if the earth is treated as a sphere of radius R and mass M, Its angular momentum about the axis of its rotation with period T, is

Options

A. `(MR^(2)T)/(2pi)`
B. `(4piMR^(2))/(5T)`
C. `(piMR^(2))/(T)`
D. `(2piMR^(2))/(T)`

Explanation

To find the angular momentum of the Earth treated as a sphere, we can follow these steps:

### Step 1: Understand the Formula for Angular Momentum
The angular momentum \( L \) of a rotating body can be expressed as:
\[ L = I \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

### Step 2: Determine the Moment of Inertia for a Sphere
For a solid sphere, the moment of inertia \( I \) is given by:
\[ I = \frac{2}{5} M R^2 \]
where \( M \) is the mass of the sphere and \( R \) is its radius.

### Step 3: Calculate the Angular Velocity
The angular velocity \( \omega \) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the period of rotation (the time taken for one complete rotation).

### Step 4: Substitute the Values into the Angular Momentum Formula
Now, substituting \( I \) and \( \omega \) into the angular momentum formula:
\[ L = I \omega = \left(\frac{2}{5} M R^2\right) \left(\frac{2\pi}{T}\right) \]

### Step 5: Simplify the Expression
Now, we can simplify the expression:
\[ L = \frac{2}{5} M R^2 \cdot \frac{2\pi}{T} = \frac{4\pi}{5} \frac{M R^2}{T} \]

### Final Answer
Thus, the angular momentum of the Earth about the axis of its rotation is:
\[ L = \frac{4\pi}{5} \frac{M R^2}{T} \]

---