A) 4 times
B) 8 times
C) 1/4 times
D) 1/8 times
Correct Answer: B
Solution :
Key Idea Keplers third law is applicable here. According to Keplers third law (law of periods), the square of the time period of revolution of a planet around the sun is directly proportional to the cube of semi-major axis of its elliptical orbit, ie, \[{{T}^{2}}\propto {{R}^{3}}\] where T is time taken by the planet to go once around the sun and R is semi-major axis (distance) of the elliptical orbit. \[\therefore \] \[{{T}^{2}}=k{{R}^{3}}\] ?(i) where k is constant of proportionality. When, R becomes 4 times let time period be T. \[\therefore \] \[{{T}^{2}}=k{{(4R)}^{3}}\] ?(i) \[\therefore \] \[\frac{{{T}^{2}}}{T{{}^{2}}}=\frac{1}{64}\] or \[\frac{T}{T}=\frac{1}{8}\] or \[T=8T\] So, time period becomes 8 times of previous value.
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