mcgeer@JI (Rick McGeer) (01/08/86)
>Date: 1986 January 08 06:27:39 PST (=GMT-8hr) >Message-Id: SU-IMSSS.REM.A132546146163.G0365 >From: Robert Elton Maas <REM@imsss.stanford.edu> >To: amdcad!cae780!weitek!mmm@ucbvax.BERKELEY.EDU >Cc: SPACE@mit-mc.arpa >Subject: If you don't climb straight up it's a much longer trip to orbit > >MT: Date: 27 Dec 85 21:36:45 GMT >MT: From: amdcad!cae780!weitek!mmm@ucbvax.berkeley.edu (Mark Thorson) >MT: Subject: Re: Skyhooks, tethers, and kites. >MT: Sender: usenet@ucbvax.berkeley.edu >MT: To: space-incoming@s1-b.arpa >MT: With all the interest in skyhooks and tethers, I'm inspired to ask "How >MT: about a kite?" ... But the kite string would be tangent to the >MT: earth, hence you could WALK into space (its only a hundred miles to LEO, >MT: any good athlete should be able to make it in a few days). > >I beg to differ. It's 100 miles up, but going out tangentially (on one >leg of a right triangle whose hypotenuse goes from the center of the >Earth to the orbital point, with the right angle at the tether-launch >point = north pole) it's about SQRT(4100**2 - 4000**2) = SQRT(810000) >= 900 miles. Some athletes indeed may be capable of walking 900 miles >with all the food and water and air they need on their back, but not >many. Your statement above "YOU could walk ..." is probably untrue for >any reader of this list except possibly Gene Salamin. (Hey Gene, care >to respond, could you or anyone you know?) > >(Pardon tardy reply but IMSSS has been down almost continuously from > Dec 30 thru Jan 06 and I'm just now catching up on replying to mail.) Well, all of this started me wondering just how hard a climb it would be. Like all good physics calculations, we ignore atmospheric resistance, storms, etc, and think of only gravity... If we assume the centre of the earth is 6500 KM from the pole, we have the distance from the centre of the earth to the traveller out on the kite string is sqrt(6500^2 + x^2), where x is the distance from the north pole along the string. The force on the traveller = cos(mu) * GMeM/(6500^2+x^2), where mu is the angle between the kite string and the radius vector to the centre of the earth. If x = 0, mu = pi/2; otherwise mu = arctan(6500/x); at 160 KM out, mu ~= 1.34. (about 78 degrees). dF/dx is then (-K sin (mu) mu' (x^2+6500^2) - K cos (mu) 2x) /( x^2 + 6500^2). mu' turns out to be -6500/(6500^2+x^2), so dF/dx turns out to be: K/(x^2 + 6500^2) * (6500 sin(mu) - 2x cos(mu)). Since I'm interested in the maximum of F, we look for the 0, and hence get to cancel out inconvenient factors., we get: 2x cos(mu) - 6500 sin(mu) = 0, or 2x = 6500 tan(mu) = 6500^2/x => 2x^2 = 6500^2, or x ~= 5000 KM, or about 3000 miles. This is comfortably outside the x = 1500 KM distance of the kite, so the maximum occurs at the kite itself, where the force is equal to .22 * G * Me* M /(6500^2+1500^2). At this point, the force is 1.97 M. How much is that? It turns out to be the equivalent of climbing a 77 degree hill on earth. Not much help from gravity here, and I doubt if even Gene Saliman could climb that hill. Rick