Re: Cavorite Travel Times

On Tue, 02 Dec 2003 02:01:05 -0800, pervect <[EMAIL PROTECTED]>
wrote:

I had the wrong model last time.  The problem has to be worked from an
inertial frame.

For the Earth-moon system, if you ignore the sun for the time being,
the proper point to consider is the center of mass of the earth-moon
system.

Looking at it from this point of view, you are nearly (but not quite)
stationary near the COM point, and the moon is circling around this
point.

Because of the cavorite you won't feel the earth's gravity,  but the
moon will.  For ease of visualization, consider the moon to be mounted
on a giant invisible arm, like a planetary orrey, and constrained to
move in (roughly) a circle about the COM point.

When you shut off the earth's gravity to your spaceship with the
cavorite, the moon continues to circle around this point.

The quickest route to the moon woulldn't be to leave the cavorite open
all the time.  For the first half-month or so, your vector would have
components in the right direction, but then you'd start to slow
yourself down again.  By the previous calculation the trip would take
at least 2 months - I.e. if the moon weren't a moving target, you'd
fall to its surface in 2.25 months under its gravity.  See the
previous post for details, this figure was calculated as 1/4 the
orbital period of an orbit around the moon with the major axis of the
orbit being the earth-moon distance.

Matching the moon's orbital velocity is still going to be a problem.

If you want to include the sun in the analysis (and assuming you shut
off the sun's gravity with the cavorite), build a bigger orrey, with
the sun in the center, a 93 million mile arm to the Earth-moon system,
and a 250,000 mile arm to the moon.

The effects of the sun's gravity will be significant, because even if
your trip takes only 2 months (the minimum) you'll have moved 1/6 of
an orbit around the sun