Einstein and GPS 1
The orbital radius of a GPS satellite is about 26,600 km and it moves with a velocity of several km/s. According to Einstein's theory of special relativity, clocks on the satellite will appear to run slower from the Earth. If we synchronize a clock on the satellite with a clock on the Earth at noon today, how much will the time difference be at noon tomorrow? Give your answer in micro-seconds, rounded to the nearest integer.
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A possible way of calculating the satellite's velocity is using the Moon as a reference. The Moon orbits the Earth in 27 days, and it has a semi-major axis of 384,400km. One can use Kepler's laws to calculate the orbital period of the satellite (it should be about 1/2 day), and the velocity of the satellite, , follows from that.
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The time difference will be , converted to micro-seconds. Here is the speed of light.
Bonus The GPS works by precision timing the radio signals and converting the time to distance by . If this timing error is left uncorrected, how much of of a distance-error is accumulated in one day?
See my other problem about the timing error explained by general relativity. That is larger and in the opposite direction.
The answer is 7.
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1 solution
Using the Moon as reference Kepler's law yields T = T M ( R M R ) ( 3 / 2 ) = 0 . 4 9 d a y . The velocity is v = 2 π T 2 π R = 3 9 0 0 m / s . The time delay comes out to be 7 . 4 μ s ≈ 7 μ s .
Bonus The distance error is 2.3 km. This looks like a huge error, but in reality the situation is a bit better. The GPS is based on comparing signals coming from several different satellites, and as long as they all suffer similar time delays, the error is not that much.
Please look at my other problem where the correction due to general relativity is considered. That is an even bigger effect, in the opposite direction.
Read more about this, and its interesting history, in "Relativity and the Global Positioning System", Neil Ashby, Physics Today, May 2002.