Consider a satellite orbiting the earth. Using the spacetime metric for a weak Newtonian gravitational field,...

Consider a satellite orbiting the earth. Using the spacetime metric for a weak Newtonian gravitational field, what is the difference in the time to complete one orbit as measured by the satellite and as measured by distant observers at rest.

Expert Solution

Assumption: The satellite will orbit earth at 1200 km.

r = 1200 + RE = 1200 + 6378 (Earth's equatorial radius) = 7578 km

V = Sqrt (Gme/r) = sqrt{(6.673*10^-11*5.974*10^24)/(7578*10^3)}=7252m/s = 7.25km/sec

T =2*pi*r/V=6562 seconds =109 minutes measured at Satellite.

As per Special Relativity, assume time runs 5000 nanoseconds per day slower for a satellite relative to us on Earth.

As per Weaker General Newtonian Gravitation Relativity it is possible to calculate that time goes faster for a satellite by 6 times minimum the Special relativity measured nanoseconds (30000 per day, due to the satellite being 1200km above the Earth (therefore in weaker gravity).

Time difference is 25,000 (30,000 – 5,000) nanoseconds faster per day for a measurement between satellite and stationary distant observers at rest on Earth.

genius_generous answered 2 years ago

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