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 $=27$ days and gravitational attraction between the satellite and the moon is neglected.

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{T}^2 \propto \mathrm{R}^3 \ & \left(\frac{\mathrm{~T}_{\mathrm{m}}}{\mathrm{T}_{\mathrm{s}}}ight)^2=\left(\frac{\mathrm{R}}{\mathrm{R} / 9}ight)^3 \ & \frac{\mathrm{~T}_{\mathrm{m}}}{\mathrm{T}_{\mathrm{s}}}=(3)^3 \ & \Rightarrow \mathrm{~T}_{\mathrm{s}}=\left(\frac{27}{27}ight)=1 	ext { day }\end{aligned}$

If a satellite orbiting the Earth is 9 times closer to the Earth than the Moon, what is the time period of…

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