From 41c470362290e4c9647a1a81548f6035f7aca998 Mon Sep 17 00:00:00 2001
From: Erik Strand <erik.strand@cba.mit.edu>
Date: Thu, 2 May 2019 12:30:00 -0400
Subject: [PATCH] Answer 14.6
---
_psets/11.md | 128 +++++++++++++++++++++++++++++++++++++++++++++++++++
1 file changed, 128 insertions(+)
diff --git a/_psets/11.md b/_psets/11.md
index b0d4958..683f311 100644
--- a/_psets/11.md
+++ b/_psets/11.md
@@ -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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