A) \[4\sqrt{2}\]years
B) \[2\sqrt{2}\] years
C) 4 years
D) 8 years
Correct Answer: B
Solution :
Given \[{{T}_{1}}=1year,{{R}_{1}}=R,{{R}_{2}}=2R\] According to Keplers IIIrd law of, planetary motion, \[{{T}^{2}}\propto {{R}^{3}}\] where R is the distance between earth and sun. \[\therefore \] \[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}={{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3}}\] \[={{\left( \frac{R}{2R} \right)}^{3}}=\frac{1}{8}\] \[\Rightarrow \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{1}{2\sqrt{2}}\] \[\Rightarrow \] \[{{T}_{2}}=2\sqrt{2}{{T}_{1}}\] \[=2\sqrt{2}\]years
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