A body is projected upwards with a velocity of `4 xx 11.2 "km s"^(-1)` from the surface of earth.What will be the velocity of the body when it escapes from the gravitational pull of earth ?

Options

A. `11.2 kms ^(-1)`
B. `2xx11.2 kms^(-1)`
C. `3xx11.2 kms^(-1)`
D. `sqrt(15)xx11.2 kms^(-1)`

Answer

D

Explanation

The minimum velocity with which a body is projected in the atmosphere so as to enable it to just overcome the gravtiational pull is known as escape velocity
initial `KE=1/2 mv^(2)=1/2xxm(4xx11.2)^(2)`
`=16xx1/2mv_(e )^(2)`
As `1/2 mv_(e )^(2)` energy is used up in coming out from the gravitatinal pull of the earth so final KE should be `15xx1/2 mv_(e)^(2)`
hence `1/2 mv^(2)=15xx1/2mv_(e )^(2)`
`rarr v^(2)=15v_(e )^(2)`
or `v'=sqrt(15)v_(e )=sqrt(15)xx11.2 kms^(-1)`