A) 1/4
B) ½
C) 1/8
D) 1/6
Correct Answer: C
Solution :
From Kepler?s third law of planetary motion, the square of period of revolution of any planet around the sun is directly proportional to the cube of the semi-major axis of its elliptical orbit. \[{{T}^{2}}\propto {{a}^{3}}\] Given, \[a'=\frac{a}{4}\] \[\therefore \] \[\frac{{{T}^{2}}}{T{{'}^{2}}}=\frac{{{a}^{3}}}{{{\left( \frac{a}{4} \right)}^{3}}}={{(4)}^{3}}\] \[\Rightarrow \] \[T'=\frac{T}{{{(4)}^{3/2}}}=\frac{T}{8}\]You need to login to perform this action.
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