STATEMENT-1`:` Radius of circular orbit of a satellite is made two times, then its areal velocity will also becomes two times.
because
STATEMENT -2 `:` Areal velocity is given as `(dA)/( dt)= ( L)/( 2m)= ( mvr)/( 2m)`.

Options

A. Statement-1 is True , Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True , Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False
D. Statement-1 is False, Statement-2 is True.

Answer

To solve the problem, we need to analyze both statements regarding the areal velocity of a satellite in a circular orbit.

### Step-by-Step Solution:

1. **Understanding Areal Velocity**:
Areal velocity (dA/dt) is defined as the rate at which area is swept out by the radius vector of a satellite orbiting a central body. It can be expressed as:
\[ \frac{dA}{dt} = \frac{L}{2m} \]
where \(L\) is the angular momentum of the satellite and \(m\) is its mass.

2. **Expression for Angular Momentum**:
The angular momentum \(L\) of a satellite in a circular orbit can be expressed as:
\[ L = mvr \]
where \(v\) is the linear velocity of the satellite and \(r\) is the radius of the orbit.

3. **Substituting Angular Momentum into Areal Velocity**:
Substituting the expression for \(L\) into the areal velocity formula gives:
\[ \frac{dA}{dt} = \frac{mvr}{2m} = \frac{vr}{2} \]

4. **Relating Areal Velocity to Radius**:
The linear velocity \(v\) of a satellite in a circular orbit is related to the radius \(r\) and the gravitational force acting on it. For a satellite in a stable orbit, the gravitational force provides the necessary centripetal force:
\[ v = \sqrt{\frac{GM}{r}} \]
where \(G\) is the gravitational constant and \(M\) is the mass of the central body.

5. **Effect of Doubling the Radius**:
If the radius \(r\) is doubled (i.e., \(r' = 2r\)), the new velocity \(v'\) becomes:
\[ v' = \sqrt{\frac{GM}{2r}} = \frac{v}{\sqrt{2}} \]
Therefore, the new areal velocity \(dA'/dt\) becomes:
\[ \frac{dA'}{dt} = \frac{v'}{2} \cdot (2r) = \frac{(v/\sqrt{2})}{2} \cdot (2 \cdot 2r) = \frac{4vr}{2\sqrt{2}} = \frac{2\sqrt{2}vr}{2} = \sqrt{2} \cdot \frac{vr}{2} \]
This shows that the areal velocity does not simply double; it actually changes based on the square root of the radius change.

6. **Conclusion**:
Since the areal velocity is proportional to the square of the radius, if the radius is doubled, the areal velocity becomes four times the original value, not two times. Therefore, Statement 1 is false, and Statement 2 is true.

### Final Answer:
- **Statement 1 is false**: Areal velocity does not become two times; it becomes four times when the radius is doubled.
- **Statement 2 is true**: The expression for areal velocity is correctly given.