Question

An artificial satellite revolves round the earth at a height of `1000 km`. The radius of the earth is `6.38xx10^(3)km`. Mass of the earth `6xx10^(24)kg,G=6.67xx10^(-11)Nm^(2)kg^(-2)`. Find the orbital speed and period of revolution of the satellite.

Answer 1

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`