Copyright © 1996 by Andy Goddard ( a DOT goddard AT strath DOT ac DOT uk). All Rights Reserved.
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There seemed to be a little confusion on the list recently about "what keeps the Beanstalks standing". I'm sorry about the length of this posting, but there's some useful equations in it for referees and interested others who want to investigate orbital physics in a bit more detail.
Firstly, there are two main forces at work on the orbital tower. There's the force due to gravity that all bodies feel:
Fg = ( G * M * m ) / ( r * r )
And then there's the force that all bodies travelling in a circle undergo:
Fc = m * w * w * r
Where:
If you substitute the options for omega in the Fc equation you get:
Fc = ( m * 4 * pi * pi * r ) / ( To * To )
and
Fc = ( m * v * v ) / r
Now for ANY stable circular orbit (including Gateway on the beanstalk), Fc is exactly equal to Fg, and using the above equations for Fc and Fg you can produce a range of results. For example:
r = Cube root of ( ( To * To * G * M ) / ( 4 * pi * pi ) )
In the above equation, putting To = 86164 and the given values for G, pi and M will give you the orbital radius of geostationary satellites. The beanstalk is a static structure, however, and Fc equals Fg ONLY at Gateway. (Starships and shuttles can independantly orbit the Earth at this altitude without lagging behind or overtaking the station.) At all altitudes, the velocity on the tower is:
Vt = 2 * pi * r / Tt
Tt = Rotation period of the tower
While the velocity in any stable, circular orbit is given by:
Vo = Square root of ( G * M / r )
For points below Gateway, Vt < Vo, and objects that fall off the Beanstalk will not have the energy ("aren't going fast enough") to occupy a circular orbit. The lower down the tower that they come off, the more elliptical their orbit becomes, with the highest altitude of this ellipse equal to the height they come off at. At a particular detachment height the lowest point of their now quite elliptical orbit will graze the planet's atmosphere, and... (I can imagine that on the passenger trips up the Beanstalk, this point would be marked in some way -- perhaps rather like the Mach meter on board Concorde jets... "you're now safe from hideously burning up in the atmosphere" signs flashing on and off...)
Above Gateway, the situation is reversed as Vt > Vo. Objects coming off here are launched into space with more energy than that needed for a circular orbit. Now the lowest points of these elliptical orbits are equal to the height at which they come off the tower, and at some point this "launch velocity" will exceed the escape velocity for this altitude, so the detached bodies will no longer be orbiting the Earth.
I think that it's intuitively obvious that the tethered asteroid, which wants to be in one of these (higher energy) elliptical orbits, is providing a force along the tower structure that can be made to be more than or exactly equal to the sum force of gravity on the tower's components. In "Red Mars", this tension on the Martian orbital tower is used (like a violin string) to oscillate the tower structure at a particular frequency to ensure that the orbits of Phobos and Deimos "miss" this far-slower moving structure. The tension's value can be set, of course, so that it's not necessary to have a mountain of tethering mass at the base of the tower (as the excellent recent engineering posting pointed out -- sorry I don't have your name here), and that in the Beta Canum case, the tower can slowly "liftoff" out of the atmosphere when severed at the base.
Phew! Next time, Hohmann transfer ellipses... (only kidding!)