Explanation

(b) According to Kelper’s third law (law of periods),
$we\ have\ \ \ \ \ \ \ \ \ \ \ \ T^{2} \propto R^{3}$
where T is time taken by the planet to go once around the sun and R is semi-major axis (distance) of the elliptical orbit.
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T^{3} = {k\ R}^{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots$(i)
Where $k$ is constant of proportionality.
When $R$ becomes 4 times let time period be $T'$.
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {T^{'}}^{2} = k(4R)^{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{T^{2}}{{T^{'}}^{2}} = \frac{1}{64}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{T}{T^{'}} = \frac{1}{8}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T^{'} = 8T$
So, time period becomes 8 times of previous value.