From: Hal Finney (hal@rain.org)
Date: Tue Dec 31 1996 - 18:52:12 MST
From: Michael Lorrey <retroman@tpk.net>
> Your numbers are wrong. The cars would be going at their very slowest at
> their highest altitude (read keplers law) and fastest at the ground. I
> am actually concerned that they wont be going fast enough at geosynch,
> as they need to be going slow enough to fall back to earth, rather than
> staying in geosynch, even though at apogee they will be tangential to
> the geosynch orbit.
Yes, sorry, you are right about my numbers. I redid my calculations
and got a lower differential, but I will wait and see what you get.
I do understand that the cars in the transfer orbit are going slower
than the geosynch orbit at apogee. But I miscalculated the speed of the
geosynch station. I now get a value of about 2800 mps. The differential
clearly can't be greater than that.
My new improved calculations suggest that orbital velocity at perigee
is 10000 mps, and since apogee is about 6 times larger, by Kepler's law
the speed must be about 1/6 as much or about 1700 mps. This would lead
to a differential of about 1100 mps, with the station going faster than
the cars in its orbital direction.
> Opening my _Introduction to Space Dynamics_ by William Tyrrell Thomson,
> I see that my deduction is right. Essentially, the cars will be
> following a Hohmann transfer orbit, with lower orbit tangential point at
> earth ground (perigee) and upper point at Geosynch station, where the
> velocity of the cars would be less than the velocity of the station, so
> the station would catch up to the cars as they reach their peaks, add a
> little momentum to keep them, and then drop them off the back, regaining
> the momentum when they are returning a car. If you drop off a car every
> time you pick up a car, you lose or gain zero momentum at the station.
> Give me a little time and I'll give you some accurate velocity
> calculations. If your equations are right, I think that you just have
> apogee and perigee mixed up.
Actually I am not sure how the geometry will work in terms of the station
"catching up" to the cars. During the transfer, the station and the
earth will have rotated to new positions. That means that each launched
car is in a different elliptical orbit (in the non-rotating frame). You
will need to launch the cars from a point counter-clockwise from the
point opposite to the station, and receive them from a point clockwise from
the opposition point. How far these offsets are will depend on how long
it takes to reach geosynch orbit, how much the earth rotates in that time.
Hal
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