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A satellite is moving in a circular orbit at a certainheight above the earth's surface. It takes 5.26xx10^3 s to complete one revolution with a centripetal acceleration equal to 9.32 m s^(-2). The height of the satellite orbit above the earth's surface is (Take radius of earth =6.37xx10^6 m ) |
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Answer» Solution :As, `T=2pisqrt(((R+h)^3)/(GM))` `T^2/(4pi^2)=(R+h)^3/(GM)` …(i) Centripetal acceleration , `a=(GM)/(R+h)^2` `(R+h)^2/(GM)=1/a` `(R+h)=T^2/(4pi^2)XXA` [Using (i)] `=(5.26xx10^3 // 2pi)^2 xx 9.32=6.54xx10^6` m `therefore h=6.54xx10^6 - R = 6.54xx10^6 - 6.37xx10^6` `= 0.17xx10^6` m = 170 km |
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