Time flow difference for satellites Clocks in satellites have to be adjusted due to the effects of relativity; but does time for satellites (GPS) flow slower, due to the relative motion, or faster, due to the weaker amount of gravitation received due to the increased distance to earth?
 A: We'll consider the relevant terms in the Schwarzschild metric, and substitute $d\phi = \frac{v}{r} dt$ with $\theta = \pi/2$ to get (assuming circular orbits):
$$d\tau^2 = \left(1-\frac{r_s}{r}\right)dt^2 - \frac{v^2}{c^2}dt^2$$
where $\tau$ is the proper time (time as measured by the object), and $v$ is the orbital speed.
The relevant quantity to measure time dilation is:
$$\frac{d\tau}{dt} = \sqrt{1 - \frac{r_s}{r} - \frac{v^2}{c^2}}$$
We need to plug in values to compare. The Schwarzschild radius of the earth is $r_s = 8.9\text{mm}$. An observer on the Earth has an angular velocity of $\frac{v}{r} = \frac{2\pi}{3600\times24}\text{rad/s} \approx 7\times10^{-5}\text{rad/s}$. Combined with a radius of $r = 6400\text{km}$, we get $v = 465\text{m/s}$:
$$\frac{d\tau_{E}}{dt} = \sqrt{1 - \frac{8.9}{6400,000,000} - \frac{465^2}{299792458^2}} \approx \sqrt{1 - 1.4\times10^{-8}}$$
where the contribution due to the speed is negligible.
For a GPS satellite, $r \approx 26800\text{km}$, and $\frac{v}{r} \approx 14\times10^{-5}\text{rad/s}$. This means that $v = 3897\text{m/s}$. Plugging this in:
$$\frac{d\tau_{G}}{dt} = \sqrt{1 - \frac{8.9}{26800,000,000} - \frac{3897^2}{299792458^2}} \approx \sqrt{1-5\times10^{-10}}$$
We see that it is the effect of the weak gravitational field that dominates, and the passage of time is therefore faster for a satellite than someone on the surface (where the notion of simultaneity is given by the $t$ coordinate).
