"Aztec50" <fasgard66@[EMAIL PROTECTED]
> wrote in message
news:e8f0a56a-3117-4e06-9b9b-929ef6e47fb6@[EMAIL PROTECTED]
> Space****p A is at L4.
>
> Space****p B is at L5.
>
> They are racing to reach the Moon. Each has the same engines, and
> will go at the same speed.
>
> Question #1: which will arrive first? Question #2: if I didn't have
> much energy (and these ****ps weren't the supersleek dragsters they
> undoubtedly are), what would be the optimal/energy-efficient orbital
> "route" from each libration point to the Moon?
>
> I.e., what I'm trying to figure out here is whether the fact that L4
> is "ahead" of the Moon's position in orbit is an advantage or
> disadvantage vis-a-vis L5's position "behind" the Moon's position in
> orbit. My first thought was that Space****p A will win easily (vis-a-
> vis Question #1), because while it moves toward the Moon, the Moon is
> moving toward it. But then I reflected that Space****p A, in going
> "backward", still has to compensate for the forces that are propelling
> it "forward." And then I reflected that Space****p A, in entering a
> retrograde orbit, will actually start to move in toward the Earth.
>
> And then I decided to put the question to all your brainiacs out there
> in the hopes that someone will actually know the answer to this #$#
> thing.
>
> Just so we're all on the same page, this is the libration point
> topography. (This map has the Sun at the center, but substitute the
> Earth for the Sun, and the Moon for the Earth, assume a
> counterclockwise lunar orbit, and we're golden.)
>
> http://en.wikipedia.org/wiki/Image:Lagrange_points.jpg
>
> thanks!
>
> FA
Assuming that as stipulated, using the least fuel is im****tant, i.e., the
rocket jockeys are not able to accelerate continuously but need to coast
most of the time, and the orbits are circular (which they aren't, but it
simplifies the thinking):
The technique to move from a Lagrangian point to the Moon (60 degrees away
in the orbit) from L4, ahead of the Moon, would be to accelerate and move
out to an orbit further from the Earth, where orbital period is longer,
then
wait for the Moon to catch up in longitude, then move back to rendezvous
(and execute an orbiting manoeuvre and powered landing like Apollo). For
the L5 space jockey the manouvre involves a similar amount of acceleration
to drop down to a lower, faster orbit, and catch up with the Moon.
Disregarding orbital eccentricity questions, accelerations caused by the
Moon during the coasting phases, etc, I would think the L5 to Moon time
would be somewhat less for the same amount of fuel, optimally expended,
because the difference in relative angular speeds would be greater. In
other words, catching up from "below" would be faster than "ascending" and
waiting for the Moon to catch up.
--
Mike Dworetsky
(Remove pants sp*mbl*ck to reply)
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