OK, let's continue with
\tau = t \; \sqrt{\left(1-\frac{r_s}{r}\right) - r^2\,\omega^2}
First we make a Taylor expansion for the square root; that means we assume
\frac{r_s}{r} \ll 1
r^2\,\omega^2 \ll 1
(again: c=1)
\tau = t \; \left(1-\frac{1}{2}\left(\frac{r_s}{r} + r^2\,\omega^2\right)\right)
Then we introduce the radius of the Earth r
E and the altitude h:
r = r_E + h
and make another Taylor expansion in h/r
E
\tau = t \; \left(1-\frac{1}{2}\left(\frac{r_s}{r_E(1+h/r_E)} + r^2\,\omega^2\right)\right) = t \; \left(1-\frac{1}{2} \left( \frac{r_s}{r_E} \left( 1 - \frac{h}{r_E} \right) + r^2\,\omega^2\right)\right)
Remember that the coordinate t is the proper time of a stationary = non-corotating observer at r → ∞. For a non-corotating observer at h=0 (she will sit at h=0 and will see a fast moving surface of the earth!) the result is
\tau(h=0,\, \omega=0) = t \; \left(1-\frac{r_s}{2r_E} \right)
So in general we have
\tau(h,\, \omega) = \tau(h=0,\, \omega=0) + \Delta\tau(h,\, \omega)
with
\Delta\tau(h,\, \omega) = \frac{t}{2} \; \left(\frac{r_sh}{r_E^2} - \frac{r^2\,\omega^2}{c^2}\right)
where I re-intruduced c to allow for explicit calculations
The missing piece is the Schwarzschild radius
r_s = \frac{2GM_E}{c^2}
In the next step we get rid of the unobservable coordinate t and express t in terms of the proper time of the non-corotating observer at h=0; in addition we use (again) the approximation
r \simeq r_E
\frac{\Delta\tau(h,\, \omega)}{\tau(0,\,0)} = \frac{1}{2} \frac{ \frac{r_sh}{r_E^2} - \frac{r^2\,\omega^2}{c^2} }{ 1-\frac{r_s}{2r_E} } \simeq \frac{1}{2} \frac{ \frac{r_sh}{r_E^2} - \frac{r_E^2\,\omega^2}{c^2} }{ 1-\frac{r_s}{2r_E} }
This is the final result for small velocities v = rω and small altitude h.
As you can see there is increasing (Delta of) proper time τ with increasing h (weaker gravitational field) and decreasing (Delta of) proper time with increasing v = rω; the second term v²/2c² is well-known from SR. The denominator is a correction due to the fact that already at h=0 we have some gravitational field which slows down time.
This formula should be applicable for an airport at h=0 and an airplane at e.g. h=10km which is small compared to r
E. Please remember that ω=0 does apply to a non-corotating observer, so in order to use the formula for airports and airplanes one would introduce the velocities
v_\text{airport} = r_E\,\omega_\text{airport}
and
v_\text{airplane} = v_\text{airport} + \Delta v_\text{airplane}
where Δv is the velocity of the airplain w.r.t. the airport, e.g. ±800km/h for westward / eastward direction; this velocity is of course small compared to c=300000km/h. So using this formula you could immediately analyse the Hafele–Keating experiment
http://en.wikipedia.org/wiki/Hafele–Keating_experiment