Question

A satellite orbits the earth near its surface. By what amount does the satellite's clock fall behind the earth's clock in one revolution? Assume that nonrelativistic analysis can be made to compute the speed of the satellite and only the time dilation is to be taken into account for calculation of clock speeds.

Answer 1

The speed of the satellite may be obtained from the equation,
$\frac{GMm}{R^2}=\frac{m{v}^2}{R}$
or, $v=\sqrt{\frac{GM}{R}}$
$={\left[\frac{(6.67\times {10}^{-11}N{m}^{2}k{g}^{-2})(6\times {10}^{24}kg)}{6400\times {10}^{3}m}\right]}^{1/2}$
$=7910m{s}^{-1}$.
Thus, $v/c=\frac{7910}{3\times {10}^8}=2.637\times {10}^{-5}$
or, $\sqrt{1-{\left(\frac{v}{c}\right)}^2}=[1-6.95\times {10}^{-10}{]}^{1/2}$
$\approx 1-3.48\times {10}^{-10}$
The time taken by the satellite to complete one revolution is
$T=\frac{2\pi R}{v}=\frac{6.28\times 6400\times {10}^{3}m}{7910m{s}^{-1}}=5080s$
The clock on the satellite will slow down as observed from the earth. If the time elapsed on the satellite's clock is t as the satellite completes one revolution (this is proper time and 5080 s is improper time).
$t=(1-3.48\times {10}^{-10})\times \left(5080s\right)$
or, $\frac{1}{5080s}=1-3.48\times {10}^{-10}$
or $\frac{(t-5080s)}{5080}=3.48\times {10}^{-10}$
or, $(t-5080)=-1.77\times {10}^{-6}s$.
The satellite's clock falls behind by $1.77\times {10}^{-6}s$ in one revolution.