Re: Cavorite Travel Times

On 2 Dec 2003 13:23:10 -0800, [EMAIL PROTECTED] (Jaak Suurpere) wrote:



>If you start at the equator and pull on the cavorite shutters all
>around you, your ship will start to rise spontaneously because the
>force of gravity is eliminated, but that of the centrifugal
>acceleration will persist. In other words, you take off at 465 m/s,
>though in a horizontal direction.

Good point.  You'll rise slowly relative to the earth (at least
initially), but as you gain altititude, the horizontal nature of your
velocity will become apparent.

>
>With that speed, it takes you about 10 days to reach Moon. Of course,
>you should  account for the Moon's revolution while you're travelling.
>
>When you're near the Moon, next step should be to get to the correct
>orbit... You will be able to approach or, if desired, pass the Moon at
>the relative speed of about 1100 m/s. Can anyone work out the details
>of the mooning maneuver? What you want is the relative speed of zero.

Is 1100 supposed to be the moon's orbital velocity?  The planetary
data fact sheet (NASA) puts is slightly lower, at 1023 m/s.

If you approach the moon at the right angle, you could approach as
slowly as 500 m/s or so, but at this point I don't see how you can
match the orbital velocity.

>
>N. B! you cannot use cavorite to take off from the near side of the
>Moon. By closing the curtains on the Moon side you will merely
>decrease the pressure you exert on the Moon, but it remains positive:
>any celestial body opposite will be attracting Moon, anyway, and the
>centrifugal force is always towards the Moon.

I don't think this is right.  You'll have the usual "centrifugal
force", plus perhaps some upward pull from the Earth's tide.

However, your whopping 4.6 m/s rotational velocity at the moon's
equator is not going to help you out significantly, unlike the 450 m/s
you got from the Earth's rotation.