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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),. ThenA. `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)` |
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Answer» Correct Answer - C 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)` |
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