Explanation

(c) From Kepler ‘s third law of planetary motion,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T^{2} \propto a^{3}$
$Given,\ \ T_{1} = 1day\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (geostationary)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{1} = a,\ a_{2} = 2a$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{a_{1}^{3}}{a_{2}^{3}}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ T_{2}^{2} = \frac{a_{2}^{3}}{a_{1}^{3}}T_{1}^{2} = \frac{(2a)^{3}}{a^{3}} \times 1 = 8$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ T_{2} = 2\sqrt{2}\ days$