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 $$\Delta t = \num{2.28e-5} \si{s}$$. -- GitLab