lunar poles, orbits, trajectories

> Forces applied to an orbit in the direction of motion have the
> effect of raising the orbit on the opposite side of the body
> orbited.

If that's not what I said, it was a typo.

> And one inevitable consequence of launch from the lunar surface
> would be that perilune would be at the location of the launcher.

That sounds as though you're talking about the orbit of the payload
as it leaves the lunar surface. I'm talking about the orbit of the
catcher, and how the catch and re-launch would affect it.

> But why are we working so hard to make catchers in relatively low
> lunar orbits work in the first place?

So that the catcher will keep coming back into position to receive
loads from a mass driver at one of the lunar poles, while not being
constrained by the requirement for nearly-horizontal launch from the
surface. The mass driver attached to the catcher can then throw
payloads any directions, as long as the combined effect leaves it in
an orbit the launcher can throw to.

--- In spacesettlers@yahoogroups.com, "Combs, Mike"
>
> From: spacesettlers@yahoogroups.com
> [mailto:spacesettlers@yahoogroups.com] On Behalf Of Dan Wylie-Sears
>
> > The way I'm trying to visualize it, it looks as though
> > excess speed at launch would have a more complicated
> > effect. The only thing I can think of that just has
> > that simple effect would be exerting a force in the
> > direction of motion at perilune.
>
> Actually, what we need is a force applied at apilune. Forces
applied to
> an orbit in the direction of motion have the affect of raising the
orbit
> on the opposite side of the body orbited. And one inevitable
> consequence of launch from the lunar surface would be that perilune
> would be at the location of the launcher.
>
> So we would need two launchers, on opposite sides of the moon.
And I'm
> not even sure that would work. I'm thinking a trajectory with
extra
> energy would be a higher trajectory which would pass over the orbit
> being aimed at, rather than what I think you're visualizing which
is a
> trajectory passing through the orbit with excess velocity. But
like
> you, I'm not equipped to do the math.
>
> But why are we working so hard to make catchers in relatively low
lunar
> orbits work in the first place? Unless our market is in low lunar
orbit
> (unlikely, I would think) we should ideally be wanting to use our
EM
> launcher to get our payloads as far up the lunar gravity well as
> possible. The L-2 point is pretty high up on the moon-Earth
gravity