A particle of mass m is projected from the surface of earth with a speed `V_(0)(V_(0) lt` escape velocity). Find the speed of particle at height `h = R` (radius of earth). (Take, `R = 6400 km` and `g = 9.8 m//s^(2))`

Options

A. `sqrt(gR)`
B. `sqrt(v_(0)^(2)-2gR)`
C. `sqrt(v_(0)^(2)-gR)`
D. None of these

Answer

C

Explanation

During motion of the particle, total mechanical energy remains constant. At the surface of earth, total mechanical energy is
`E_(i) = (GmM)/(R) +(1)/(2)mv_(0)^(2)`
`=- (GM)/(R^(2)) mR +(1)/(2)mv_(0)^(2)`
`=- gmR +(1)/(2)mv_(0)^(2)`
Total mechanical energy at height `h = R` is
`E_(t) = (-GmM)/(2R) +(1)/(2)mv^(2)`
`=- (gmR)/(2) +(1)/(2)mv^(2)`
According to conservation principle of energy
`E_(i) = E_(t)`
`:. -gmR +(1)/(2)mv_(0)^(2) =- (gmR)/(2) +(1)/(2)mv^(2)`
or `-2gR +v_(0)^(2) =- gR +v^(2)`
`:. v = sqrt(v_(0)^(2)-gR)`