lew@ihuxr.UUCP (10/26/83)
The "natural" acceleration of a geostationary object at a distance r from the earth is given by: a = a(gravity) - a(centrifugal) = g*(Re/r)^2 - w^2 * r ; Re = 6.366e6m ; w = 2pi/24hrs ; g = 9.8m/sec2 a = w^2*R * ( r/R - (R/r)^2 ) ; R^3 = g*Re^2/w^2 At r = R = 6.6*Re this expression is zero. An anchor station for a space elevator must provide tension in the elevator cable to hold up the cable and the elevator cars, so we must have r > R. The following table shows the requirements for the mass of the anchor station as a function of r. The first column is the distance, r, normalized by the geostationary orbit radius (R). The second column gives the natural acceleration at r, normalized to g, (9.8 m/sec2), the third column gives the natural acceleration of a cable anchored at r whose other end is at Re. The fourth column gives the mass of the anchor station required to hold up the cable, normalized by the mass of the cable (several thousand tons at least). Finally, the fifth column gives the extra mass required to support a unit mass at Re; this is just the reciprocal of column two. The values of column three were obtained by integrating over the expression for 'a' from Re to r, and dividing by (r-Re). This gives the expression: A = w^2 * ( (r+Re)/2 - R^3/(r*Re) ) Note that at r=3.5*R, the cable becomes self-supporting, but it still takes a lot of extra mass to balance a car near the ground. r/R a/g A/g -A/a g/a 1 0 -.188 inf inf 1.5 .024 -.081 3.4 41.6 2 .040 -.051 1.28 25.1 2.5 .053 -.031 .566 18.8 3.0 .065 -.014 .219 15.2 3.5 .078 -.0015 .02 12.8 I think these figures are rather sobering, but of course, they may be open to question, if anyone wants to check them out. Lew Mammel, Jr. ihnp4!ihuxr!lew