Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/5/84; site ssc-vax.UUCP Path: utzoo!linus!decvax!tektronix!uw-beaver!ssc-vax!eder From: eder@ssc-vax.UUCP (Dani Eder) Newsgroups: net.space,net.columbia Subject: Re: Sixty Cents a Pound? Message-ID: <200@ssc-vax.UUCP> Date: Mon, 12-Nov-84 13:49:37 EST Article-I.D.: ssc-vax.200 Posted: Mon Nov 12 13:49:37 1984 Date-Received: Tue, 13-Nov-84 07:04:59 EST References: <1062@inuxc.UUCP> Organization: Boeing Aerospace Co., Seattle, WA Lines: 42 > I heard the end of a segment on NPR this morning where > > to escape velocity today. However if we were to switch to > electromagnetic launchers (mass drivers?) the cost of > electricity required to accelerate a pound of material to > escape velocity would be $0.65. > > Do these numbers sound > reasonable to the space wizards on the net or are they science fiction. > What are the problems with electromagnet launches, i.e. payloads must > be designed to withstand 3000000G and must be launched in a restricted > direction, etc.etc.? > Fred Mendenhall > > The velocity for circular orbit at 296 km (the nominal Shuttle orbit) is 7728 meters/second. The kinetic energy of a payload moving at this velocity is (1/2)(mass*velocity*velocity). If you divide by mass, you have per kilogram energy requirements. This is 29.861 MJ/kg. Now add the potential energy of raising a kilogram from ground to 296km, which is (acceleration of gravity*height)= 2.903 MJ/kg. Total is 32.764 MJ/kg. One kiloWatt-hour(kWh)= 3.6 MJ, so we have 9.1 kWh/kg. Around here (Pacific Northwest), electric rates are 3.5 cents/kWh, so that's 32 cents/kg, or 14 cents/pound. Since you have drag losses going through the atmosphere, inefficiencies in the accelerator, and payments on the money you borrowed to build the accelerator, 65 cents/lb is in the right ballpark. Note that to make it pay off, you need to have a lot of traffic. An acclelerator is like an oil pipeline, efficient but high volume. As for acceleration required, a=(velocity*velocity)/(2*distance). So if you have 20000 meters to accelerate in (as in the west slope of Hawaii Island) and you want a muzzle velocity of 6000 meters/second, which is 3/4 orbital velocity, then a=900 m/s*s= 92 g's. You can use a small rocket to make up the remaining velocity to orbit. Going at less than orbital velocity reduces the drag problem, and launching off Hawaii gets you up to 12000 feet at the muzzle, which helps because you skip the thickest part of the atmosphere. To further reduce air drag, you make your vehicle long and thin, like a telephone pole. Dani Eder / Boeing Aerospace Company / ssc-vax!eder / (206)773-4545