From many other questions posted here and in other forums about orbits around black holes, I arrived at a contradiction. From these parameters:
$$a = 1.00 - 1.33\times10^{-14}$$
$$M = 10^8 M_\odot$$
$$r = 1.00003763264895$$
I calculated the orbital period seen by a distant stationary observer ($P$), the orbital period measured locally ($τ$) and the orbital velocity measured locally ($v_{local}$). All of this was possible thanks to these great responses I linked.
$$P=12,379\:s$$ $$τ=0.2\:s$$
These results make sense cause $P/τ=dt/dτ=61,368$; which is the correct time dilation. But now, the local velocity (ZAMO's frame) is:
$$v_{local}=0.5\:c=149,900\:km/s$$
The contradiction is that when I apply the time dilation factor to the local velocity to calculate the speed seen by a distant stationary observer, the result is too small. Too little to make sense with the orbital period ($P$) seen by that distant observer.
$$ ???\:v=2.86\:km/s\:??? $$
One way to solve this is to calculate it simply by dividing the circumference of the orbit ($2πr$) by the orbital period ($P$). This was suggested by @Yukterez here.
$$v=\frac{2πrGM}{Pc^2}=149,897\:km/s$$
The curious thing is that if I plot this function in a velocity-r graph, local velocity and velocity seen by distant observer converge at 0.5c in an extremal rotating black hole. This solves the contradiction.
Plot of the orbital velocities when a=1. I added the escape local and distant-observed escape velocities in red.
Another clue as to why I think this is correct is that when a=0, the plot matches exactly this plot made in the Schwarzschild metric.
Is this how the orbital velocity around a rotating black hole seen by a distant observer is calculated?
