I am trying to determine the time dilation onboard a satellite (say GPS @ 20,000km) w.r.t an observer on the earth. I have already determined the special relativity component using:

$$ t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} $$

And I got the correct answer for the time dilation simply due to relative motion (7 microseconds after 24 hrs). Im not sure if determining the component due to GR is this simple, but my first attempt was to evaluate it using the equation to determiine gravitational time dilation outside a non rotating sphere in a circular orbit using the Schwarzschild metric.

$$ t' = t\sqrt{1-\frac{3GM}{rc^2}} $$

I dont seem to be getting the correct answer (45 microseconds after 24 hrs). Any ideas?


1 Answer 1


The equation you quote:

$$ t' = t\sqrt{1-\frac{3GM}{rc^2}} \tag{1} $$

gives the time relative to an observer at infinity. You want the time relative to an observer on the Earth's surface. You need to calculate:

$$ t_\text{satellite} = t\sqrt{1-\frac{3GM}{r_\text{satellite}c^2}} $$


$$ t_\text{Earth} = t\sqrt{1-\frac{2GM}{r_\text{Earth}c^2}} $$

where $r_\text{Earth}$ is the radius of the Earth and $r_\text{satellite}$ is the radius of the satellite's orbit (measured from the centre of the Earth). The relative time dilation is then the ratio of these two times.

Note that equation (1) combines the special and general relativity contributions to the time dilation i.e. it includes both the gravitational time dilation and the effect of the orbital velocity. The observer on the Earth's surface isn't in a circular orbit so the equation is slightly different (a factor of 2 in the square root rather than 3).

Incidentally, when the gravitational field is weak (like the Earth's) we can use the weak field approximation for time dilation:

$$ \frac{dt_B}{dt_A} = \sqrt{1 - \frac{2(\Phi_A - \Phi_B)}{c^2}} \tag{1} $$

The quantity $\Phi_A - \Phi_B$ is the difference in the Newtonian gravitational potential energy between $A$ and $B$, and $dt_B/dt_A$ is the time dilation of $B$'s clock relative to $A$'s clock.

  • $\begingroup$ The observer on earth is in a couple of different orbits - once around the Sun, and once every 24 hours around the center of Earth. The former orbit is "shared" by the satellite, while the latter is "very slow". Is it worth computing the correction needed for the latter (which is dependent on latitude)? $\endgroup$
    – Floris
    Commented Nov 20, 2015 at 8:48
  • $\begingroup$ @Floris: no :-) $\endgroup$ Commented Nov 20, 2015 at 8:49
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    $\begingroup$ See also Sagnac distortion. It results in an error "on the order of hundreds of nanoseconds, or tens of meters in position". $\endgroup$
    – Floris
    Commented Nov 20, 2015 at 8:56
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    $\begingroup$ @Floris No (well, of course it depends on your application): look at John's equation (1): if the Newtonian approximation is mildly affected by an effect, then the error in a Newtonian potential difference becomes that error divided by $c^2$ in the weak field time dilation factor. $\endgroup$ Commented Nov 20, 2015 at 9:02
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    $\begingroup$ @Kafros: I've just done the calculation to check and I get 38us per day, which is exactly right. For the Earth I get t'/t = 0.999999999305 and for the satellite I get t'/t = 0.999999999975. The ratio of these two is 1.000000000445. Subtract 1 off this and multiply by the 86400 seconds in a day and the answer is 38us. $\endgroup$ Commented Nov 20, 2015 at 16:40

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