lunar poles, orbits, trajectories

Thank you for the reply. This in particular makes sense of it:

> B. Passing through at an angle sufficiently close to right angles
to the
> line drawn through the Earth and moon such that the biggest part
of the
> small remaining momentum is expended along this vector. The entire
> operation is contingent on the unique stability properties of the
L-2
> (or L-1) point.

It sounds as though if some greatly improved catcher technology came
along, it could change things fairly dramatically.

> We also can't launch to a catcher orbiting the moon without the
> catcher losing altitude with each capture, which would have to be
> made up with some kind of thruster (which would probably wreck the
> economics).

Isn't most or all the mass being sent on somewhere else via a mass
driver attached to the catcher? I was assuming that it depended on
that to make up for a lot, since you can throw stuff any direction
you feel like, even though it means some of what goes off the moon
can't go to your preferred destination.

--- In spacesettlers@yahoogroups.com, "Combs, Mike"
>
> From: spacesettlers@yahoogroups.com
> [mailto:spacesettlers@yahoogroups.com] On Behalf Of Dan Wylie-Sears
>
> > In the thread titled "Re: solar power sats", people have
> > been saying you can't get mass-driver payloads to L1 and
> > L2 from the lunar poles.
> >
> > Why not? What are the constraints? Would a lunar mass-driver
> > have to launch nearly horizontally? Nearly vertically?
>
> Nearly horizontally. I would question if a near-vertical launcher
could
> be built. If we're talking 1,800 G's, maybe one could dig a
tunnel or
> construct a tower long enough for the main part of the launcher.
But
> the assumed accuracy of the mass-driver is contingent on a series
of
> electrostatic deflection plates, and those are miles downrange of
the
> main part of the launcher.
>
> > Is
> > the maximum speed you can launch with only slightly greater
> > than lunar escape velocity, either because of outright
> > limitations or because of severely decreased efficiency at
> > higher velocity?
>
> I corresponded with somebody who could do the math who concluded
that
> you need ever-so-slightly greater than escape velocity.
Intuitively, I
> would have guessed slightly less.
>
> > It looks on this diagram
> > http://map.gsfc.nasa.gov/m_mm/ob_techorbit1.html
> >
> > as though you can see both L1 and L2 from the pole, sitting on
> > the horizon. If you were launching your payloads with huge
> > speed, so that the effects of gravity and the motion of the
> > moon were both negligible, you could shoot them straight
> > through along line-of-sight.
>
> Line-of-sight doesn't enter into it. The illustration I posted
earlier
> shows this best: http://ssi.org/assets/images/Ch08p150.gif
>
> > At the next level of not-quite-so-crude, I still don't see
> > any absolute reason why a trajectory from a lunar pole couldn't
> > pass through L1 or L2. But I'll leave that until I see what I
> > get in response to this.
>
> Just passing through the point is insufficient. What's required
is:
>
> A. Passing through the point with nearly all of the initial
velocity
> gone due to the ascent up the gravity well.
>
> B. Passing through at an angle sufficiently close to right angles
to the
> line drawn through the Earth and moon such that the biggest part
of the
> small remaining momentum is expended along this vector. The entire
> operation is contingent on the unique stability properties of the
L-2
> (or L-1) point.
>
> We can't launch to just any old Lagrange point. Nor can we launch
to
> any arbitrary point because action/reaction delta-V's in the
catcher
> would have to be overcome. We also can't launch to a catcher
orbiting
> the moon without the catcher losing altitude with each capture,
which