I'm going through an old relativity assignment, and I've been asked to calculate the time dilation for a satellite which orbits the earth in 12 hours at 26000km from the surface, and travels at a constant speed. The given radius of the Earth is 6400km. In the solutions, the author has calculated the velocity traveled by the satellite to be $\dfrac{3.24\times 10^7\times 2\pi}{12\times 3600} \text{ms}^{-1}\simeq 4500 \text{ms}^{-1} $. I'm fine with the calculation (if I've gone wrong with the numbers somewhere, I don't mind). What I'm not so OK with is that they proceeded to use $v=4500$ in the formula $\gamma=\dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}}$, which is needed to calculate time dilation. As far as I am aware, this formula only holds for observers whose relative velocity is $v$. However, in this case where there is an observer on the Earth's surface and a satellite moving in circular motion around the center of the earth, their relative velocity isn't constant!

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    $\begingroup$ Have you figured the degree of difference (hint: it is non-trivial but not particularly large)? Have you considered averaging over the largest and smallest $\gamma$ that could be involved? How big a difference is it going to make in the final result? $\endgroup$ Commented Jul 17, 2014 at 0:51
  • $\begingroup$ I've actually confused myself to the point of not understanding my original question! I'm no longer convinced there is a difference at all, as the relative magnitude of the velocity is constant. But now I've got the problem that I don't know what difference you were referring to! Sorry to be so confusing $\endgroup$
    – Lammey
    Commented Jul 17, 2014 at 1:26
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    $\begingroup$ The relative velocity can change. In fact does change. But even at the equator the velocity of the observer is less than 500 m/s relative the center of the Earth, so the range of relative velocities is between 4000 and 5000 m/s. Compute the $\gamma$ for each of those and ask yourself if it matters. (Use the binomial approximation for gamma or a calculating tool with a lot of figures.) $\endgroup$ Commented Jul 17, 2014 at 1:35

1 Answer 1


The time dilation due to motion in a circle, relative to an observer at the centre, is just the usual Lorentz time dilation due to the velocity of the motion. If you're interested, in my answer to Is gravitational time dilation different from other forms of time dilation? I showed how this is derived from the metric.

Anyhow, as you say, the time dilation relative to a second observer also in circular motion will be a complex function of time as the relative velocity of the two observers changes. However you can calculate the time dilation for both observers relative to the centre, and then take the ratio to get the average (over many orbits) time dilation between your two observers. If we represent the velocity of the satellite as $v_s$ and the velocity of the observer on the Earth as $v_e$, then the time dilation of the satellite relative to the surface will be:

$$ t_r = \frac{\gamma_e}{\gamma_s} = \frac{\sqrt{1 - v_s^2/c^2}}{\sqrt{1 - v_e^2/c^2}} \tag{1} $$

If you attempt this calculation you'll find your calculator doesn't have enough significant figures to avoid rounding errors (well, probably, at least mine doesn't) but we can use a binomial expansion to approximate $\gamma$:

$$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \approx 1 + \frac{1}{2}\frac{v^2}{c^2} $$

or alternatively:

$$ \frac{1}{\gamma} \approx 1 - \frac{1}{2}\frac{v^2}{c^2} $$

Put these approximations into equation (1) and you get:

$$\begin{align} t_r &\approx (1 + \frac{1}{2}\frac{v_e^2}{c^2})(1 - \frac{1}{2}\frac{v_s^2}{c^2}) \\ &\approx 1 - \frac{v_s^2}{2c^2}\left(1 - \frac{v_e^2}{v_s^2} \right) \end{align}$$

So we have a correction factor of $1 - v_e^2/v_s^2$. In your example the velocity of the satellite is 4500m/s and the velocity of the faster moving bit of the Earth's surface (at the equator) is 464m/s, so the correction factor is about 1%. The time dilation of the satellite as observed from the equator will be about 1% less than as observed from centre of the Earth.

  • $\begingroup$ Thanks for the clear and detailed response! The only part I don't understand is why the ratio of the two $\gamma$ factors represents an average time dilation? $\endgroup$
    – Lammey
    Commented Jul 17, 2014 at 18:12
  • $\begingroup$ Making up some numbers to illustrate, suppose $1/\gamma_e = 0.8$ for the Earth's surface and $1/\gamma_s = 0.6$ for the satellite. That means when 100 seconds pass for the stationary observer 80 seconds pass on Earth and 60 seconds pass on the satellite. So as seen from Earth 60 seconds pass on the satellite for 80 seconds on Earth, and the time dilation is therefore 60/80, which is simply $\gamma_e/\gamma_s$. $\endgroup$ Commented Jul 18, 2014 at 5:14

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