As time passes more slowly in a region of space close to the source of a gravitational field, shouldn't the moon, which always has one side facing towards the earth, have a higher fraction of radioactive isotopes on that side than on the far side? Could the fraction be used to determine the time tidal locking occured?

Also, shouldn't the fraction of radioactive isotopes be different in the earth depending on how deep you dig? Gravitation drops off with $r^2$, whereas the gravitationally relevant mass increases with $r^3$ (only the mass of the earth's sphere lying below the radioactive atom results in gravitation due to Gauss' law) , so outer layers should have more time dilation and therefore a higher fraction of radioactive isotopes. Can this be observed?


In the limit of weak-field gravity, time dilates according to (relative to an observer with zero gravitatainal potential)${}^{1}$:

$t = \frac{t_{0}}{\sqrt{1-\phi}}$

Taking the accepted values for the mass of the moon of the earth, the radius of the moon, and the earth-moon distance from google, you can calculate that time on the near side of the moon progresses at $1 - 1.15\times 10^{-11}$ the rate that time on the far side of the moon does. Therefore, if you look at the relative abundances of even something like Uranium, which has a half life of ~4 billion years, you're still stuck with a time difference of the order of milliseconds, which is likely undetectable, but given the energy that we've devoted as a society to assaying and enriching uranium, perhaps I'm wrong.

${}^{1}$: note that $\phi$ is unitless in units of $G=c=1$

  • $\begingroup$ The situation on the surface of the moon is presumably the same as on the surface of the earth: it's approximately an equipotential, so the effect vanishes completely. $\endgroup$ – Ben Crowell Aug 7 '13 at 21:52
  • $\begingroup$ @BenCrowell: was the moon phase locked when it hardened? I would assume the answer is no. $\endgroup$ – Jerry Schirmer Aug 7 '13 at 22:07
  • $\begingroup$ I don't think it matters. Only for very small bodies such as asteroids do you get shapes that aren't equipotentials. Rocks can roll downhill, the body can exhibit plasticity, etc. $\endgroup$ – Ben Crowell Aug 7 '13 at 22:14
  • $\begingroup$ @BenCrowell: the question is an equipotential relative to what. If the moon was not phase locked when it hardened, then it's initial configuration would have been an equipotential relative to the Earth's potential, but then that shape would move relative to the Earth. Although maybe that would also determine which part of the moon faces the Earth in the end. I don't think the answer is clear or trivial, though. $\endgroup$ – Jerry Schirmer Aug 7 '13 at 22:55
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    $\begingroup$ The moon is guaranteed to be an equipotential in its current state. Bodies that big conform themselves to equipotentials, and if the equipotential changes, the body's shape changes on a relatively short timescale. It doesn't matter if it hardened, because hardening isn't absolute. Only very small bodies are rigid enough to avoid conforming to the equipotential on relatively short time-scales. $\endgroup$ – Ben Crowell Aug 7 '13 at 23:34

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