REM@MIT-MC@sri-unix.UUCP (07/26/83)
From: Robert Elton Maas <REM @ MIT-MC> Sigh, I was hoping somebody on this list had already computed it or had access to NASA info whereby somebody there had. But since everybody seems to be guessing, here's my try at actual calculation: I'm assuming Earth's radius is 3800 miles and Geosynch orbit is 22,300 miles from the surface, or 26,100 miles from the center. I'm assuming the Earth's equator is circular and STS orbits 100 miles up (50 miles is the official start of "outer space" and STS wants to be up a bit higher). If any of these figures are wrong please send me corrections or recompute it yourself and send the calculations. Draw tangent from TDRS in geosynch to Earth, and radius lines to both tangent pint and TDRS, yielding right triangle. THETA = angle of half-Earth as viewed from TDRS. sin(THETA) = 3800/26100 = 0.14559, THETA = 8 deg 22 in = 8.367 deg. PHI = angle between tangent point and TDRS as viewed from center of Earth = 90-8.367 deg = 81.633 deg. Thus that portion of surface that can see TDRS (assuming no atmosphere) is 81.633*2/360 = 0.4535, a little less than half as expected. Now extend the ray from TDRS to tangent point past the tangent point to intersect the orbit of STS, and draw radius line to this new point. The right triangle including tangent point, STS point, and center of Earth, has a center angle OMEGA with cos(OMEGA) = 3800/3900 = 0.97435, OMEGA=13 deg. Adding OMEGA and PHI we get 94.633 deg, so STS can see TDRS (assuming atmosphere transparent to microwave) 94.633*2/360 = 0.5257, i.e. more than half the time. Lowering STS to 50 miles, cos(OMEGA) = 3800/3850 = 0.98701, OMEGA = 9 deg 15 min = 9.25 deg, OMEGA+PHI = 90.883, so even in very low Earth orbit STS would just barely exceed 50% of the time seeing TDRS. I've assumed STS is in equatorial orbit. I don't think the inclination affects whether 50% is exceeded or not, just by how much it's exceeded or not-reached. (Argument, consider the point of STS orbit nearest TDRS, even though it isn't at the same longitude as TDRS is. Forget about latitude and longitude here. Now consider the points exactly one quarter of an STS orbit in each direction. These points are in identical places in space regardless of inclination of STS's orbit providing we change the inclination in only the direction that affects its visibility from TDRS. The above calculations show these two points are slightly above the Earth-horizon as viewed from TDRS.)