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Correct option is A)

Kepler's law applies to planetary orbits, whether they be of circular, or elliptical shape. It says that $T_{2}T_{2} =R_{3}R_{3} $* where T is the period of an orbit and R*** is its semi-major axis. The semi-major axis is the average of the planet's maximum and minimum distances from the sun.**

Let the earth's mean radius be $R_{1}$.**.**This straight fall can be considered 1/2 of a degenerate elliptical orbit with major axis equal to $R_{1}$. Its semi-major axis is $2R_{1} $ (the average of *R1* and zero). Its period will be designated $T_{2}$.

So: $T_{1}T_{2} =R_{1}(2R_{1} )_{3} =(21 )_{3}$

And therefore, $T_{2}=0.353$** year***, and the time to fall into the sun is **the
time to fall into the sun is **,**64.52 days*

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