Explanation

(c) According to Kepler’s third law, we have
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T^{2} \propto R^{3}$
$Hence,\ \ \frac{T_{A}^{2}}{T_{B}^{2}} = \left( \frac{4R}{R} \right)^{3} = \frac{64}{1}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{T_{A}}{T_{B}} = \frac{8}{1}$
$or\ \ \ \ \ \ \ \ \ \ \ \frac{2\pi\omega_{B}}{2\pi\omega_{A}} = \frac{8}{1}$
$or\ \ \ \ \ \ \frac{v_{B} \times 4R}{R \times v_{A}\ } = \frac{8}{1}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{v_{B}}{3v} = 2$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{B} = 6v$