diff --git a/_psets/11.md b/_psets/11.md index 683f3113a1115fb6908260fc09fdfa5e1dc82aaf..f25e4b37b965d3a95c12ac024d76babe4faea8f2 100644 --- a/_psets/11.md +++ b/_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*}