% Bernd Berg sep 21 2017.
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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2017 
Solutions} \medskip

\centerline{\bf Set 4} 

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\item{\bf 12. Proper time under rotation.}
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$$ c^2\,d\tau^2 = c^2\,dt^2 - \omega^2 R^2\,dt^2 \Rightarrow
   \frac{d\tau}{dt} = \sqrt{1-\frac{\omega^2 R^2}{c^2}}\ .$$
Now $g=v^2/R=\omega^2\,R$, where $g=9.81\,[m/s^2]$ and $v$ is the 
velocity of the satellite.
For $R=(6,378+160)\,[km]$ (equatorial radius plus $160\,[km]$) we get 
$\omega=0.0012249\,[s^{-1}]$, which correpsonds to a period of $T=2\pi
/\omega=85.5\,[minutes]$. For the time dilation we find
$$ 1-\frac{d\tau}{dt} = \frac{1}{2}\,\left(\frac{\omega\,R}{c}\right)^2 
   = 0.356321\times 10^{-9}\,. $$
Compared to this the time dilation of a clock on the equator is with
$$ 1-\frac{d\tau}{dt} = \frac{1}{2}\,
   \left(\frac{\omega\,R}{c}\right)^2 = 0.119517\times 10^{-11} $$
negligle. For gravity we obtain along the lines of the previous
problem~4
$$ 1-\frac{d\tau}{dt} = -\, 0.17422\times 10^{-10}\,, $$
which reduces the the SR effect by about 5\%.
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