For a spinning space station such as in 2001, A Space Odyssey, what would be the time slowing in the perimeter of the spinning space station with respect to the center axis of the station?

The perimeter is moving at a speed such that the acceleration is $g=9.81\text{ m/s}$. Combining $g=\frac{{v}^{2}}{R}_{s}$ with $\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$ gives dilation factor $$\sqrt{1-\frac{g\,{R}_{s}}{{c}^{2}}}$$ Assuming the radius ${R}_{s}$ of the space station is 500 meters, a perimeter clock would lose about 1e-6 seconds per year with respect to a clock in the center axis.

Now since the clock at the perimeter is subject to acceleration g, by the equivalence principle it would seem that the gravitational time dilation would apply, which is $$\sqrt{1-\frac{2R_e g}{c^2}}$$ where $R_e=6.38\times 10^6\text{ m}$ (source). This would make the perimeter clock slow by about 0.02 seconds per year.

So do I add the two dilation factors to get the total dilation factor? The gravitational dilation factor $\sqrt{1-\frac{2R_e g}{c^2}}$ is a function of both acceleration $g$ and radius $R_e$, unlike the formula for velocity dilation $\sqrt{1-\frac{v^2}{c^2}}$, which is only a function of velocity. So I suppose applying the principle of equivalence has some subtleties. My knowledge of special relativity is way ahead of my knowledge of general relativity.