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GPS satellites move at about \SI{4}{\kilo\meter\per\second}. Show that a good GPS receiver needs to correct for time dilation if it is to produce results consistent with atomic clocks accurate to one part in \numprint{e13}.
Let's calculate the magnitude of the time dilation effect: \begin{align} \Delta t &= \frac{1}{\sqrt{1\frac{v^2}{c^2}}} \cdot \Delta t_0\\ &= \frac{1}{\sqrt{1\numprint{1.8e10}}} \cdot \Delta t_0 \end{align} Since the accuracy of most pocket calculators is not within $10^{10}$, the calculator will say $\Delta t = \Delta t_0$. We need binomial expansion to get results: The binomial expansion ($(1\pm x)^n \approx 1+nx$ for $x\ll 1$ leads with $n=\frac12$ to: \begin{align} \Delta t &\approx (1\frac12\cdot \numprint{1.8e10}) \cdot \Delta t_0 \end{align} The time error divided by the time interval is: \begin{align} \frac{\Delta t\Delta t_0}{\Delta t_0} \approx \numprint{e10} \end{align} Time dilation, if not accounted for, would introduce an error of about 1 part in \numprint{e10}, which is \numprint{1000} times greater than the precision of atomic clocks. Not correcting for time dilation means a receiver could give a much poorer position accuracy. GPS devices must make other corrections as well, including effects associated with General Relativity.
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