The radius of orbit of a planet is two times that of the earth. The time period of planet is

Options

A. `1yr`
B. `2.8yr`
C. `4yr`
D. `8yr`

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. This can be mathematically expressed as:

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

### Step-by-Step Solution:

1. **Identify the relationship from Kepler's Third Law**:
\[ \frac{T^2}{T_0^2} = \frac{r^3}{r_0^3} \]
where \(T_0\) and \(r_0\) are the time period and radius of the Earth's orbit, respectively.

2. **Set the values for Earth**:
Let the radius of the Earth’s orbit be \(r_0 = r\) and the time period be \(T_0 = T\) (1 year).

3. **Determine the radius of the planet's orbit**:
The radius of the planet’s orbit is given as \(r = 2r_0\).

4. **Substitute the values into Kepler's Law**:
\[ \frac{T^2}{T_0^2} = \frac{(2r_0)^3}{r_0^3} \]
Simplifying the right side:
\[ \frac{T^2}{T_0^2} = \frac{8r_0^3}{r_0^3} = 8 \]

5. **Express the time period of the planet**:
\[ T^2 = 8T_0^2 \]

6. **Take the square root of both sides**:
\[ T = T_0 \sqrt{8} = T_0 \cdot 2\sqrt{2} \]

7. **Substitute \(T_0 = 1\) year**:
\[ T = 2\sqrt{2} \text{ years} \]

8. **Calculate the numerical value**:
Approximating \(\sqrt{2} \approx 1.414\):
\[ T \approx 2 \times 1.414 \approx 2.828 \text{ years} \]

9. **Round to the nearest option**:
The closest whole number to 2.828 years is approximately 3 years, but since the options may vary, we can conclude that the nearest option is 2 years.

### Final Answer:
The time period of the planet is approximately **2 years**.