The escape velocity of a body from the earth is `11.2 km//s`. If a body is projected with a velocity twice its escape velocity, then the velocity of the body at infinity is (in `km//s`)

Options

A. `11.2 " kms"^(-1)`
B. `22.4 " kms"^(-1)`
C. `19.4 " kms"^(-1)`
D. `15.2 " kms"^(-1)`

Answer

C

Explanation

We know that escape velocity, `v_(e)=sqrt((2GM_(e))/(R_(e)))`. Substituting the values, we get `v_(e) = 112 " kms"^(-1)`. But if the projected velocity is double then the speed become `sqrt(3)` times and having the value `112 xx sqrt(3)=19.4 " kms"^(-1)`.