(a) From the observation that clock in its rest frame goes slower by \(\gamma\) we have
\begin{equation*}
\Delta t_{S} = \gamma \Delta t_{S'},
\end{equation*}
where we can expand \(\gamma\) in powers of \((v/c)^2\text{.}\)
\begin{equation*}
\gamma = \left( 1 - (v/c)^2 \right)^{-1/2} = 1 - (-1/2) (v/c)^2 + ...
\end{equation*}
Dropping higher powers gives the answer. Similarly,
\begin{equation*}
\frac{1}{\gamma} \approx \left( 1 +(1/2) (v/c)^2 \right)^{-1} \approx 1 - \frac{1}{2}\frac{v^2}{c^2}.
\end{equation*}
(b) We place a clock at surface of Earth. So, \(t_s\) is the time in the \(S'\) frame of (a). Hence, with \(v = \omega R\) since moving in a circle of radius \(R\) with angular speed \(\omega\text{.}\)
\begin{equation*}
t_s = \frac{t_0}{\gamma} = t_0 \left( 1 - \frac{1}{2}\frac{\omega^2 R^2}{c^2} \right).
\end{equation*}
(c) Similar to (b), with speed \(v=\omega R + u\text{,}\) we get time in the easward moving plan
\begin{equation*}
t_{ae} = t_0 \left( 1 - \frac{1}{2}\frac{(\omega R + u)^2}{c^2} \right).
\end{equation*}
For flight in the opposite direction \(u\) will change sign with respec to \(\omega R\text{.}\) Therefore,
\begin{equation*}
t_{aw} = t_0 \left( 1 - \frac{1}{2}\frac{(\omega R - u)^2}{c^2} \right).
\end{equation*}
(d) The difference will be
\begin{equation*}
t_{aw} - t_{ae} = \frac{2 \omega R u}{c^2}\; t_0.
\end{equation*}
(e) Now, we use the numerical values. From \(\omega= 2\pi/24\text{hr}\) and \(R=6.37\text{ km}\text{,}\) we get \(t_0\) from \(t_s= 48\text{ hr}\text{.}\) First we have
\begin{equation*}
\frac{\omega R}{c} = \frac{2\pi \times 6.37 \text{ km}}{24\text{hr} \times 3\times 10^5\times 3600 \text{ km/hr}} = 1.544\times 10^{-9}.
\end{equation*}
Then, we calculate
\begin{align*}
t_0 \amp = t_s \left( 1 + \frac{1}{2} (\omega R/c)^2 \right) \\
\amp = 48\text{ hr}\, \left( 1 + \frac{1}{2} (0.02)^2 \right) = 48.0096\text{ hr}.
\end{align*}
Now, we evaluate the difference
\begin{align*}
t_{aw} - t_{ae} \amp = \frac{2 \omega R u}{c^2}\; t_0 \\
\amp = 2\times 1.5432\times 10^{-9} \frac{u}{c}\; t_0 = 4.13\times 10^{-10}\text{ s}.
\end{align*}
