\begin{problem}
\Part{a} How long (in Earth time) does it take for a plane to circle
the Earth at low altitude (average radius $6,370\U{km}$) going
600\U{mph}?  \Part{b} How much less time passes according to a clock
on the plane?

For the purpose of this problem, you may assume the Earth is
stationary and not rotating.
\end{problem}

\begin{solution}
\Part{a}
\begin{align}
  600\U{mph} \cdot \frac{1.6\U{km}}{1\U{mile}}\cdot\frac{1\U{hour}}{3600\U{s}}
     &= 267\U{m/s} \\
  v\Delta t &= \Delta x = 2 \pi r \\
  \Delta t &= \frac{2 \pi r}{v}
    = \frac{2 \pi \cdot 6.37\E{6}\U{m}}{267\U{m/s}}
    = \ans{150\U{ks}} = \ans{41.7\U{hours}}
\end{align}

\Part{b}
This is pretty much a twin-paradox, with the clock on the
constantly-accelerating plane running slower than the clock on Earth.
\begin{align}
  \Delta t' &= \frac{\Delta t}{\gamma} = \sqrt{1-\p({\frac{v}{c}})^2} \Delta t
    = (1-3.95\E{-13}) \Delta t \\
  \Delta t - \Delta t' &= 3.95\E{-13} \Delta t = \ans{59.3\U{ns}}
\end{align}
\end{solution}
