How does $g$-force relate to time dilation? In the imdb.com goofs page of the movie Interstellar, I found this statement:

How does time dilation affect $g$-force? If this could be explained in Laymen's terms it would be much appreciated.
 A: The radial coordinate for the innermost stable prograde orbit at which the planet could orbit the black hole would in natural units of $\rm G=M=c=1$ be at
$$\rm r = 3 + Z_2 - \sqrt{(3-Z_1)(3+Z_1+2Z_2)}$$
with the terms
$$\rm Z_1 = 1 + \sqrt[3]{1-a^2} \left( \sqrt[3]{1+a} + \sqrt[3]{1-a} \right) \ , \ \ Z_2 = \sqrt{3a^2 + Z_1^2}$$
For a black hole with a spin parameter of $\rm a=0.99999999999999$ which they used according to this source that would be
$$\rm r = 1.000034191427736$$
where the gravitational component of the time dilation is
$$\sqrt{g^{\rm tt}} = \rm \surd\left({\frac{4 \left(\left(a^2+r^2\right)^2-a^2 \left(a^2+(r-2) r\right) \sin ^2 \theta \right) \left(a^2 \cos ^2 \theta +r^2\right)}{\left(a^2+(r-2) r\right) \left(a^2 \cos (2 \theta )+a^2+2 r^2\right)^2}}\right)$$
which for our spin parameter and in the equatorial plane where $\theta=\pi/2$ gives
$$\sqrt{g^{\rm tt}} = 58494.69347667821$$
The prograde orbital velocity relative to a ZAMO is
$$\rm v_{+} = \frac{a^2-2 a \sqrt{r}+r^2}{\sqrt{a^2+(r-2) r} \left(a+r^{3/2}\right)}$$
which is in units of $c$
$$\rm v_{+}=0.500012818545588$$
so the total time dilation would be
$$\dot{\rm t}=\sqrt{\frac{g^{\rm tt}}{\rm 1-v_{+}^2}} = 67544.43127396276$$
that means you get $7.71$ years per hour, as it is claimed here. In the older version of the answer I wrote before I found out which spin parameter they used in they movie I went under the assumption that the black hole spun at the Thorne limit of $\rm a=0.998$, but in the movie they used a much larger spin parameter which can only be achieved by artificially spinning up the black hole, but not naturally by accretion. The Imdb argument about the g-force is gish gallop though, since in orbit you do not feel any g-force.
