Where are the time dilatational effects of orbital motion and gravitational acceleration equal?

Question

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.

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.