Problem — GPS, the quantitative version

Problem

A GPS satellite orbits at altitude $h=20\,200$ km with orbital velocity $v\approx 3{,}87$ km/s. Compare the special-relativistic dilation effect (due to $v$) with the gravitational dilation effect (due to the $\Delta\phi$ relative to the ground): which of the two is larger? By how much would a satellite clock drift relative to an Earth clock in $24$ hours without corrections? Data: $g=9{,}8$ m/s², $R_T=6378$ km, $GM_T\approx 4{,}0\cdot 10^{14}$ m³/s².

Solution

Special-relativity effect (velocity). The satellite is moving, so its clock slows down relative to the ground. To first order the fraction is $-v^2/(2c^2)$. With $v = 3870$ m/s: $\frac{\Delta t_\text{SR}}{t} = -\frac{v^2}{2c^2} = -\frac{(3870)^2}{2\cdot 9\cdot 10^{16}} = -\frac{1{,}50\cdot 10^{7}}{1{,}8\cdot 10^{17}} \approx -8{,}3\cdot 10^{-11}$ Over one day ($86\,400$ s): $\Delta t_\text{SR} \approx -8{,}3\cdot 10^{-11}\cdot 86\,400\ \text{s} \approx \ev{-7{,}2\ \mu\text{s/day}}$

General-relativity effect (gravitational potential). The satellite is higher up, in a less negative potential, so its clock speeds up. The fraction is $\Delta\phi/c^2$ with $\Delta\phi = GM_T\!\left(\dfrac{1}{R_T} - \dfrac{1}{R_T+h}\right)$. With $R_T = 6{,}378\cdot 10^6$ m and $R_T+h = 6{,}378\cdot 10^6 + 2{,}02\cdot 10^7 = 2{,}658\cdot 10^7$ m: $\Delta\phi = 4{,}0\cdot 10^{14}\left(\frac{1}{6{,}378\cdot 10^6} - \frac{1}{2{,}658\cdot 10^7}\right) = 4{,}0\cdot 10^{14}\cdot 1{,}19\cdot 10^{-7} \approx 4{,}77\cdot 10^{7}\ \text{m}^2/\text{s}^2$ $\frac{\Delta t_\text{GR}}{t} = \frac{\Delta\phi}{c^2} = \frac{4{,}77\cdot 10^{7}}{9\cdot 10^{16}} \approx +5{,}3\cdot 10^{-10}$ Over one day: $\Delta t_\text{GR} \approx +5{,}3\cdot 10^{-10}\cdot 86\,400\ \text{s} \approx \ev{+45{,}7\ \mu\text{s/day}}$

Comparison and net drift. The gravitational effect (GR) is larger in magnitude than the velocity effect (SR) and has the opposite sign. The satellite overall runs faster than the ground: $\Delta t_\text{net} = \Delta t_\text{GR} + \Delta t_\text{SR} \approx +45{,}7 - 7{,}2 \approx \ev{+38{,}5\ \mu\text{s/day}}$

Interpretation. Without corrections, GPS clocks would run ahead by about $38{,}5$ μs per day. Since light travels $\sim 30$ cm in a nanosecond, an error of $38\,500$ ns would correspond to a position error of about $11$ km per day: GPS would be unusable within a few hours. Relativity is not a theoretical luxury, but an indispensable engineering ingredient.

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Topics: Special relativity Concepts: Time dilation Skills: Symbolic setup · Limiting-case and sci-fi-physics analysis