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## Do GPS satellites move in the sky?

Following on from my blog “Is Tim Peake getting younger or older?” , a bit of fun to work out whether time was passing more slowly or more quickly for Tim Peake in the ISS than it is for us on the ground, it got me thinking about the global positioning system (GPS) that so many of us use on a daily basis. Whether it is using a SATNAV in our car, or a GPS-enabled watch to measure how far and fast we have run, or using maps on a smartphone, GPS must be one of the most-used satellite developments of the last few decades.

As I blogged about here, communication satellites need to be at a particular height above the Earth’s surface so that they orbit the Earth in the same time that it takes the Earth to rotate. In addition to their altitude, communication satellites can only orbit the Earth about the equator, no other orientation will allow the satellite to hover in the same place relative to a location on Earth.

But what about the satellites used in GPS? What kind of an orbit are they in?

## The GPS satellites’ orbits

It turns out that the GPS satellites are not in a geo-stationary orbit, but are in fact in an orbit which leads to their orbiting the Earth exactly twice in each sidereal day (for a definition of sidereal day see my blog here).

The GPS system consists of 31 satellites in orbit around the Earth

We can work out what radius from the Earth’s centre this needs to be by remembering that the speed of orbit is given by

$v = \sqrt { \frac{ GM }{ r } } \text{ (1) }$
where $v$ is the speed of orbit, $G$ is the universal gravitational constant, $M$ is the mass of the Earth and $r$ is the radius of orbit from the centre of the Earth (not from its surface).

A sideral day is 23 hours and 56 minutes, which in seconds is $8.6160 \times 10^{4}$ seconds. So, half a sidereal day is $4.308 \times 10^{4}$ seconds. We will call this the period $T$. The speed of orbit, $v$ is related to the period via the equation
$v = \frac{ 2 \pi r }{ T }$
where $r$ is the radius of the orbit, the same $r$ as in equation (1), and $2 \pi r$ is just the circumference of a circle. So, squaring Equation (1), we can write
$v^{2} = \frac{ GM }{ r } = \left( \frac{ 2 \pi r }{ T } \right)^{2}$
So, in terms of $r$ we can write
$r^{3} = \frac{ G M T^{2} }{ 4 \pi^{2} } \rightarrow r = \sqrt[3]{ \frac{ G M T^{2} }{ 4 \pi^{2} } }, \; \text{ so } r = 26.555 \times 10^{6} \text{ m}$
In terms of height above the Earth’s surface, we need to subtract off the radius of the Earth, so the altitude, which I will call $a_{gps}$, is going to be
$a_{gps} = 26.555 \times 10^{6} - 6.371 \times 10^{6} = 20.184 \times 10^{6} \text{ m } \text{ or } \boxed{ 20.2 \text{ thousand kilometeres} }$

## Why are GPS satellites in this kind of an orbit?

As I didn’t know what kind of an orbit GPS satellites were in before I wrote this blog, the next obvious question is – why are they in an orbit which is exactly half a sidereal day? It is clearly not coincidental! To answer this question, we need to first of all discuss how GPS works.

GPS locates your position by measuring the time a signal takes to get to your GPS device from at least four satellites. Your device can identify from which satellites it gets a signal, and the system knows precisely the position of these satellites. By measuring the time the signals take to you reach you from each of the satellites, it is able to calculate how far each one is from you, and then by using triangulation it can work our your location. There are currently 31 satellites in the system, so often there are more than four visible to your GPS device. The current 31 satellites have all been launched since 1997, the original suite of 38 satellites launched between 1978 and 1997 are no longer in operation.

As I mentioned in my blog about geostationary satellites, a satellite in a geostationary orbit can only orbit above the Earth’s equator. This would clearly be no good for a GPS system, as all the satellites would lie to the south of someone in e.g. Europe or North America. As I said above, there are currently 31 operational satellites; the 31 are divided into 6 orbital planes. If there were 30 satellites this would be 5 in each orbit. The orbits are inclined at $55^{\circ}$ to the Earth’s equator. Each orbit is separated from the other one by 4 hours (equivalent to $60^{\circ}$) in longitude.

As one can see approximately 6 hours in right ascension to both the east and west of one’s location, this means that there will be at least 3 of the orbits above the horizon, and sometimes more. If there were 5 satellites in each orbit this would mean that each one would pass a particular latitude 4 hours before the next one. So, at any particular time there should be some satellites further north than one’s location and some further south, as well as some further east and some further west. This configuration allows for the necessary triangulation to obtain one’s location.

The orbits are inclined at $55^{\circ}$ to the equator and separated by 4 hours (equivalent to $60^{\circ}$) in right ascension, as this diagram attempts to show

## Is the time-dilation effect due to SR or GR more important for these satellites?

We already showed in this blog that, for the International Space Station, the time-dilation due to Special Relativity (SR) has a greater effect on the passage of time than the time-dilation due to General Relativity (GR). What about for the GPS satellites?
The speed of orbit for the GPS satellites at a radius of $26.555 \times 10^{6}$ from the Earth’s centre is, using Equation (1),
$v = \sqrt{ \frac{ GM }{ r } } = 3.873 \times 10^{3} \text{ m/s}$
As we showed in my blog about Tim Peake, the speed of someone on the Earth’s surface relative to the centre of the Earth is $v_{se} = 463.35 \text{ m/s}$, so the relative speed between a GPS satellite and someone on the Earth’s surface is given by
$v = 3.873 \times 10^{3} - 463.35 = 3.410 \times 10^{3} \text{ m/s}$
Compare this to the value for the ISS, which was $7.4437 \times 10^{3}$, it is less than half the speed.

This value of $v$ leads to a time dilation factor $\gamma$ in SR of
$\gamma = \frac{ 1 }{ \sqrt{ 0.9999999999} } \approx 1$
which means that the time dilation due to SR is negligible.
The time dilation due to GR is given by (see my blog here on how to calculate this)
$\left( 1 - \frac{ gh }{ c^{2} } \right) = (1 - 2.2 \times 10{-9}) = 0.9999999978$, or 22 parts in $10^{10}$. Compare this to the ISS, where it was about 1 part in $10^{11}$. Clearly the GR effect for GPS satellites is greater, by about a factor of 5, than it was for the ISS. But, conversely, the SR time-dilation effect has become negligible.

To conclude, the time dilation for GPS satellites is nearly entirely due to General Relativity, and not due to Special Relativity. Time is passing more quickly for the clocks on the GPS satellites than it is for us on Earth, the converse of what we found for the ISS, which is in a much lower orbit.

Because the timings required for GPS to work are so precise, the time dilation effect due to GR needs to be taken into account, and is one of the best pieces of evidence we have that time dilation in GR actually does happen.