Explanation

(a) From Kepler’s third law of planetary motion
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T^{2} \propto R^{3}$
$Given,\ \ R_{p} = 2R_{e}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{T_{e}^{2}}{T_{p}^{2}} = \frac{R_{e}^{3}}{R_{p}^{3}}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{T_{e}^{2}}{T_{p}^{2}} = \frac{R_{e}^{3}}{\left( 2R_{e} \right)^{3}}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{T_{e}}{T_{p}} = \left( \frac{1}{2} \right)^{3/2}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{p} = 2\sqrt{2}\ T_{e}$
$Since,\ \ T_{e} = 365\ days = 1\ year,\ we\ have$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{p} = 2\sqrt{2} \times 365\ days$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{p} = 1032.37$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{p} = 1032\ days.$