Define orbital velocity and the time period of a satellite. Derive expressions for these

Explanation

Orbital velocity: It is the velocity given to a satellite, so that it starts orbiting around the earth. Consider a satellite of mass m revolving around the earth in an orbit. The radius of the orbit is `(R+h)`. Where R= radius of earth and h= height of satellite from earth.s surface
Let `v_(0)` be the velocity of satellite. The gravitational attraction between satellite and earth provides the centripetal force to the satellite to move around earth in orbit. i.e., Gravitational force= Centripetal force. `(GMm)/((R+h)^(2))= (mv_(0)^(2))/((R+h)) rArr v_(0)= ((GM)/(R+h))^(1//2)`
As `(GM)/(R^(2))=g rArr GM= gR^(2)`
So `v_(o)= ((gR^(2))/(R+h))^(1//2)` ...(i)
If sateilite is very close to earth.s surface, then R+h= R
`rArr v_(0)= (gR)^(1//2)= sqrt(gR)`

Time Period: It is the time taken by the satellite to complete one revolution around the earth. It is denoted by T.
Time period, `T= ("circumference of the orbit")/("orbital velocity") T= (2pi (R+h))/(v_(0))` ...(ii)
Using (i) in (ii), we get `T= (2pi (R+h))/((gR^(2)//R+h)^(½))= (2pi)/(R ) [((R+h)^(3))/(g)]^(½)`
If `R gt gt h`, then `R+h=R`
therefore, `T= (2pi)/(R ) xx (R^(3//2))/(g^(½))= 2pi sqrt((R)/(g))`