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Wikipedia pegs the mass of Earth at $5.972\cdot 10^{24}\,\text{kg}$ or $5.972\cdot 10^{21}\,\text{metric tons}$.

Assuming Earth accumulates approximately $30,000\,\text{metric tons}$ annually. To illustrate, since man's first known predecessor walked earth about $4.5\cdot 10^6$ years ago, it has gained about $1.35\cdot 10^{11}\,\text{metric tons}$.

Similarly Luna gains a proportional amount of mass too.

Does gain in mass of the participating bodies bring Earth/Luna tidal-lock closer ever faster as the masses increase & time goes by? Can the acceleration (if my assumption is valid) be approximated?

Further, once tidal lock is achieved, if the bodies continue to gain mass are they likely to eventually pull each other apart and coalesce again?

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    $(3 \times 10^{4}) / (6 \times 10^{21}) = 5 \times 10^{-16}$ which implies twenty trillion years to make a 1% difference... That's around 7000ish times the current age of the universe. – dmckee --- ex-moderator kitten Oct 15 '12 at 18:46
  • In the above equation, I understand (3 \times 10^{4}) is the annual gain, and (6 \times 10^{21}) is the calculated mass. Could you elaborate on the rest please, or perhaps post it as an answer? – Everyone Oct 15 '12 at 19:06
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    I've just restated some of your own numbers. $5\times 10^{-16}$ is the annual fractional gain in mass in this era. Then I asked "How long would that have to go on to make a 1% difference in the mass of the Earth?", and got an answer many times the age of the universe. That was implicit in what you did above: $10^{11}$ tons seems like a big number, but it is less than one part in 10 billion of the mass of the Earth. Those kinds of ratios matter. – dmckee --- ex-moderator kitten Oct 15 '12 at 19:10
  • Understood. But it's not just the mass of the one body, both gain mass ... and probably gain more mass faster as they grow more massive. – Everyone Oct 15 '12 at 19:13
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    It doesn't really matter when the ratios are that steep. The problem is that for any acceleration to be noticed you have to make a noticeable difference in the starting quantity, which we have already shown takes a very long time. It is certainly worth spending some time puzzling over this, if only to convince yourself that you can learn something from this kind of analysis. Think of it as a compound interest problem where the interest is taken down to 0.000000000000001%. – dmckee --- ex-moderator kitten Oct 15 '12 at 19:28

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