An asteroid of mass `m` is approaching earth, initially at a distance `10R_(E)` with speed `v_(i)`. It hits earth with a speed `v_(f)` (`R_(E)` and `M_(E)` are radius and mass of earth),. Then

Options

A. `v_(f)^(2) = v_(i)^(2) + (2GM)/(R_(e )) (1 + (1)/(10))`
B. `v_(f)^(2)= v_(i)^(2) + (2GM_(e ))/(R_(e )) (1 + (1)/(10))`
C. `v_(f)^(2)= v_(i)^(2)+ (2GM_(e))/(R_(e )) (1- (1)/(10))`
D. `v_(f)^(2)= v_(i)^(2)+ (2GM)/(R_(e )) (1- (1)/(10))`

Answer

C

Explanation

Initial energy, of the asteroid is `E_(i)=K_(i) + U_(i) = (1)/(2) mv_(i)^(2)-(GM_(e)m)/(10R_(e))`
Final energy of the asteroid, `E_(f)= (1)/(2) mv_(f)^(2)- (GM_(e )m)/(R_(e ))`
According to law of conservation of energy, `E_(i)= E_(f)`
`(1)/(2) mv_(i)^(2) - (GM_(e)m)/(10R_(e))= (1)/(2) mv_(f)^(2) - (GM_(e)m)/(R_(e))`
`v_(f)^(2)- (2GM_(e))/(R_(e))= v_(i)^(2) - (2GM_(e))/(10R_(e))`
`rArr v_(f)^(2)= v_(i)^(2) + (2GM_(e))/(R_(e)) (1- (1)/(10))`