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?
Asked
Active
Viewed 385 times
1 Answers
1
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).
AV23
- 3,253