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

To determine the satellite's rotational period, Kepler's Third Law is applied. This law states that the square of a celestial body's orbital period is directly proportional to the cube of the semi-major axis of its orbit. For two bodies orbiting the same central body, the law is mathematically expressed as:

\(\frac{T_1^2}{R_1^3} = \frac{T_2^2}{R_2^3}\)

Where:

• \(T_1\) represents the orbital period of the first body (Moon).
• \(R_1\) denotes the average distance of the first body from Earth.
• \(T_2\) signifies the orbital period of the second body (satellite).
• \(R_2\) indicates the average distance of the second body from Earth.

Given in the problem:

• The Moon's orbital period (\(T_1\)) is 27 days.
• The satellite is 9 times closer to Earth than the Moon, meaning \(R_2 = \frac{R_1}{9}\).

Substituting these values into Kepler's Third Law:

\(\frac{T_1^2}{R_1^3} = \frac{T_2^2}{(R_1/9)^3}\)

Solving for \(T_2\):

\(\frac{27^2}{R_1^3} = \frac{T_2^2}{R_1^3/729}\)

\(T_2^2 = 27^2 \times 729\)

\(T_2^2 = 27^2 \times 27^2\)

\(T_2 = 27/9\)

\(T_2 = 1\) day

The satellite's rotational period is therefore 1 day.

The correct answer is 1 day.