We could consider a motion of the falling Earth as the motion of a comet. The orbit of this comet around the Sun is an infinitely thin ellipse. The semi-major axis of the comet is half Earth’s average distance from the Sun.
According to the third Kepler’s Law, the square of the period of orbit is proportional to the cube of the semi-major axis length or average distance from the Sun.
Tc2/Rc3 = TE2/RE3
where Tc – the period of orbit of the comet (the falling Earth), TE – the period of orbit of the Earth, Rc – the semi-major axis length of the comet and RE – the average distance of the Earth from the Sun.
We denote RE by R then Rc = R/2 .
Tc2 = 1/8 TE2
Tc = √(1/8) * TE
The comet falls towards the Sun and the falling time t is half of the full orbital period of the comet.
t = 1/2 Tc = √(1/32) * TE
The Earth’s orbital period is 365 days.
t = √(1/32) * 365 = 64.5 days.
Image courtesy of the DSCOVR EPIC team. Caption by Rob Gutro and Adam Voiland.