If the earth suddenly stopped in its orbit (assume orbit to be circular) the time that would elapse before it falls into the sun is

$\frac{1}{4 \sqrt{2}} T$

The time t₀ taken by earth to fall into the sun can be calculated by considering a very elongated ellipse having major axis equal to the radius of orbit of earth about sun (r).

⇒ 2a = r

According to Kepler's Third Law,

$\left(2 t_0\right)^2 \propto a^3$

$\Rightarrow\left(2 t_0\right)^2 \propto\left(\frac{r}{2}\right)^3$

Since, $T^2 \propto r^3$

$\Rightarrow \frac{2 t_0}{T}=\frac{1}{\sqrt{8}}$

$\Rightarrow t_0=\frac{T}{4 \sqrt{2}} \frac{365}{4 \sqrt{2}}$ = 64.53 days