I think you need to check your math here. A speed of 5.6 cm/yr for
4.5 x 10^9 yr corresponds to a distance of 252,000 km (significantly less
than 384,403 km). To cover a distance of 384,403 km in 21.53 X 10^6 yr
requires an average recession speed of 17.85 m/yr (which is 319 times faster
than the actual current recession speed quoted by Alan).
The current recession speed of the moon is significantly greater than the
average recession speed for the past billion years or so. This is because the
current configuration of the ocean basins, gulfs, coastlines, bays, etc. allow
for a greater than average amount of frictional dissipation of tidal energy
because a number of oscillation modes of the oceans, seas and bays are close
to being in resonance with the fundamental driving frequency. In the past
other continental and ocean shapes would not have had a resonant coupling to
the driving frequency causing less a weaker tidal response, causing less
frictional dissipation, causing less angular momentum transfer to the moon's
orbital motion, causing a slower lunar recession speed. There is, however a
compensating effect which tends to speed up the recession speed, esp. in the
very distant past. When the moon was closer to the earth its greater
gravitational field gradient in the vicinity of the earth would create a
greater tidal stress on the earth and raise higher tides even if the basins
are off-resonance. Since the strength of the tidal stress exerted on the
earth is inversely proportional to the cube of the earth-moon separation
distance, we see that this effect becomes most significant when the moon was
significantly closer than it is now.
David Bowman
dbowman@gtc.georgetown.ky.us