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

Hint:

For satellites orbiting the same planet, the time period depends only on the orbital radius and follows Kepler’s third law.

Answer 1

The Correct Option is B

Concept:
For a body revolving around the same central mass, Kepler’s third law applies: \[ T^2 \propto r^3 \] or \[ T \propto r^{3/2} \] where \(T\) is the time period and \(r\) is the radius of orbit.
Step 1: Let the distance of the Moon from the Earth be \(r_m\) and the distance of the satellite be \(r_s\). Given: \[ r_s = \frac{r_m}{9} \]
Step 2: Using Kepler’s third law: \[ \frac{T_s}{T_m} = \left(\frac{r_s}{r_m}\right)^{3/2} \]
Step 3: Substituting the values: \[ \frac{T_s}{27} = \left(\frac{1}{9}\right)^{3/2} \] \[ \frac{T_s}{27} = \frac{1}{27} \]
Step 4: Solving for \(T_s\): \[ T_s = 1\,\text{day} \]