A) 4.2T years
B) 2.8 T years
C) 5.6Tyears
D) 8.4Tyears
Correct Answer: B
Solution :
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 i.e. , \[{{T}^{2}}\,\propto \,{{R}^{3}}\] Given, \[{{R}_{2}}=2{{R}_{1}}\,and\,\,{{R}_{1}}=R\] \[\therefore \] \[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}={{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3}}={{\left( \frac{R}{2R} \right)}^{3}}\] \[\Rightarrow \] \[\frac{T}{{{T}_{2}}}={{\left( \frac{1}{2} \right)}^{3/2}}\] \[\Rightarrow \] \[{{T}_{2}}=2.8T\,\,years.\] Note: Larger the distance of a planet from the sun, larger will be its period of revolution around the sun.
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