If a planet was suddenly stopped in its orbit supposed to be circular, If it falls onto the sun in a time `K times the period of the planet's revolution , then K is equal to

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 $T^{\text{'}}$ according to Kepler's law will be given by

  $\left(\frac{T\text{'}}{T}\right)={\left(\frac{r^{\text{'}}}{r}\right)}^{\frac{3}{2}}$

or    $T=T{\left(\frac{1}{2}\right)}^{\frac{3}{2}}$

$\therefore$ Time taken by the planet to fall onto the sun is

     $t=\frac{T^{\text{'}}}{2}=\frac{T}{2}{\left(\frac{1}{2}\right)}^{\frac{3}{2}}\Rightarrow t=\frac{\sqrt{2}}{8}T$