Question

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.

• A. 1 day
• B. 81 days
• C. 27 days
• D. 3 days

Hint:

Kepler's third law helps in finding the time period of satellites based on their distance from the central body.

Answer 1

The Correct Option is A

The time period of a satellite is related to the distance from the Earth by Kepler’s third law, which states: \[ T^2 \propto r^3. \] If the satellite is 9 times closer to the Earth than the Moon, the distance ratio is: \[ r_{\text{satellite}} = \frac{r_{\text{moon}}}{9}. \] Thus, the time period of the satellite is related to the time period of the Moon by: \[ T_{\text{satellite}} = T_{\text{moon}} \cdot \left( \frac{1}{9} \right)^{3/2}. \] Since the Moon’s time period is 27 days: \[ T_{\text{satellite}} = 27 \cdot \frac{1}{3} = 1 \, \text{day}. \]