the angular momentum of the moon is
m*v*r=(1 lunar mass)*(86400 km/day)*384,400 km = 3.32 * 10^10
the angular momentum of Earth is
(2/5)*m*v*r=(2/5)*(80 lunar masses)*(40,000 km/day)*6380 = 8.166 * 10^9
for the Earth moon system as a whole, the sum of the Earth's and the moons angular momentum must be constant.
at the present time the vast majority of the angular momentum of the Earth moon system is in the moon.
If the moon was very close to the Earth then the vast majority of the angular momentum of the Earth moon system would be in the earth.
we assume as a first approximation that when the moon was very close to the Earth that the Earth spun around 4 times as fast and a day was 5.9 hours
then we need to determine where the moon would be if it were tidally locked at 6 hours
period of an orbit:
T^2 = Kr^3
t for an orbit just above the surface of the Earth is about 90 min
so if the moon were tidally locked with the Earth at 5.9 hours then
(6 hours)^2/r^3 = (1.5 hours)^2/(6380 km)^3
r^3 = (5.9/1.5)^2 * 2.597*10^11
r = (16*2.597*10^11)^(1/3)
r = 16075 km[/color]
r = 2.5 Earth radii
At this orbital radius the angular momentum of the moon at that time would be:
m*v*r=(1 lunar mass)*(5*2pi*16075/day)*16075 = 8.1182 * 10^9
therefore the moon would still have significant angular momentum.
the Earth's angular momentum at that time would be:
present angular momentum of Earth + present angular momentum of moon - angular momentum of moon at that time
8.166 * 10^9 + 3.32 * 10^10 - 8.1182 * 10^9
which is about 4 times what it is now
so the Earth would indeed be spinning 4 times as fast
A day would indeed be 6 hours
thus agreeing with our assumed approximation
the roche limit for the moon is
fluid=18,261 km
rigid=9,496 km[/color]
conclusion:
the moon could never have 'touched us'.