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

To solve the problem of finding the time period of rotation of the satellite that is 9 times closer to the Earth than the Moon, we can use Kepler's Third Law of Planetary Motion. Kepler's Third Law states:

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

Where:

• \(T_1\) is the time period of the satellite.
• \(T_2\) is the time period of the Moon.
• \(R_1\) is the distance of the satellite from Earth.
• \(R_2\) is the distance of the Moon from Earth.

Given that the satellite is 9 times closer to the Earth than the Moon, we have:

• \(R_1 = \frac{1}{9}R_2\)

We are given:

• \(T_2 = 27 \text{ days (Moon's time period)}\)

Substituting the given values into Kepler’s Third Law equation:

\(\frac{T_1^2}{27^2} = \frac{\left(\frac{1}{9}R_2\right)^3}{R_2^3}\)

Simplify the equation:

\(\frac{T_1^2}{27^2} = \frac{1}{9^3}\)

\(\frac{T_1^2}{729} = \frac{1}{729}\)

Solve for \(T_1\):

\(T_1^2 = 1\)

\(T_1 = \sqrt{1}\)

Thus, \(T_1 = 1 \text{ day}\).

The correct answer is 1 day.

This calculation shows that when a satellite is 9 times closer to Earth compared to the Moon, its time period of rotation is 1 day, as per Kepler's Third Law.