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 the two statements provided and determine their validity based on the principles of physics related to circular motion and areal velocity.

### Step-by-Step Solution:

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

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

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

4. **Relating Speed to Radius**:
For a satellite in a circular orbit, the centripetal force is provided by gravitational force:
\[ mg = \frac{mv^2}{r} \]
From this, we can derive the speed \(v\):
\[ v^2 = rg \quad \Rightarrow \quad v = \sqrt{rg} \]

5. **Substituting Speed into Areal Velocity**:
Now substituting \(v\) into the areal velocity equation:
\[ \frac{dA}{dt} = \frac{r \sqrt{rg}}{2} = \frac{r^{3/2}g^{1/2}}{2} \]

6. **Effect of Doubling the Radius**:
If the radius \(r\) is doubled (i.e., \(r' = 2r\)):
\[ \frac{dA'}{dt} = \frac{(2r)^{3/2}g^{1/2}}{2} = \frac{2^{3/2}r^{3/2}g^{1/2}}{2} = \frac{2\sqrt{2}r^{3/2}g^{1/2}}{2} = \sqrt{2} \cdot r^{3/2}g^{1/2} \]
This shows that the new areal velocity is not simply double the original areal velocity, but rather it is multiplied by \(\sqrt{2}\).

7. **Conclusion**:
Therefore, **Statement 1** is false because the areal velocity does not become two times when the radius is doubled; it actually becomes \(\sqrt{2}\) times the original value.

**Statement 2** is true as it correctly describes the relationship between areal velocity and angular momentum.

### Final Answer:
- **Statement 1** is false.
- **Statement 2** is true.
- The correct option is **D**: Statement 1 is false and Statement 2 is true.