Nearly four years ago, upon hearing of the observation of time dilation in two optical atomic clocks at an elevation one metre apart, due to acceleration towards earths centre of gravity by Chou, C. W.; Hume, D. B.; Rosenband, T.; Wineland, D. J. I wondered where the point would be, at which time dilation caused by acceleration toward the earth gives way to time dilation caused by acceleration around the earth, via orbital velocity. (I mean the point at which the two effects are at equilibrium with each other.)

I asked a friend who is a physicist, he got back to me almost immediately, fascinated by his back of an envelope calculations that this point is the Clarke orbit, or the geosynchronous orbit.

If for instance the rotation velocity of earth were a slightly different speed, the result would be different, and no relationship apparent. I think there may be a very interesting relationship waiting to be revealed.

Is there a physical or mathematical relationship? And if so, what is it? Not being a mathematician (I'm an artist) I'm not capable of doing the math, but this has fascinated me ever since.

  • $\begingroup$ "where the point would be, at which time dilation caused by acceleration toward the earth gives way to time dilation caused by acceleration around the earth, via orbital velocity." What you mean by this is not clear. There are 2 effects that change the rate of clocks on satellites relative to us. One is related to special relativity and is due to the velocity of the satellite, the other is due to general relativity and is due to the difference in position in the earth's gravity well. These 2 counteract each other. Is your question "at which point do these cancel out?" ? $\endgroup$
    – ticster
    Jul 13, 2014 at 12:19
  • $\begingroup$ Yes. That was the initial point I made to my friend, that there should be a point at which one cancels the other out. When I learned it was Clarke Orbit, I was very intrigued. $\endgroup$
    – user55372
    Jul 13, 2014 at 12:31
  • $\begingroup$ Comment to the question (v2): It would be good if OP (or someone else) could find a more descriptive and useful title. $\endgroup$
    – Qmechanic
    Jul 13, 2014 at 12:59
  • $\begingroup$ I'm in the process of writing up an answer. For now OP, know that what you say is not true for geosynchronous orbit. $\endgroup$
    – ticster
    Jul 13, 2014 at 13:03

1 Answer 1


Let $\Delta_S$ and $\Delta_G$ be the time dilation effects due to General Relativity (gravity) and Special Relativity (motion) respectively (i.e. the clock rate on the satellite due to SR and GR is $1 - \Delta_S + \Delta_G$, signs chosen for simplicity). If these are small, they can be approximated as :

\begin{eqnarray} \Delta_S &=& 1 - \sqrt{1 - \frac{v^2}{c^2}} &\approx& \frac{v^2}{2c^2} \\ \Delta_G &=& \sqrt{1 - \frac{2GM}{R c^2}} - \sqrt{1 - \frac{2GM}{r c^2}} &\approx& \frac{G M}{R c^2} - \frac{G M}{r c^2} \end{eqnarray}

Where $v$ is the orbital velocity, $r$ the orbital radius, $R$ the radius of the earth, and $M$ its mass. Note that $\Delta_S$ and $\Delta_G$ act in opposite directions. Setting the 2 equal, and recalling that for a circular orbit we have $v^2 = \frac{GM}{r}$, we obtain :

\begin{eqnarray} \frac{G M}{R c^2} - \frac{G M}{r c^2} &=& \frac{GM}{2 r c^2} \\ r &=& \frac{3 R}{2} \end{eqnarray}

This is about $3,000$ km higher than the earth's surface, well below the $35,000$ km of geostationary orbits. By then, the GR effect largely dominates over the SR one.

tl;dr There is such an orbit but it is actually much lower than the Clarke orbits you suggested.

  • $\begingroup$ Thanks for that. My mind was boggled by the notion that it was the same altitude as the Clarke Orbit. I couldn't imagine why that should be, but since it isn't, (my friend must have done his calculations wrongly) there isn't a mystery after all. Many thanks. $\endgroup$
    – user55372
    Jul 13, 2014 at 17:38
  • $\begingroup$ No problem. Indeed if it had been the Clarke Orbit that would have one hell of an astronomical coincidence. Then again these things happen. The angular size of the moon being that close to that of the sun is what inspired countless civilizations to think the moon was a counterpart of the sun, when in reality the only connection is coincidence. Btw, if the answer is satisfying to you, you should mark it as so. $\endgroup$
    – ticster
    Jul 13, 2014 at 17:43
  • $\begingroup$ It most certainly is satisfying. duly marked as such. : ) $\endgroup$
    – user55372
    Jul 13, 2014 at 17:54

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