A space klship is launched into a circular orbit close earth's surface. What additional velocity has now to imparted to the spaceship in the orbit to overcome gravitational pull?
(Radius of earth =6400 km,g=`9.8m//s^(2)`)

Answer

A, B, C

Explanation

The speed of the spaceship in a circular orbit close to the earth's surface is given by ,
`nu_(0) = sqrt(gR)`
and escape velocity is given by, `nu_(e) = sqrt(2gR)`
`:.` Additional velocity of required to escape
` nu_(e) - nu_(0) = sqrt (2gR)-sqrt(gR) = (sqrt2-1) sqrt (gR)`
Substituting the values of g and R, We get `nu_(e) -nu_(o) = 3.278 xx 10^(3) m//s`.