Answer 14.5

_psets/11.md

@@ -189,6 +189,62 @@ Let's zoom in on those spikes.
 If a ship traveling on the equator uses one of John Harrison’s chronometers to navigate, what is the
 error in its position after one month? What if it uses a cesium beam atomic clock?
 
+To solve this we need to know how many seconds are in a month.
+
+$$
+1 \si{month} \cdot \frac{30 \si{days}}{1 \si{month}} \cdot \frac{86400 \si{s}}{1 \si{day}}
+= \num{2.59e6} \si{s}
+$$
+
+Harrison's chronometer has a relative error of $$10^{-5}$$. So after a month it will be off by
+
+$$
+\begin{align*}
+\Delta t &= 10^{-5} \cdot \num{2.59e6} \si{s} \\
+&= 25.9 \si{s}
+\end{align*}
+$$
+
+In Harrison's time the only reference available to sailors at sea were the stars. Using a sextant
+you can measure the angle between a known astronomical object (say Alpha Andromedae) and the
+horizon. This tells you your angle relative to the stars. You want your angle relative to the Earth
+(i.e. longitude). So the missing information is the angle of the Earth relative to the stars. This
+you can compute if you just know what time it is (and the time of year).
+
+But if your time measurement is off by $$\Delta t$$, your estimate of the Earth's angle will be off
+by $$\Delta t \cdot v$$ where $$v$$ is the speed of the surface of the Earth due to rotation.
+Assuming you did take a really accurate measurement of your own angle relative to the stars, this
+will be the same error in your estimate of your position relative to the Earth.
+
+So first let's find $$v$$ (at the equator).
+
+$$
+\begin{align*}
+v &= r \omega \\
+&= r \cdot \frac{1 \si{rotation}}{1 \si{day}} \cdot \frac{2 \pi \si{rad}}{1 \si{rotation}}
+    \cdot \frac{1 \si{day}}{86400 \si{s}} \\
+&= \frac{2 \pi \cdot \num{6.37e6} \si{m}}{86400 \si{s}} \\
+&= 463.3 \si{m/s}
+\end{align*}
+$$
+
+Then the error in the ship's position measurement is
+
+$$
+\begin{align*}
+\Delta t \cdot v &= 25.9 \si{s} \cdot 463.3 \si{m/s} \\
+&= 12 \si{km}
+\end{align*}
+$$
+
+If we instead use a Cesium clock, with a relative error of $$10^{-12}$$, we'll only be off by
+
+$$
+10^{-12} \cdot \num{2.59e6} \si{s} \cdot 463.3 \si{m/s} = 1.2 \si{mm}
+$$
+
+
+
 
 ## (14.6)
 
@@ -256,20 +312,9 @@ $$
 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
+We have already found the velocity of the satellite. As we found in the previous problem, the
+velocity of a clock on the surface of the Earth is $$463.3 \si{m/s}$$. Thus the respective [Lorentz
+factors](https://en.wikipedia.org/wiki/Lorentz_factor) are
 
 $$
 \begin{align*}