`(i)` Define orbital velocity and establish an expression for it.
`(ii)` Calculate the value of orbital velocity for an artifical satellite of earth orbiting at a height of `1000km` (Mass of the earth `=6xx10^(24)kg`, radius of the earth `=6400km`).

Explanation

`(i)` `(i)` Definition : It is the horizontal velocity that has to be imported to a satellite at the determined height so that it makes a circular orbit around the planet is called orbital velocity.
`(ii)` Let us assume that a satellite of mass `'m'` moves around the earth in a circular orbit of radius `'r'` with uniform speed `'v'`. Let the satellite be at a height `'b'` from the surface of the earth. Here `r-R+h`, where `'R'` is the radius of the earth.
`(iii)` The centripetal force required to keep the satellite in circular orbit is `F=(mv^(2))/(r )`
`=(mv_(0)^(2))/(R+h)`
`(iv)` The gravitational force between the earth and the satellite is `F=(GMm)/(r^(2))=(GMm)/((R+h)^(2))`
for stable motion.
`(mv_(0)^(2))/(R+h)=(GMm)/((R+h)^(2))`
`v_(0)=sqrt((GM)/(R+h))`
`:'g=(GM)/(R )impliesv_(0)=sqrt((gR^(2))/(R+h))`
If satellite is at a height of few hundred kilometers (i.e. say `200km`) `R+h` will be `R`.
`:.v_(0)=sqrt(gR)`
`(ii)` Given : height `h=1000km`
Mass of Earth `M=6xx10^(2)kg`
Radius of Earth `R=6400km`
Orbital velocity `v_(0)=?`
Formula : `v_(0)=sqrt((GM)/(R+h))`
`v_(0)=sqrt((6.67xx10^(-11)xx6xx10^(24))/(6400+1000))`
`=sqrt((40.02xx10^(13))/(7400))`
`=sqrt((40.02xx10^(13))/(7.4xx10^(3)))=sqrt((40.02xx10^(10))/(7.4))`
`v_(0)=sqrt((40.02)/(7.4))xx10^(5)=(6.32613xx10^(5))/(2.7202)`
`=232561.209`
`=2.325xx10^(5)km//s`
`=2.325xx10^(5)xx10^(3)m//s`
`=2.325xx10^(8)m//s`.