A) \[4\sqrt{2}yr\]
B) \[2\sqrt{2}yr\]
C) 4 yr
D) 8 yr
Correct Answer: B
Solution :
From Keplers third law of planetary motion \[{{T}^{2}}=k\,{{R}^{3}}\] where T is time period and R the distance between earth and sun. \[\therefore \] \[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}={{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3}}\] Given, \[{{R}_{2}}=2R\], \[{{R}_{1}}=R\] \[\therefore \] \[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}={{\left( \frac{1}{2} \right)}^{3}}=\frac{1}{8}\] \[\Rightarrow \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{1}{2\sqrt{2}}\] \[\Rightarrow \] \[{{T}_{2}}=2\sqrt{2}\,{{T}_{1}}\] Given, \[{{T}_{1}}=1yr\] \[\therefore \] \[{{T}_{2}}=2\sqrt{2}\,yr\]
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