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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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% \end{figure} \centerline{Potential as function of $z$.}

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\item{\bf 10. Proper time under rotation.}
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\noindent 
The velocity $v$ of the satellite is related to the angular frequency 
$\omega$ by $v=\omega\,r$. Therefore,
$$ c^2\,d\tau^2 = c^2\,dt^2 - \omega^2 r^2\,dt^2 \Rightarrow
   \frac{d\tau}{dt} = \sqrt{1-\frac{\omega^2\,r^2}{c^2}}\ .$$

\medskip\noindent
For the gravitational acceleration $a_g$ we have the equation
$$ a_g = \frac{G\,m_e}{r_s^2} = \frac{v^2}{r_s}\,, $$
where $r_s$ is introduced for the radius of the stationary orbit. 
Eliminating $v$ gives
$$ \frac{G\,m_e}{r_s^2}=\omega^2\,r_s=r_s\,\left(\frac{2\pi}{T}\right)^2 
   ~\Rightarrow~r_s^3=G\,m_e\,\left(\frac{T}{2\pi}\right)^2 \,,       $$
i.e., Kepler's 3rd law. For a stationary satellite, as well as for a 
point on the equator, we have $T=24\times 3600\,[s]$ for the period. 
Crunching the numbers, the result for $r_s$ is
$$ r_s = \left[G\,m_e\,\left(\frac{T}{2\pi}\right)^2\right]^{1/3}
     = 4.22\times 10^7\,[m]\,,       $$
which is about $6.6$ times the earth radius.

\medskip\noindent
From the first equation we find
\begin{eqnarray} \nonumber
  \frac{d\tau_s}{d\tau_e}&=&\frac{d\tau_s}{d\,t}\,\frac{d\,t}{d\tau_e}
   = \frac{\sqrt{1-\omega^2\,r_s^2/c^2}}
          {\sqrt{1-\omega^2\,r_e^2/c^2}}
   \\ \nonumber &\approx& \left(
   1-\frac{1}{2}\,\frac{\omega^2\,r_s^2}{c^2}\right)\,\left(
   1+\frac{1}{2}\,\frac{\omega^2\,r_e^2}{c^2}\right)
   \approx 1-\frac{1}{2}\,\frac{\omega^2\,r_s^2}{c^2}
            +\frac{1}{2}\,\frac{\omega^2\,r_e^2}{c^2}\,.
\end{eqnarray}
The numerical values are
$$ \frac{1}{2}\,\frac{\omega^2\,r_s^2}{c^2}=5.249\times 10^{-11} ~~{\rm 
and}~~\frac{1}{2}\,\frac{\omega^2\,r_e^2}{c^2}=1.194\times 10^{-12}\,.$$
The combined, measurable, effect is therefore \vfil\eject

$$ 1-\frac{d\tau_s}{d\tau_e}=\frac{1}{2}\,\frac{\omega^2\,r_s^2}{c^2}-
   \frac{1}{2}\,\frac{\omega^2\,r_e^2}{c^2}=5.130\times 10^{-11}\,,$$
i.e., the clock on the satellite ticks slower.

\medskip\noindent
From gravity we obtain 
$$ 1-\frac{d\tau_s}{d\tau_e} = - \frac{G\,m_e}{c^2}\,\left(\frac{1}{r_s} 
   - \frac{1}{r_e}\right) = - 5.911\times 10^{-10}\,, $$
which is about ten times larger than the SR effect.
\bigskip

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