If the distance between the earth and sun becomes `1//4^(th)`then its period of revolution around the sun wi ll become

Options

A. 6 hr
B. 8 hr
C. 16 hr
D. 3 hr

Answer

To solve the problem of how the period of revolution of the Earth around the Sun changes when the distance between them is reduced to one-fourth, we can use Kepler's Third Law of planetary motion. This law 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. Mathematically, this is expressed as:

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

### Step-by-Step Solution:

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

2. **Determine the New Distance**:
- The new distance is given as one-fourth of the original distance, so:
\[ R_2 = \frac{R}{4} \]

3. **Apply Kepler's Third Law**:
- According to Kepler's Third Law, we can write:
\[ \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \]
- Substituting the values:
\[ \frac{T_1^2}{T_2^2} = \frac{R^3}{\left(\frac{R}{4}\right)^3} \]

4. **Simplify the Right Side**:
- The right side simplifies as follows:
\[ \left(\frac{R}{4}\right)^3 = \frac{R^3}{64} \]
- Thus:
\[ \frac{T_1^2}{T_2^2} = \frac{R^3}{\frac{R^3}{64}} = 64 \]

5. **Relate the Periods**:
- Now we have:
\[ \frac{T_1^2}{T_2^2} = 64 \]
- Taking the square root of both sides gives:
\[ \frac{T_1}{T_2} = 8 \]
- Therefore:
\[ T_2 = \frac{T_1}{8} \]

6. **Calculate the New Period**:
- Substitute \( T_1 = 365 \) days:
\[ T_2 = \frac{365}{8} \]
- Performing the division:
\[ T_2 = 45.625 \text{ days} \]

### Final Answer:
The new period of revolution of the Earth around the Sun when the distance is reduced to one-fourth will be approximately **45.6 days**.