If a satellite orbiting the Earth is 9 times | 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 orbital radius $(R)$:

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$1 	ext{ day}$

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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 orbital radius $(R)$: $T^2 \propto R^3$.Let $T_m$ and $R_m$ be the time period and orbital radius of the Moon, and $T_s$ and $R_s$ be the time period and orbital radius of the satellite.Given: $R_s = R_m / 9$ and $T_m = 27 	ext{ days}$.Using the ratio formula: $\left(\frac{T_m}{T_s}ight)^2 = \left(\frac{R_m}{R_s}ight)^3$.Substituting the values: $\left(\frac{27}{T_s}ight)^2 = \left(\frac{R_m}{R_m / 9}ight)^3 = (9)^3 = 729$.Taking the square root on both sides: $\frac{27}{T_s} = \sqrt{729} = 27$.Therefore, $T_s = \frac{27}{27} = 1 	ext{ day}$.

If a satellite orbiting the Earth is $9$ times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon $= 27 \text{ days}$ and gravitational attraction between the satellite and the moon is neglected.

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