A navigation satellite moves on a circular orbit of radius from the Earth's centre

PHYSICS · Multi-part · Fermi University Physics · Relativity & Gravitation · Special relativity: spacetime and kinematics · The equivalence principle and gravity as geometry

Relativity and Gravitation, Ch Experimental foundations: free fall, Lorentz invariance and clock tests

Problem

A navigation satellite moves on a circular orbit of radius

r = 2.96\times10^{7}\ \mathrm{m}

from the Earth's centre. A student is asked how much its onboard clock gains or loses per day on a clock at the Earth's surface, and writes: "The satellite clock is moving, so it runs slow. Its orbital speed is

v=\sqrt{GM/r}

, and the fractional rate loss is

v^{2}/2c^{2}

. Over one day this is

6.47\ \mu\mathrm{s}

lost. A receiver that ignored this would be wrong by

c\times 6.47\ \mu\mathrm{s} = 1.94\ \mathrm{km}

after a day." Take

M = 5.972\times10^{24}\ \mathrm{kg}

for the Earth,

R = 6.371\times10^{6}\ \mathrm{m}

for its radius,

G = 6.674\times10^{-11}\ \mathrm{N\,m^{2}\,kg^{-2}}

and

c = 2.998\times10^{8}\ \mathrm{m\,s^{-1}}

. The Earth is treated as spherically symmetric, the orbit as exactly circular, and the ground clock's motion due to the Earth's rotation is ignored. Work to first order in

\Phi/c^{2}

and in

v^{2}/c^{2}

. ((a)) Name the physical effect the student has omitted, and say which way it acts. ((b)) Compute the correct net offset of the satellite clock, in

\mu\mathrm{s}

per day, and the daily position error

c\,\Delta t

in km that ignoring it would produce.

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