What would be the duration of the year, if the distance between the earth and the sun gets doubled ? [Assume the earth's orbit to be a circular one]

Options

A. 1032 days
B. 365 days
C. 129 days
D. 556 days

Explanation

To determine the duration of the year if the distance between the Earth and the Sun is doubled, we can use Kepler's Third Law of Planetary Motion, which states that the square of the period of revolution (T) of a planet is directly proportional to the cube of the semi-major axis (r) of its orbit. The law can be mathematically expressed as:

\[ T^2 \propto r^3 \]

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Let the initial distance (radius) between the Earth and the Sun be \( r_1 \).
- The initial period of revolution (duration of the year) is \( T_1 = 365 \) days.

2. **Determine New Distance**:
- If the distance is doubled, the new distance becomes \( r_2 = 2r_1 \).

3. **Apply Kepler's Third Law**:
- According to Kepler's Third Law, we have:
\[ \frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3} \]
- Substitute \( r_2 = 2r_1 \):
\[ \frac{T_2^2}{T_1^2} = \frac{(2r_1)^3}{r_1^3} \]

4. **Simplify the Equation**:
- Calculate \( (2r_1)^3 \):
\[ (2r_1)^3 = 8r_1^3 \]
- Thus, the equation becomes:
\[ \frac{T_2^2}{T_1^2} = \frac{8r_1^3}{r_1^3} = 8 \]

5. **Solve for \( T_2 \)**:
- Rearranging gives:
\[ T_2^2 = 8T_1^2 \]
- Taking the square root of both sides:
\[ T_2 = T_1 \sqrt{8} \]

6. **Substitute \( T_1 \)**:
- Substitute \( T_1 = 365 \) days:
\[ T_2 = 365 \sqrt{8} \]

7. **Calculate \( \sqrt{8} \)**:
- Since \( \sqrt{8} = 2\sqrt{2} \approx 2.828 \):
\[ T_2 \approx 365 \times 2.828 \approx 1032 \text{ days} \]

### Final Answer:
The duration of the year, if the distance between the Earth and the Sun is doubled, would be approximately **1032 days**.