A) 32 days
B) 18 days
C) 8 days
D) 9 days
Correct Answer: C
Solution :
Larger the distance of planet from the sun, larger will be its period of revolution around the sun. Keplers third low 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 \[\therefore \] \[{{T}^{2}}\propto {{R}^{3}}\] \[\frac{{{T}_{s}}}{{{T}_{c}}}={{\left( \frac{{{R}_{s}}}{{{R}_{c}}} \right)}^{3/2}}\] given \[{{R}_{s}}=4{{R}_{c}}\] \[\frac{{{T}_{s}}}{{{T}_{v}}}={{\left( \frac{4{{R}_{c}}}{{{R}_{c}}} \right)}^{3/2}}=8\] for \[{{T}_{c}}=1day\] \[{{T}_{s}}=8days\]
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