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Commit 41c47036 authored by Erik Strand's avatar Erik Strand
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Answer 14.6

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......@@ -200,19 +200,147 @@ GPS satellites orbit at an altitude of 20,180 km.
{:.question}
How fast do they travel?
We need a relationship between orbital altitude and velocity. Consider a particle moving in a
circle of radius $$r$$, with angular velocity $$\omega$$. We can model its trajectory as
$$
\begin{align*}
x(t) &= r e^{i \omega t} \\
\dot{x}(t) &= i r \omega e^{i \omega t} \\
\ddot{x}(t) &= -r \omega^2 e^{i \omega t}
\end{align*}
$$
The magnitude of its acceleration is $$r \omega^2$$, so it experiences a force
$$
\begin{align*}
F &= ma \\
&= m r \omega^2 \\
&= m \frac{v^2}{r}
\end{align*}
$$
where the last line follows since $$v = r \omega$$.
In this case we know $$r$$ (recall that the radius of the Earth is $$\num{6.37e6} \si{m}$$), and we
can assume the only force is gravitational. Thus
$$
\begin{align*}
\frac{G M m}{r^2} &= m \frac{v^2}{r} \\
v &= \sqrt{\frac{GM}{r}} \\
&= \sqrt{\frac{\num{6.67e-11} \si{m^3 kg^{-1} s^{-2}} \cdot \num{5.97e24} \si{kg}}
{\num{6.37e6} \si{m} + \num{2.02e7} \si{m}}} \\
&= \num{3.87e3} \si{m/s}
\end{align*}
$$
### (b)
{:.question}
What is their orbital period?
$$
\begin{align*}
T &= \frac{2 \pi r}{v} \\
&= \frac{2 \pi \left( \num{6.37e6} \si{m} + \num{2.02e7} \si{m} \right) }{\num{3.87e3} \si{m/s}} \\
&= \num{4.31e4} \si{s} \\
&= \num{11.96} \si{hours}
\end{align*}
$$
### (c)
{:.question}
Estimate the special-relativistic correction over one orbit between a clock on a
GPS satellite and one on the Earth. Which clock goes slower?
We have already found the velocity of the satellite. The velocity of a clock on the surface of the
Earth is
$$
\begin{align*}
v &= r \omega \\
&= r \cdot \frac{1 \si{rotation}}{1 \si{day}} \cdot \frac{2 \pi \si{rad}}{1 \si{rotation}}
\cdot \frac{1 \si{day}}{86400 \si{s}} \\
&= \frac{2 \pi \cdot \num{6.37e6} \si{m}}{86400 \si{s}} \\
&= 463.3 \si{m/s}
\end{align*}
$$
Thus the respective [Lorentz factors](https://en.wikipedia.org/wiki/Lorentz_factor) are
$$
\begin{align*}
\gamma_\text{Earth} &= \frac{1}{\sqrt{1 - (463.3 \si{m/s})^2 / (\num{3e8} \si{m/s})^2}} \\
&= 1 + \num{1.19e-12} \\
\gamma_\text{orbit} &= \frac{1}{\sqrt{1 - (\num{3.87e3} \si{m/s})^2 / (\num{3e8} \si{m/s})^2}} \\
&= 1 + \num{8.33e-11}
\end{align*}
$$
So one day on the surface of the Earth is longer than one day at the center of the Earth by
$$
\begin{align*}
\Delta t &= (\gamma_\text{Earth} - 1) \cdot 1 \si{day} \\
&= \num{1.19e-12} \cdot 86400 \si{s} \\
&= \num{1.03e-7} \si{s}
\end{align*}
$$
And one day in orbit is longer than one day at the center of the Earth by
$$
\begin{align*}
\Delta t &= (\gamma_\text{orbit} - 1) \cdot 1 \si{day} \\
&= \num{8.33e-11} \cdot 86400 \si{s} \\
&= \num{7.20e-6} \si{s}
\end{align*}
$$
Similarly one orbital period on the surface of the Earth is longer than one orbital period at the
center of the Earth by
$$
\begin{align*}
\Delta t &= \num{1.19e-12} \cdot \num{4.31e4} \si{s} \\
&= \num{5.13e-8} \si{s}
\end{align*}
$$
And one orbital period in orbit is longer than one orbital period at the center of the Earth by
$$
\begin{align*}
\Delta t &= \num{8.33e-11} \cdot \num{4.31e4} \si{s} \\
&= \num{3.59e-6} \si{s}
\end{align*}
$$
The latter is also a good approximation of the difference between a clock in orbit vs the surface of
the Earth over one orbit. Technically speaking the surface of the Earth isn't an inertial frame of
reference, but relative to orbit it's close.
### (d)
{:.question}
What is the general-relativistic correction over one orbit? Which clock goes
slower?
The general relativistic correction indicates that time passes more quickly in orbit than on Earth
because the Earth's gravitational field is stronger on the surface than in orbit. The difference
over one day is
$$
\begin{align*}
\Delta t &= \left(\frac{1 - \frac{GM}{rc^2}}{1 - \frac{GM}{r'c^2}} - 1 \right) \cdot 1 \si{day} \\
&= \left(\frac{1 - \frac{\num{3.99e14} \si{m^3/s^2}}{(\num{6.37e6} \si{m} + \num{2.02e7} \si{m})
\cdot (\num{3e8} \si{m/s^2})^2}}{1 - \frac{\num{3.99e14} \si{m^3/s^2}}{\num{6.37e6} \si{m}
\cdot (\num{3e8} \si{m/s^2})^2}} - 1\right) \cdot 86400 \si{s} \\
&= \num{4.57e-5} \si{s}
\end{align*}
$$
Similarly the difference over one orbital period is $$\Delta t = \num{2.28e-5} \si{s}$$.
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