Question

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 1

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`.