A planet is at R distance from sun and its time period of revolution is T. What will be its new time period of revolution if it brought 0.5 R distance closer to sun ?

Options

A. `(4T)/(sqrt(2))`
B. `(sqrt(2)T)/(4)`
C. `2sqrt(2)T`
D. `sqrt(2)T`

Answer

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

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

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- Initial distance from the sun: \( R \)
- Initial time period of revolution: \( T \)

2. **Apply Kepler's Third Law**:
- According to Kepler's law, we can write the relationship for the initial conditions:
\[ T^2 = k R^3 \]
where \( k \) is a constant of proportionality.

3. **Determine the new distance**:
- The new distance after moving 0.5R closer to the sun is:
\[ R' = R - 0.5R = 0.5R \]

4. **Apply Kepler's Third Law to the new distance**:
- For the new distance \( R' \), the new time period \( T' \) can be expressed as:
\[ T'^2 = k (R')^3 = k (0.5R)^3 \]

5. **Calculate \( (0.5R)^3 \)**:
- Simplifying \( (0.5R)^3 \):
\[ (0.5R)^3 = 0.5^3 R^3 = \frac{1}{8} R^3 \]

6. **Substitute back into the equation for \( T'^2 \)**:
\[ T'^2 = k \left(\frac{1}{8} R^3\right) \]

7. **Relate \( T'^2 \) to \( T^2 \)**:
- Since \( T^2 = k R^3 \), we can substitute this into our equation:
\[ T'^2 = \frac{1}{8} T^2 \]

8. **Solve for \( T' \)**:
- Taking the square root of both sides gives:
\[ T' = T \sqrt{\frac{1}{8}} = T \cdot \frac{1}{\sqrt{8}} = \frac{T}{2\sqrt{2}} \]

### Final Answer:
The new time period of revolution when the planet is brought 0.5R closer to the sun is:
\[ T' = \frac{T}{2\sqrt{2}} \]