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