When an object is in free fall from rest at an infinite radius, is its potential energy proportional to its clock rate? The total energy $E$ of an object in free fall from rest at an infinite radius is known to be
$$
E = m c^2
$$
As the object falls, its energy doesn't change since the object is merely moving freely through curved spacetime.
For a distant observer, is $E$ the sum of the potential energy $U$ and the kinetic energy $E_k$ such that
$$
U = \left(1 - \frac{r_0}{r}\right) mc^2
$$
and 
$$
E_k = \left(\frac{r_0}{r}\right) m c^2
$$
?
 A: I'll assume that the energy is measured by a stationary observer ($dr=0$).
Taking the spherically symmetric Schwarzschild metric as an example and radial infall from infinity at rest, then we know that.
$$\left(1 - \frac{r_s}{r}\right) \frac{dt}{d\tau} = 1$$
So this expression gives the time dilation inferred by an observer at infinity for an infalling object, which I assume is what you mean by the clock rate - the proper time interval  divided by the coordinate time interval.
$$ d\tau = dt (1-r_s/r)$$
This leads to expressions for $dr/dt$, a velocity in coordinate space time
$$\frac{dr}{dt}=-\left(1 - \frac{r}{r_s}\right)\left(\frac{r_s}{r}\right)^{1/2}$$
and a velocity according to a stationary observer at $r$
$$v = -\left(\frac{r_s}{r}\right)^{1/2}$$
Where, in each case, $r_s$ is the Schwarzschild radius and $c=1$ units are used.
The kinetic energy measured by the stationary observer is $(\gamma -1)mc^2$ and thus
$$ E_K = \left(\frac{1}{(1 - r_s/r)^{1/2}} -1\right)mc^2 $$
According to coordinate space time, the equivalent kinetic energy expression would be
$$E_K = \left(\frac{1}{[1 - (1   - r_s/r)^2 (r_s/r)]^{1/2}}-1\right ) mc^2$$
which also doesn't simplify to what you suggest, since you see that this approaches zero at the Schwarzschild radius.
I think that in GR, this simple separation between potential and kinetic energy is not possible. 
