caught?
Here, we are not assuming that $R \gg 2M$ or that $R_1 - R \ll R$. Also, I want the physical speed here, $v$, as would be measured, e.g. by a radar gun, not a coordinate speed, such as $dr/dt$.

Since the Schwarzschild metric is invariant under time translations $t \mapsto t + \Delta t$, we get a Killing field $\partial_t$ and thus we know $g(V, \partial_t)$ is conserved along a geodesic with velocity field $V$. This quantity can be written
$$ -Q = g(V, \partial_t) = V^i g_{it} = -(1-2M/r) V^t,$$
and can be thought of as an energy of some kind. We can use this conservation law at the top and bottom of the tower to determine the velocity. Since the ball is at rest in the Earth's frame when it is dropped, the initial velocity is purely in the time direction; so if we choose our geodesic to be parametrised such that $g(V,V) = -1$ then we get $V^t=1$ at the initial time, and of course $r=R_1$ there; so 
$$ Q = (1-2M/R_1).$$
Thus at the bottom of the tower we must have $$Q = (1-2M/R)V^t = (1-2M/R_1),$$ so an observer in the Earth's frame sees $$\gamma = V^t = \frac{1 - 2M/R_1}{1-2M/R}.$$ Since $\gamma = 1/\sqrt{1-v^2}$, the observed speed is 
$$ v=\sqrt{1- \frac 1 {\gamma^2}} = \sqrt{1-\left(\frac{1-2M/R}{1-2M/R_1}\right)^2}$$
times the speed of light.
