Re: Orbital Towers.

From: Michael S. Lorrey (
Date: Thu Mar 02 2000 - 21:46:22 MST

Spike Jones wrote:

> Darin Sunley wrote:
> > It was my impression that if one wanted a reasonably circular orbit, the
> > only altitude one could jump into orbit from is the altitude of
> > goesynchronous orbit.
> Thats a good question, one that I want to work on. If one went
> up about an earth diameter on a cable and let go, what would
> be the shape of that orbit? Also, I have sharpened my calcs a
> bit and found that the whole question of climbing a cable and
> letting go to get into a minimal orbit might be wrong: I found a
> flaw in my reasoning today. I have probably already been
> pummeled by Doug Jones on this, but havent gotten that far
> in my messages. {8^D spike

12000 miles from the center of the earth * 2 * 3.14159 / 24 hours = 3141.59 mph
downrange speed at apogee (isn't that interesting) sitting on top of the

12000 miles * 5280ft/mi = 6960000 ft
 6960000 ft = (32 ft/s^2)(t^2)/2
t^2 = 435000 s^2 = 11 minutes falling time from top of beanstalk at 1 g
 11 minutes gets you about 18% of 3141.59 miles laterally, or about 600 miles
downrange. Better get yourself something to keep accelerating you. Of course I
could be way off here, doing back of the envelope calculations with equations
and constants off the top of my head. Somebody please tell me how wrong I am....


Michael S. Lorrey Owner, Lorrey Systems ArtLocate.Com Director, Grafton County Fish & Game Assoc. Member, Extropy Institute Member, National Rifle Association "Live Free or Die, Death is not the Worst of Evils." - General John Stark

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