The escapoe speed of a body on the earth's surface is `11.2kms^(-1)`. A body is projected with thrice of this speed. The speed of the body when it escape the gravitational pull of earth is

Options

A. `11.2kms^(-1)`
B. `22.4sqrt(2)kms^(-1)`
C. `22.4/sqrt(2)kms^(-1)`
D. `22.4sqrt(3)kms^(-1)`

Answer

B

Explanation

Let `v` be the speed of the body when it escape the gravitational pull of the earth and `u` be speed of projection of the body form the earth's surface. According to law of conservation of mechanical energy,
`1/2mv^(2)-(GM_(E)m)/(R_(E))=1/2mv^(2)-0`
Where `m` and `M_(E)` be masses of the body and earth respectively and `R_(E)` is the radius of the earth.
or `v^(2)=v^(2)-(2GMm)/(R_(E)`
`v^(2)=u^(2)-v_(e)^(2) ( :' v_(e)=sqrt((2GM_(E))/R_(E))`
`v=sqrt(u^(2)-v_(e)^(2))=sqrt((3v_(e))^(2)-v_(e)^(2))`
`sqrt(8)v_(e)=11.2sqrt(8)kms^(-1)=22.4sqrt(4)kms^(-1)`