A) \[32\,\,day\]
B) \[18\,\,day\]
C) \[8\,\,day\]
D) \[9\,\,day\]
Correct Answer: C
Solution :
Key Idea : Larger che distance of planet from the sun, larger will be its period of revolution around the sun. From Kepler's third law of planetary motion, the Square of the period of revolution of any planet around the sun is directly proportional to the cube of the semi-major axis of its elliptical orbit. \[\therefore \] \[{{T}^{2}}\,\propto \,{{R}^{3}}\] \[\therefore \] \[\frac{{{T}_{s}}}{{{T}_{c}}}={{\left( \frac{{{R}_{s}}}{{{R}_{c}}} \right)}^{3/2}}\] Given, \[{{R}_{s}}=4{{R}_{c}}\] \[\therefore \] \[\frac{{{T}_{s}}}{{{T}_{c}}}={{\left( \frac{4{{R}_{c}}}{{{R}_{c}}} \right)}^{3/2}}=8\] For \[{{T}_{c}}=1\,\,day\] \[{{T}_{s}}=8\,\,days.\]
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