


I thought this might be of some interest, as it pretains to weather and I just finished a term paper on it. We will use Kepler's third law rs3/Ts2 = rm3/Tm2 where Ts = peroid of orbit of Satellite = 1 day Tm = period of orbit of Moon = 27.3 days rm = distance from center of Moon to center of Earth = 3.84 x 108 m and rs = distance from Satellite to center of Earth Solving: rs3/Ts2 = rm3/Tm2 for rs gives us rs = rm (Tm/Ts)2/3 or rs = 0.11 rm hence (a) the satellite is situated about 1/10 th of the distance to the moon (b) rs = 0.11 rm = 42.35 x 106 m and since R = 6.38 x 106 m then the satellite is 6.64 earth radii from the earth's centre or 5.64 earth radii above the ground at the equator. (c) its height above the ground is 5.64 R or 5.64 6.38 x 106 m or 35,970 km above the earth at the equator. For those that don't feel that my artimetic is right, 2/3 means square it and then divide by the cube root of the squared. Looks pretty hard, but it isn't! 