The distances of Neptune and Saturn from the | vedclass

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(A) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the semi-major axis $R$ of th

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$10\sqrt{10}$

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(A) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the semi-major axis $R$ of the orbit: $T^2 \propto R^3$.Given:$R_1 = 10^{13} \ m$ (Neptune)$R_2 = 10^{12} \ m$ (Saturn)The ratio of the time periods is given by:$\frac{T_1}{T_2} = \left( \frac{R_1}{R_2} \right)^{3/2}$Substituting the values:$\frac{T_1}{T_2} = \left( \frac{10^{13}}{10^{12}} \right)^{3/2}$$\frac{T_1}{T_2} = (10)^{3/2}$$\frac{T_1}{T_2} = 10^1 \cdot 10^{1/2} = 10\sqrt{10}$

The distances of Neptune and Saturn from the Sun are nearly $10^{13} \ m$ and $10^{12} \ m$ respectively. Assuming that they move in circular orbits,their periodic times will be in the ratio:

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