A body is projected with a velocity equal to twice the escape velocity `(v_(e))` from the surface of the earth. With what velocity it will travel in interplanetary space?

Options

A. `sqrt(3GM)/( R)`
B. `sqrt(6GM)/( R)`
C. `sqrt(5GM)/( R)`
D. `sqrt(GM)/( R)`

Answer

B

Explanation

Let `v_(0)` be its velocity in the interplanetary space and v be the velocity of projection `= 2v_(e)`
Then `(1)/(2)mv_(0)^(2)=(1)/(2)mv^(2)-(1)/(2)mv_(e)^(2)`
`=(1)/(2)m(2v_(e))^(2)-(1)/(2)mv_(e)^(2)`
`v_(0)^(2)=4v_(e)^(2)-v_(e)^(2)=3v_(e)^(2)`
`therefore v_(0)=sqrt(3)v_(e)=sqrt(3)sqrt((2GM)/(R ))=sqrt((6GM)/(R ))`