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 the rotational time period of Moon = 27 days and gravitational attraction between the satellite and the moon is neglected.

To determine the time period of the satellite, we will use Kepler’s Third Law, which states:

$$T^2 \propto R^3$$

where:

• $T$ is the time period of the orbit,
• $R$ is the orbital radius.

Step 1: Define the Given Data

• The time period of the Moon: $T_m = 27$ days.
• The satellite is 9 times closer to Earth than the Moon.
  • This means the orbital radius of the satellite is $R_s = \frac{R_m}{9}$, where $R_m$ is the Moon’s orbital radius.

Step 2: Apply Kepler’s Third Law

$$\left(\frac{T_s}{T_m}\right)^2 = \left(\frac{R_s}{R_m}\right)^3$$

Substituting $R_s = \frac{R_m}{9}$:

$$\left(\frac{T_s}{27}\right)^2 = \left(\frac{1}{9}\right)^3$$

 $$\left(\frac{T_s}{27}\right)^2 = \frac{1}{729}$$ $$T_s = 27 \times \sqrt[2]{\frac{1}{729}}$$ $$T_s = 27 \times \frac{1}{27}$$ $$T_s = 1 \text{ day}$$

Final Answer:

The time period of the satellite is 1 day.