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A satellite revolves around the earth at a height of 1000 km. The radius of the earth is `6.38 xx 10^(3)`km. Mass of the earth is `6xx10^(24)kg and G=6.67xx10^(-14)"N-m"^(2)kg^(-2)`. Determine its orbital velocity and period of revolution. |
Answer» Given, `h= 1000 km = 10^(6) m` `R=6.38xx10^(3)"km"=6.38xx10^(6)"m"` `rArr h+R=7.38xx10^(6) "m",M=6xx10^(24) kg` `:.` Orbital velocity, `v_(0)=sqrt((GM)/(R+h))` `=sqrt((6.67xx10^(-11)xx6xx10^(24))/(7.38xx10^(6)))` `=7364 "ms"^(-1)` `:.` Period of revolution, `T=sqrt((2pi(R+h))/(v_(0)))=(2pixx7.38xx10^(6))/(7364)` `rArr T=6297 s`. |
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