The distance of Neptune and Saturn from the S | vedclass

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(C) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is directly proportional to the cube of the semi-major axi

Vedclass · Accepted Answer

$10\sqrt{10}$

Vedclass · Answer

(C) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is directly proportional to the cube of the semi-major axis $(R)$ of the orbit: $T^2 \propto R^3$.Given:Distance of Neptune,$R_1 = 10^{13} \ m$Distance of Saturn,$R_2 = 10^{12} \ m$Using the ratio formula:$\frac{T_1^2}{T_2^2} = \left( \frac{R_1}{R_2} ight)^3$Substitute the values:$\frac{T_1^2}{T_2^2} = \left( \frac{10^{13}}{10^{12}} ight)^3 = (10)^3 = 1000$Taking the square root on both sides:$\frac{T_1}{T_2} = \sqrt{1000} = \sqrt{100 	imes 10} = 10\sqrt{10}$

The distance of Neptune and Saturn from the Sun is 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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