Kinematic delay of GPS clocks

Problem

GPS satellites orbit at $v\approx 4$ km/s. Estimate the delay their clock accumulates in one day relative to a clock on Earth, due to kinematic time dilation alone. (Effect: the satellite’s clock runs slow relative to Earth.) Compare it with the much larger general-relativistic effect ($\sim +45\;\mu$s/day due to altitude).

Solution

Dilation fraction. For $v \ll c$ the Lorentz factor expands as

$\gamma = \frac{1}{\sqrt{1-\beta^2}} \approx 1 + \frac{1}{2}\beta^2.$

The moving clock runs slow by a fraction $\approx \tfrac{1}{2}\beta^2$ per unit time. With $v = 4\cdot 10^3$ m/s and $c = 3\cdot 10^8$ m/s:

$\beta = \frac{4\cdot 10^3}{3\cdot 10^8} \approx 1{,}33\cdot 10^{-5}, \qquad \frac{1}{2}\beta^2 \approx \frac{1}{2}(1{,}33\cdot 10^{-5})^2 \approx 8{,}9\cdot 10^{-11}.$

Delay in one day. In one day there are $T = 86\,400$ s:

$\Delta t = \frac{1}{2}\beta^2 \cdot T \approx 8{,}9\cdot 10^{-11} \times 86\,400 \approx 7{,}7\cdot 10^{-6}\ \text{s}.$

$\ev{\Delta t \approx 7\ \mu\text{s/day (the satellite's clock runs slow)}}$

Comparison with general relativity. The kinematic (special-relativistic) effect makes the satellite’s clock run slow by $\approx -7\ \mu$s/day. General relativity, on the other hand, due to weaker gravity at altitude, makes it run fast by $\approx +45\ \mu$s/day. The two effects have opposite sign and do not cancel: the net balance is $\approx +38\ \mu$s/day. Without correcting for it, GPS would accumulate position errors of kilometres per day.

Links

Topics: Relatività ristretta Concepts: Dilatazione dei tempi Skills: Stime di Fermi