1.000000000445 Seconds Pass in GPS Satellite, in 1 Earth-Second
Do GPS satellites actually need atomic clocks on board?
We discussed how time dilation is “written in stone” in modern technologies:
Let’s elaborate on the magnitude of time flow discrepancies on the Earth and on GPS satellites. Two known causes for time dilation (time slowing down): gravitational time dilation, which happens near massive objects like the Earth, and relativistic time dilation, which happens in fast moving objects (like on the Earth surface, which spins around its axis at about 1,000 miles per hour). Formula for gravitational time dilation
sqrt(1–2GM/Rc²) ~ 1–GM/Rc²,
where M is the Earth’s mass and R its radius, means that for 1 second of time passed far from the Earth, 1–GM/Rc² seconds pass on the Earth surface (time slows down because of the Earth’s mass).
GPS satellite orbit’s radius is about 4 times the Earth’s radius, therefore, for 1 second of time passed far from the Earth, 1– GM/4Rc² pass on the satellite. Thus, gravitational dilation difference between on the GPS orbit and on the Earth is:
(1–GM/4Rc²)–(1–GM/Rc²)= ¾ GM/Rc² .
Relativistic time dilation caused by high velocity of the Earth surface spinning, or high velocity of GPS satellites rotation, the formula is:
sqrt(1–v²/c²) ~ 1–0.5v²/c² .
Earth surface velocity v comes from a 24-hour Earth’s rotation period:
v = 2πR/(24×3600).
GPS satellite period is about 12 hours, so on its 4R radius orbit, velocity v is: v = 8πR/(12×3600) = 16πR/(24×3600).
Thus, relativistic dilation difference between inside the GPS satellite and on the Earth’s surface is:
1–0.5(16πR/(24×3600))²/c² — (1–0.5(2πR/(24×3600))²/c²) = -127π²R²/(24×3600×c)².
Combined difference of gravitational and relativistic time dilation is:
¾ GM/Rc² –127π²R²/(24×3600×c)² .
With G = 6.67/10 ¹¹, M = 5.97×10 ²⁴, R = 6.37×10 ⁶, c = 3×10 ⁸, π = 3.14, the result is:
5.2/10 ¹⁰–0.75/10 ¹⁰ = 0.445/10 ⁹,
which is about half nanosecond. That means for 1 second passed on the Earth, 1.000000000445 seconds pass inside a GPS satellite. Relativistic effect –0.75/10 ¹⁰ is much less than gravitational effect +5.2/10 ¹⁰.
If not accounted for such difference in time flow, the error in location triangulation is 0.445/10 ⁹ ×c= 0.445/10 ⁹×300,000,000 m/sec = 0.133 m/sec, and in one day the error accumulates to 0.133 m/sec×24×3600 sec ~ 11.5 km/day ~ 7 miles/day => 350 km/month, and so forth …
An alternative to such an expensive and sophisticated GPS system of 24 satellites with big atomic clocks and powerful power sources on the board to maintain continuous broadcasting for many years, delivered to a very high orbit, is a marine buoy: no problems with huge weight to deliver on high orbit, no time dilation (the buoy is on the Earth’s surface), no problems with maintenance. Buoys do not even need atomic clocks, because the same chip that is used in our smartphones can derive precise local time. And they do not need to broadcast continuously every/every other second. They can be engaged remotely when needed only — sending a response to a request, a response that is a powerful transmission overpowering any noise/interference around. To avoid a horizon between the receiver and the transmitters, the receiver should be elevated and the transmitters should be not far away: for a receiver at 100 meters (330 ft.) above sea level, the transmitter(s) should be no farther than 36 kilometers (22 miles) from the receiver.
Transmitters can be elevated too: there are 5,000 lightweight Starlink/SpaceX satellites at 500-kilometer low orbit, they are transceivers: radio transmitters and receivers combined. They can receive GPS satellite transmissions, recalculate local time and current position (like our smartphones do) and retransmit altered signals.
Do GPS satellites actually need atomic clocks on board? Why not use a few dozen ground or ocean-based stations broadcasting their atomic time and location, and GPS satellites computing their time and location, like smartphones do? At a 4R-radius orbit, the horizon should not be a problem. Primary transmitters on the Earth → GPS satellites as transceivers → everything else as receivers.
P.S. For those who wonder where these approximations come from:
sqrt(1–2GM/Rc²) ~ 1–GM/Rc² and sqrt(1–v²/c²) ~ 1–0.5v²/c².
From the Taylor series: f(1+X) = f(1) + f’(1)×X + f’’(1)/2!×X² + f’’’(1)/3!×X³ + …
For small values X: f(1+X) ~ f(1) + f’(1)×X — the rest is ignored as X²–small.
For the function f(x)=sqrt(x), f’(x) = 1/2 / sqrt(x), and f’(1) = 0.5 .
For X = –2GM/Rc² : sqrt(1–2GM/Rc²) ~ f(1) + f’(1)×(–2GM/Rc²) = 1 — GM/Rc²,
and for X = –v²/c² :sqrt(1–v²/c²) ~ f(1) + f’(1)×(–v²/c²) = 1–0.5v²/c² .
