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A satellite is launched into a circular of radius R around the earth.While a second satellite launched in an orbit of radius 1.01R.The period of the second satellite is longer than the first one by approximately(please express in %)
Dear Sunny,
according to Kepler's third law T2 isproportional to R3. Thus t1=kR3/2,t2=kR3/2(1+0.01)3/2=kR3/2(1+(0.01*1.5)). Thus the time period of the second satellite is longer than that of the first one by approximately 1.5%.
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