If a planet was suddenly stopped in its orbit supposed to be circular, it will fall onto the sun in a time $\frac{1}{\sqrt{2^x}}$  times the period of the planet’s revolution. Find $x.$  

Consider an imaginary planet moving along a strongly extended flat ellipse, the extreme points of which are located on the planet’s orbit and at the centre of the sun, the semi-major axis of the orbit of such a planet would apparently be half the semi-major axis of the planet’s orbit. So, the time period of the imaginary planet T’ according to Kepler’s Law will be given by :
$\begin{array}{l}\left(\frac{T\text{'}}{T}\right)={\left(\frac{r\text{'}}{r}\right)}^{3/2}\\ \Rightarrow T\text{'}=T{\left(\frac{1}{2}\right)}^{3/2}\end{array}$

So, time taken by the planet to fall onto the sun is
$t=\frac{T\text{'}}{2}=\frac{T}{2}{\left(\frac{1}{2}\right)}^{3/2}$


$\begin{array}{l}\Rightarrow t=\frac{\sqrt{2}}{8}T=\sqrt{\frac{2}{64}T}=\sqrt{\frac{1}{32}}=\frac{1}{\sqrt{2^5}}\\ \Rightarrow x=5\end{array}$