Doug... wrote in message ...
Ummmm... OK, I'm not visualizing the setup very well, here. For a
tether to drop straight down from an orbiting craft, that craft would
have to be in selenosynchronous orbit, wouldn't it? And 20 km is a LOT
less than the height of a SSO, isn't it?

From a vehicle orbiting 20 km above the surface of the Moon, how do you
propose to completely null out the orbital vehicle's velocity from the
scoop end of the tether?

We are talking about a rotating space tether. It is rotating
in the plane of its orbit with the part closer to the moon going
backwards relative to the orbital motion. Say the orbital motion
is 1.6 km/sec and the tip speed is 1.6 km/sec, so near the surface
of the moon these cancle. On the high side they add together, so
the tether can pickup something down low and then toss it fast up high.

It does not drop exactly straight down, but for a guy sitting on
the moon who only sees it when it is nearby it will mostly
seem to be coming down from one side and then going up the
other.

This general idea is called a "momentum exchange tether". You
will also hear "bolo" and "rotovator" with rotovator being the
case that touches a moon/planet. I like "rotating tether" as
it seems clearer. See:

http://tethers.com/MXTethers.html

My simulator is a Java applet, so it should be easy to run.
You can just click on samples and run them to help you
visualize what is going on. It is at:

http://spacetethers.com/spacetethers.html

-- Vince

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Vincent Cate Space Tether Enthusiast
http://spacetethers.com/
Anguilla, East Caribbean http://offshore.ai/vince
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~

You have to take life as it happens, but you should try to make it
happen the way you want to take it. - German Proverb