In the earth-moon system, if `T_(1)` and `T_(2)` are period of revolution of earth and moon respectively about the centre of mass of the system them

Options

A. `T_(1) gt T_(2)`
B. `T_(1) = T_(2)`
C. `T_(1) lt T_(2)`
D. Insufficient data

Answer

To solve the problem regarding the Earth-Moon system and the periods of revolution \( T_1 \) and \( T_2 \) about their center of mass, we can follow these steps:

### Step 1: Understand the System
The Earth and Moon revolve around their common center of mass (barycenter). The center of mass is located closer to the Earth due to its significantly larger mass compared to the Moon.

### Step 2: Define Angular Velocities
Let \( \omega_1 \) be the angular velocity of the Earth and \( \omega_2 \) be the angular velocity of the Moon. Since both bodies are in orbit around the center of mass, they will have the same angular velocity relative to the center of mass.

### Step 3: Relate Angular Velocities to Periods
The angular velocity is related to the period of revolution by the formula:
\[ \omega = \frac{2\pi}{T} \]
Thus, for the Earth:
\[ \omega_1 = \frac{2\pi}{T_1} \]
And for the Moon:
\[ \omega_2 = \frac{2\pi}{T_2} \]

### Step 4: Set Angular Velocities Equal
Since both the Earth and Moon revolve around the same center of mass, we have:
\[ \omega_1 = \omega_2 \]
This leads to:
\[ \frac{2\pi}{T_1} = \frac{2\pi}{T_2} \]

### Step 5: Simplify the Equation
By canceling \( 2\pi \) from both sides, we obtain:
\[ \frac{1}{T_1} = \frac{1}{T_2} \]

### Step 6: Conclude the Relationship Between Periods
From the above equation, we can conclude:
\[ T_1 = T_2 \]
This means the periods of revolution of the Earth and the Moon about their center of mass are equal.

### Final Answer
Thus, the correct option is that \( T_1 \) is equal to \( T_2 \).

---