Explanation

(a) The velocity with which satellite is orbiting around the earth is the orbital velocity $(v_{o})$ and that required to escape out of gravitational pull of earth is the escape velocity $(v_{e})$.
We know that
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{e} = \sqrt{2gR}\ \ and\ v_{o} = \sqrt{gR}$
$\therefore\ \ Increase\ in\ velocity\ required\ $
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{v_{e} - v_{o}}{v_{o}} = \frac{\sqrt{2gR} - \sqrt{gR}}{\sqrt{gR}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sqrt{2} - 1 = 0.414$
Percent increase in velocity required
$= 0.414 \times 100 = 41.4\%$