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Question

Physics Question on Keplers Laws

The semi-major axis of the orbit of Saturn is approximately nine times that of Earth. The time period of revolution of Saturn is approximately equal to

A

81 years

B

27 years

C

729 years

D

813\sqrt[3]{81} years

Answer

27 years

Explanation

Solution

Given, Semi-major axis of the orbit of saturn =grE=g r_{E} Where, rE=r_{E}= semi major axis of earth According to Kepler's law, T2r3T^{2} \propto r^{3} Let the time period of revolution of saturn around the sun is TsT_{s} TS2TE2=(9rErE)3\therefore \frac{T_{S}^{2}}{T_{E}^{2}}=\left(\frac{9 r_{E}}{r_{E}}\right)^{3} TS2=TE2(9)3T_{S}^{2} =T_{E}^{2}(9)^{3} TS=TE2(9)3T_{S} =\sqrt{T_{E}^{2}(9)^{3}} =93/2×1=9^{3 / 2} \times 1 year 27\approx 27 years