A spaceship is launched into a circular orbit close to earth's surface. The additional velocity that should be imparted to the spaceship in the orbit to overcome the gravitational pull is
(Radius of earth 6400 km and `g = 9.8 m s^(-2)`)

Explanation

Let `Delta K` be the additional kinetic energy imparted to the spaceshi[ to overcome the gravitation pull then `Delta K=-` (total energy of spaceship) `=(GMm)/(2R)`
Total kinetic energy `=(GMm)/(2R) +Delta K= (GMm)/(2R)+(GMm)/(2R)=(GMm)/R`, then `1/2 mv_(2)^(2) =(GMm)/R implies v_(2) = sqrt((2GM)/R)`
But `v_(1)=sqrt((GM)/R)`. So additional velocity required `=v_(2)-v_(1)=sqrt((2GM)/R)-sqrt((GM)/R)=(sqrt(2)-1) sqrt((GM)/R)`
Alternate solution :
Additional velocity = Escape velocity - Orbital velocity
`=v_(es)-v_(0)`
`=sqrt((2GM)/R)-sqrt((GM)/R)`
`=(sqrt(2)-1) sqrt((GM)/R)`