A) 4
B) 9
C) 64
D) 27
Correct Answer: B
Solution :
From Keplers third law of planetary motion. The square of the period of revolution\[(T)\]of any planet around the sun is directly proportional to the cube of the semi major axis\[(X)\]of its elliptical orbit \[ie,\] \[{{T}^{2}}\propto {{R}^{3}}\] Given, \[{{T}_{p}}=27{{T}_{e}},\] \[\frac{T_{e}^{2}}{T_{p}^{2}}=\frac{R_{e}^{3}}{R_{p}^{3}}\] \[\frac{T_{e}^{2}}{{{(27{{T}_{e}})}^{2}}}=\frac{R_{e}^{3}}{R_{p}^{3}}\] \[\frac{{{R}_{p}}}{{{R}_{e}}}={{(27)}^{2/3}}\] \[\frac{{{R}_{p}}}{{{R}_{e}}}={{3}^{2}}\] \[{{R}_{p}}=9{{R}_{e}}\]
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