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.
Text Solution
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Here `h=1000 km=1000xx10^(3)m=10^(6)m` `r=R+h=6.38xx10^(6)+10^(6)=7.38xx10^(6)m` Orbital speed, `v_(0)=sqrt((GM)/(R+h))=sqrt((6.67xx10^(-11)xx6xx10^(24))/(7.38xx10^(6)))=7364ms^(-1)` Time period `T=(2pir)/(v_(0))=(2xx(22/7)xx(7.38xx10^(6)))/(7.364xx10^(3))=6297s`
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