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 of its orbit (R). Mathematically, this can be expressed as:

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

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial distance from the sun, \( R_1 = R \)
- Initial time period of revolution, \( T_1 = T \)
- New distance from the sun, \( R_2 = R - 0.5R = 0.5R \)

2. **Apply Kepler's Third Law:**
According to Kepler's law, we can write:
\[ \frac{T_1^2}{R_1^3} = \frac{T_2^2}{R_2^3} \]

3. **Substitute the Known Values:**
Substitute \( R_1 = R \) and \( R_2 = 0.5R \):
\[ \frac{T^2}{R^3} = \frac{T_2^2}{(0.5R)^3} \]

4. **Simplify the Right Side:**
Calculate \( (0.5R)^3 \):
\[ (0.5R)^3 = 0.125R^3 = \frac{1}{8}R^3 \]
So the equation becomes:
\[ \frac{T^2}{R^3} = \frac{T_2^2}{\frac{1}{8}R^3} \]

5. **Cross-Multiply to Solve for \( T_2^2 \):**
\[ T^2 \cdot \frac{1}{8}R^3 = T_2^2 \cdot R^3 \]
Dividing both sides by \( R^3 \) gives:
\[ \frac{T^2}{8} = T_2^2 \]

6. **Solve for \( T_2 \):**
Taking the square root of both sides:
\[ T_2 = \sqrt{\frac{T^2}{8}} = \frac{T}{\sqrt{8}} = \frac{T}{2\sqrt{2}} \]

7. **Final Expression:**
We can express \( T_2 \) as:
\[ T_2 = \frac{\sqrt{2}}{4} T \]

### Conclusion:
The new time period of revolution when the planet is brought 0.5R closer to the sun is:
\[ T_2 = \frac{\sqrt{2}}{4} T \]