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.
Show Hint
- 1 day
- 81 days
- 27 days
- 3 days
The Correct Option is A
Approach Solution - 1
To determine the time period of rotation of the satellite, we will use Kepler's Third Law, which states that the square of the period of any planet (or satellite) is proportional to the cube of the semi-major axis of its orbit.
Mathematically, for two bodies orbiting the same celestial body, the law is expressed as:
\(\frac{T_1^2}{R_1^3} = \frac{T_2^2}{R_2^3}\)
Where:
- \(T_1\) is the time period of the first body (Moon).
- \(R_1\) is the average distance of the first body from the Earth.
- \(T_2\) is the time period of the second body (satellite).
- \(R_2\) is the average distance of the second body from the Earth.
According to the question:
- The Moon's time period (\(T_1\)) is 27 days.
- The satellite is 9 times closer to the Earth than the Moon, hence \(R_2 = \frac{R_1}{9}\).
Substituting 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
Thus, the time period of rotation of the satellite is 1 day.
Hence, the correct answer is 1 day.
Approach Solution -2
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