The satellites when launched from the earth are not given the orbital velocity initially, a multistage rocket propeller carries the spacecraft up to its orbit and during each stage rocket has been fired to increase the velocity to acquire the desired velocity for a particular orbit. The last stage of the rocket brings the satellite in circular/elliptical (desired) orbit.
Consider a satellite of mass `150 kg` in a low circular orbit. In this orbit, we cannot neglect the effect of air drag. This air opposes the motion of satellite and hence the total mechanical energy of earth-satellite system decreases. That means the total energy becomes more negative and hence the orbital radius decreases which causes the increase in `KE` When the satellite comes in the low enough orbit, excessive thermal energy generation due to air friction may cause the satellite to burn up. Based on the above information, answer the following questions.
If due to air drag, the orbital radius of the earth decreases from `R` to `R-/_\R, /_\Rlt ltR`, then the expression for increase in the orbital velocity `/_\v` is

Options

A. `(/_\r)/2 sqrt((GM)/R^3)`
B. `-(/_\R)/2 sqrt((GM)/2)`
C. `/_\R sqrt((GM)/(R^(3)))`
D. `-/_\Rsqrt((GM)/(R^(3)))`

Answer

A

Explanation

We have `v=sqrt((GM)/R)`
Taking log on both the sides and then differentiating
`(dv)/v=(-1xxdR)/(2R)`
`(/_\v)/v=(/_\R)/(2R)[:'/_\R=-dR]`
`/_\v=(/_\R)/2sqrt((GM)/(R^(3)))`