Not much velocity required for the return
(I sent the following to the 'corrections' link, but it looks as if there's some
interest here, too...)
If we'd had to get to 11.2 km/s to leave earth, then do it all over again to
get back, I don't think anyone would have gotten to the moon and back.
The escape speed for the moon is a mere 2.4 km/s. What you'd want to do,
assuming a minimum-fuel maneuver, is to apply a bit more speed so that,
when you go flying away from the moon, you're departing at its orbital
speed around the earth, which is about one km/s. That is to say, you leave
the moon, but kill your speed relative to earth in the process, so it's
as if you're just "hanging" 385000 km or so above the earth.
Since you are now at rest relative to the earth, you'll fall back toward
it, re-enter at 11.2 km/s, and someone in China will pick you up.
I'll spare you the derivation, but it works out to sqrt(2.4^2 + 1^2) =
sqrt( 6.76) = 2.6 km/s.
Oh, %^*! it... that assumes you're lifting off from the moon and have
to get to escape speed plus the extra bit to cancel out the moon's orbital
speed around the earth. _This_ object is in orbit around the moon, and
is already moving at 70% of escape speed, or 1.68 km/s. So we just need
to add 0.92 km/second, and about three days later, we fall back to earth.
This, incidentally, is a bit like the way the Apollo missions worked.
They had to get up to about 11.2 km/s to get as high as the moon. If they
hadn't done anything, the moon would have swept by them at about a km/s.
Instead, they fell into the moon's gravity well, speeding up to about
2.6 km/s at perilune or periselene or whatever it's called. They would
have gone right by it, except they decelerated by 0.92 km/s, and ended
up in lunar orbit, circling the moon at a low-lunar-orbit speed of 1.68
km/s.
Next, two guys crawled into the LEM, fired its engines to cancel
out that 1.68 km/s, walked around on the moon, got back into the LEM,
and fired the ascent engine to recover the 1.68 km/s so they could go back
to the command module. (Which was a nice solution; it meant there was
a heck of a lot of hardware to get back to earth and re-enter that
could stay in lunar orbit. You didn't have to use up 3.36 km/s worth
of fuel ferrying it down to the surface and bringing it back to lunar
orbit. I must confess, I've never run the numbers before and had
missed the importance of this point.)
Anyway. The return trip was just the same thing in reverse --
speed up by 0.92 km/s to escape the moon and kill speed relative to
the earth; wait to drop back down to earth, accelerating to 11.2 km/s
as you do so; then kill that speed... fortunately by atmospheric
braking, rather than needing more bucket-loads of rocket fuel.
-- Bill