The earth revolves round the sun in one year. If the distance between them becomes double, the new period of revolution will be

Options

A. `4sqrt(2)` years
B. `2sqrt(2)` years
C. 4 years
D. 8 years

Answer

To solve the problem, we will use Kepler's Third Law of planetary motion, which states that the square of the time period (T) of a planet is directly proportional to the cube of the semi-major axis (r) of its orbit. Mathematically, this can be expressed as:

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

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The initial time period \( T_1 \) of the Earth's revolution around the Sun is 1 year.
- The initial distance (radius) \( r_1 \) is the current average distance from the Earth to the Sun.
- The new distance (radius) \( r_2 \) is double the initial distance: \( r_2 = 2r_1 \).

2. **Apply Kepler's Third Law:**
- According to Kepler's Third Law, we can write the relationship between the two time periods and the two distances as:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]

3. **Substitute the Known Values:**
- We know \( T_1 = 1 \) year and \( r_2 = 2r_1 \). Therefore, we can substitute:
\[ \frac{1^2}{T_2^2} = \frac{r_1^3}{(2r_1)^3} \]

4. **Simplify the Right Side:**
- The right side simplifies as follows:
\[ (2r_1)^3 = 8r_1^3 \]
So we have:
\[ \frac{1}{T_2^2} = \frac{r_1^3}{8r_1^3} = \frac{1}{8} \]

5. **Cross-Multiply to Solve for \( T_2^2 \):**
- From the equation \( \frac{1}{T_2^2} = \frac{1}{8} \), we can cross-multiply to find:
\[ T_2^2 = 8 \]

6. **Find \( T_2 \):**
- Taking the square root of both sides gives us:
\[ T_2 = \sqrt{8} = 2\sqrt{2} \text{ years} \]

### Final Answer:
The new period of revolution when the distance between the Earth and the Sun is doubled is \( 2\sqrt{2} \) years.

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