A) \[5\frac{1}{2}times\]
B) \[4\,\,times\]
C) \[3\,\,times\]
D) \[None\,\,of\,\,these\]
Correct Answer: B
Solution :
Key Idea : Kepler's third law provides the relation between time period of revolution with distance r. 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 (Kepler's third law) \[{{T}^{2}}\,\propto \,{{a}^{3}}\] Given, \[{{T}_{x}}=8{{T}_{y}}\] \[\therefore \] \[\frac{T_{x}^{2}}{T_{y}^{2}}=\frac{R_{x}^{3}}{R_{y}^{3}}\] \[\Rightarrow \] \[\frac{{{\left( 8{{T}_{y}} \right)}^{2}}}{T_{y}^{2}}=\frac{R_{x}^{3}}{R_{y}^{3}}\] \[\Rightarrow \] \[\frac{64}{1}=\frac{R_{x}^{3}}{R_{y}^{3}}\] \[\Rightarrow \] \[{{R}_{x}}=4{{R}_{y}}\] Therefore, distance of planet \[x\] is four times that of planet\[y\]. Note: Larger the distance of a planet from the sun. Larger will be its period of revolution around the sun.
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