The ratio of the earth's orbital angular momentum (about the Sun) to its mass is `4.4 xx 10^(15) m^(2) s^(-1)`. The area enclosed by the earth's orbit is approximately-___________m^(2).

Options

A. `6.94xx10^(22)`
B. `6.94xx10^(23)`
C. `7.94xx10^(22)`
D. `7.94xx10^(23)`

Explanation

To solve the problem, we need to find the area enclosed by the Earth's orbit around the Sun using the given ratio of the Earth's orbital angular momentum to its mass.

### Step-by-Step Solution:

1. **Understand the relationship between angular momentum, mass, and area**:
The aerial velocity (rate of area swept out) of an object in orbit is given by:
\[ \frac{dA}{dt} = \frac{L}{2m} \]
where \( L \) is the angular momentum and \( m \) is the mass of the Earth.

2. **Integrate to find the total area**:
We can rearrange the equation to find the area \( A \):
\[ dA = \frac{L}{2m} dt \]
Integrating both sides gives:
\[ A = \frac{L}{2m} \int dt = \frac{L}{2m} \cdot t \]
Here, \( t \) is the time taken for one complete revolution of the Earth around the Sun, which is one year.

3. **Substitute the known values**:
We know from the problem that:
\[ \frac{L}{m} = 4.4 \times 10^{15} \, \text{m}^2/\text{s} \]
Therefore, we can write:
\[ A = \frac{L}{2m} \cdot t = \frac{4.4 \times 10^{15}}{2} \cdot t \]

4. **Calculate the time for one revolution**:
The time \( t \) for one complete revolution (one year) is:
\[ t = 365 \, \text{days} \times 24 \, \text{hours/day} \times 3600 \, \text{seconds/hour} \]
Calculating this gives:
\[ t = 365 \times 24 \times 3600 = 31,536,000 \, \text{s} \]

5. **Calculate the area**:
Now substituting \( t \) into the area formula:
\[ A = \frac{4.4 \times 10^{15}}{2} \cdot 31,536,000 \]
Simplifying this:
\[ A = 2.2 \times 10^{15} \cdot 31,536,000 \]
Performing the multiplication:
\[ A \approx 6.94 \times 10^{22} \, \text{m}^2 \]

6. **Final result**:
Therefore, the area enclosed by the Earth's orbit is approximately:
\[ A \approx 6.94 \times 10^{22} \, \text{m}^2 \]