Problem — Estimating the GPS relativistic correction

Problem

Estimate the fractional relativistic correction $(\gamma - 1)$ for a GPS satellite in orbit at $v \approx 3{,}9$ km/s. Compare with the precision required ($\sim 10^{-9}$ per day) for the correct operation of the system.

Solution

Velocity ratio.

$\beta = \frac{v}{c} = \frac{3{,}9\cdot 10^{3}\;\text{m/s}}{3\cdot 10^{8}\;\text{m/s}} \approx 1{,}3\cdot 10^{-5}$

Approximation for small velocities. For $\beta \ll 1$ we expand in series:

$\gamma = \frac{1}{\sqrt{1 - \beta^2}} \approx 1 + \tfrac{1}{2}\beta^2 \quad\Rightarrow\quad \gamma - 1 \approx \tfrac{1}{2}\beta^2$

Estimate.

$\gamma - 1 \approx \tfrac{1}{2}(1{,}3\cdot 10^{-5})^2 \approx \tfrac{1}{2}\cdot 1{,}7\cdot 10^{-10} \approx 8{,}5\cdot 10^{-11}$

$\ev{\gamma - 1 \sim 10^{-10}}$

Comparison. The time-dilation correction due to velocity is of order $10^{-10}$, about a factor of 10 smaller than the precision required per day ($\sim 10^{-9}$). It is therefore not negligible: accumulated over time (and added to the gravitational effect of general relativity, which is even larger), it would produce position errors of kilometres if left uncorrected.

Links

Topics: Special relativity Concepts: Time dilation Skills: Fermi estimation