Explanation

(b) The period of revolution of geostationary satellite is the same as that of the earth.
$Orbital\ velocity\ \ \ \ \ \ \ \ \ \ \ v_{o} = \sqrt{gR_{e}}$
$Escape\ velocity\ \ \ \ \ \ \ \ \ \ v_{e} = \sqrt{2gR_{e}}$
where $R_{e}$ is radius of earth.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \%\ increase = \frac{v_{e} - v_{o}}{v_{o}} \times 100$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \%\ increase = \frac{\sqrt{2gR_{e}}\ - \ \sqrt{gR_{e}}}{\sqrt{gR_{e}}} \times 100$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left( \sqrt{2} - 1 \right) \times 100$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (1.141 - 1) \times 100 = 41.4\%$