Explanation

(a) If the mass of sun is $M$ and radius of the planet’s orbit is $r,$
then as $v_{0} = \sqrt{GM/r}$
$T = \frac{2\pi r}{v_{0}} = 2\pi r\sqrt{\frac{r}{GM}},\ ie,\ T^{2} = \frac{4\pi^{2}r^{2}}{GM}\ $ …(i)
Now, if the planet (When stopped in the orbit) has velocity $v$ when it is at a distance $x$ from the sun, by conservation of mechanical energy,
$\frac{1}{2}mv^{2} + \left( - \frac{GMm}{x} \right) = 0 - \frac{GMm}{r}$
or $\left( - \frac{dx}{dt} \right)^{2} = \frac{2GM}{r}\left\lbrack \frac{r - x}{x} \right\rbrack,$
$ie,\ - \frac{dx}{dt} = \sqrt{\frac{2GM}{r}}\sqrt{\frac{(r - x)}{x}}$
or $\int_{0}^{t}{dt = - \sqrt{\frac{r}{2GM}} \times}\int_{r}^{0}{\left\lbrack \frac{x}{(r - x)} \right\rbrack dx}$
Substituting $x\ r\sin^{2}\theta$ and solving the RHS,
$T = \sqrt{\frac{r}{2GM}} \times \left( \frac{\pi r}{2} \right)$
In the light of Eq. (i) reduces to
$t = \frac{1}{\sqrt{4}\sqrt{2}}T,\ ie,\ t = \left( \frac{\sqrt{2}}{8} \right)T$