Escape velocity for a projectile at earth's surface is `V_(e)`. A body is projected form earth's surface with velocity `2 V_(e)`. The velocity of the when it is at infinite distance from the centre of the earth is :-

Options

A. `v_(e)`
B. `sqrt(2)v_(e)`
C. `sqrt(3)v_(e)`
D. `sqrt(5)v_(e)`

Answer

C

Explanation

`U_(i)+K_(i)=U_(f)+K_(f)`
`rArr-(GMm)/(R)+(1)/(2)m(2v_(e))^(2)=0+(1)/(2)mv^(2)`
or `-(GM)/(R)+2v_(e)^(2)=(1)/(2)v^(2)`
or `-(2GM)/(R)+(8GM)/(R)=v^(2)`
or `v=sqrt((6GM)/(R))=sqrt(3((2GM)/(R)))=sqrt(3(2gR))=sqrt(3)v_(e)`.