If A is the areal velocity of a planet of mass M, its angular momentum is

Options

A. M/A
B. 2MA
C. `A^2M`
D. `AM^2`

Answer

To find the angular momentum of a planet with mass \( M \) and areal velocity \( A \), we can follow these steps:

### Step 1: Understand Areal Velocity
Areal velocity \( A \) is defined as the area swept out by the radius vector per unit time. Mathematically, it can be expressed as:
\[ A = \frac{\text{Area swept}}{\text{Time taken}} = \frac{\pi r^2}{t} \]

### Step 2: Relate Areal Velocity to Angular Velocity
From the definition of areal velocity, we can also relate it to angular velocity \( \omega \). The area swept out in time \( t \) is also given by:
\[ \text{Area} = \frac{1}{2} r^2 \theta \]
where \( \theta \) is the angular displacement in radians. Therefore, we can express \( t \) in terms of \( r \) and \( \omega \):
\[ t = \frac{\text{Area}}{A} = \frac{1}{2} r^2 \theta / A \]
Using the relationship \( \theta = \omega t \), we can derive:
\[ t = \frac{r^2 \omega}{2A} \]

### Step 3: Express Angular Momentum
The angular momentum \( L \) of a planet is given by:
\[ L = I \cdot \omega \]
where \( I \) is the moment of inertia. For a point mass \( M \) at a distance \( r \) from the axis of rotation, the moment of inertia \( I \) is:
\[ I = M r^2 \]
Thus, substituting for \( L \):
\[ L = M r^2 \cdot \omega \]

### Step 4: Substitute for Angular Velocity
From our previous relationship, we have:
\[ \omega = \frac{2A}{r^2} \]
Substituting this into the angular momentum equation:
\[ L = M r^2 \cdot \frac{2A}{r^2} \]

### Step 5: Simplify the Expression
The \( r^2 \) terms cancel out:
\[ L = M \cdot 2A \]
Thus, the angular momentum \( L \) of the planet is:
\[ L = 2MA \]

### Final Answer
The angular momentum of the planet is:
\[ L = 2MA \]