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.

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Kepler's third law: $T^2\propto r^3$. $T=27\left(\frac19\right)^{3/2}=\frac{27}{27}=1$ day. Answer (2).

Q. 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.

Explanation

Reviewed by Dr. Abubacker

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