Consider that GPS satellites circle the Earth at an altitude of about 20,000 km

Question

Consider that GPS satellites circle the Earth at an altitude of about 20,000 km above the surface of the Earth. What is the time difference from general relativity between a clock on board the GPS satellite and a clock on the Earth. If GPS satellte measures 1 day (= 24 hrs), how much time will be measured by the clock on Earth ? (I.e., calculate the difference in time between the GPS satellite’s clock and a clock on the Earth given that 24 hrs pass as measued by the GPS satellite.) Give your answer in micro-seconds, µs (where 106µs = 1 second).

Explanation / Answer

Me = 5.98*10^24 kg

Re = 6.37*10^6 m

h = 20000 km

= 2*10^7 m

Then orbital speed of stellite,

v = sqrt(G*M/(Re + h))

Put values

= sqrt(6.67*10^-11*5.98*10^24/(6.37*10^6 + 2*10^7))

= 3890 m/s

now use time daialation equation

to = 24 hours.

t' = to/sqrt(1 - (v/c)^2)

put values

= 24*60*60/sqrt(1 - (3890/(3*10^8))^2)

= 86400.000007263 s

the difference in time between the GPS satellite’s clock and a clock on the Earth = t' - to

t'=86400.000007263 s

to=24*60*60

put values

= 86400.000007263 - 24*60*60

= 7.26*10^-6 s

= 7.26 micro s Answer

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