The mean radius of the earth's orbit round the sun is `1.5xx10^(11)` . The mean radius of the orbit of mercury round the sun is `6xx10^(10)m`. The mercury will rotate around the sun in

Answer

To solve the problem of determining the time period of Mercury's orbit around the Sun using Kepler's Third Law of planetary motion, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mean radius of Earth's orbit (R1) = \(1.5 \times 10^{11} \, \text{m}\)
- Mean radius of Mercury's orbit (R2) = \(6 \times 10^{10} \, \text{m}\)

2. **Use Kepler's Third Law:**
According to Kepler's Third Law, 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:
\[ \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \]
Here, \(T_1\) is the time period of Earth, and \(T_2\) is the time period of Mercury.

3. **Substitute the Known Values:**
We know that the time period of Earth (T1) is 1 year. Therefore, we can write:
\[ \frac{1^2}{T_2^2} = \frac{(1.5 \times 10^{11})^3}{(6 \times 10^{10})^3} \]

4. **Calculate the Radii Cubes:**
- Calculate \(R_1^3\):
\[ R_1^3 = (1.5 \times 10^{11})^3 = 3.375 \times 10^{33} \, \text{m}^3 \]
- Calculate \(R_2^3\):
\[ R_2^3 = (6 \times 10^{10})^3 = 2.16 \times 10^{32} \, \text{m}^3 \]

5. **Set Up the Equation:**
Substitute the values into the equation:
\[ \frac{1}{T_2^2} = \frac{3.375 \times 10^{33}}{2.16 \times 10^{32}} \]

6. **Simplify the Right Side:**
\[ \frac{3.375 \times 10^{33}}{2.16 \times 10^{32}} = \frac{3.375}{2.16} \times 10^{1} \approx 1.56 \times 10^{1} \approx 15.6 \]
Therefore,
\[ \frac{1}{T_2^2} = 15.6 \]

7. **Solve for \(T_2^2\):**
\[ T_2^2 = \frac{1}{15.6} \]

8. **Calculate \(T_2\):**
\[ T_2 = \sqrt{\frac{1}{15.6}} \approx \sqrt{0.0641} \approx 0.253 \, \text{years} \]

### Final Answer:
The time period for Mercury to rotate around the Sun is approximately **0.253 years**.