Q. The period of revolution of planet A around the sun is 8 times that of B. The distance of A from the sun is how many times greater than that of B from the sun?

Solution:

(a) : Period of revolution of planet $ A (T_A) = 8T_B $
According to Kepler's III law of planetary motion
$T^2 \propto R^3 $
Therefore $\bigg( \dfrac{ r_A}{r_B } \bigg)^3 = \bigg( \dfrac{ T_A}{T_B } \bigg)^2= \bigg( \dfrac{ 8 T_A}{T_B } \bigg)^2 = 64 $
or $ \dfrac{ r_A}{r_B} = 4 \, \, \, \, or \, \, \, \, r_A = 4 r_B $

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