The distance of a planet from the sun is 5 times the distance between the earth and the sun. The time period of the planet is

Options

A. `5^(3//2)` years
B. `5^(2//3)` years
C. `5^(1//3)` years
D. `5^(1//2)` years

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 around the sun. Mathematically, this can be expressed as:

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

### Step-by-Step Solution:

1. **Identify the distances**:
Let \( r_1 \) be the distance of the Earth from the Sun, and \( r_2 \) be the distance of the planet from the Sun. According to the problem, \( r_2 = 5r_1 \).

2. **Apply Kepler's Third Law**:
According to Kepler's Third Law, we can write the relationship between the time periods of the Earth and the planet as:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]

3. **Substitute the distances**:
Substitute \( r_2 = 5r_1 \) into the equation:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{(5r_1)^3} \]

4. **Simplify the equation**:
This simplifies to:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{125r_1^3} = \frac{1}{125} \]

5. **Cross-multiply to find the relationship between T1 and T2**:
\[ T_1^2 = \frac{1}{125} T_2^2 \]
Taking the square root of both sides gives:
\[ \frac{T_1}{T_2} = \frac{1}{\sqrt{125}} = \frac{1}{5\sqrt{5}} \]

6. **Express T2 in terms of T1**:
Rearranging gives:
\[ T_2 = 5\sqrt{5} T_1 \]

7. **Substituting the known value of T1**:
We know that the time period of the Earth \( T_1 \) is 1 year. Therefore:
\[ T_2 = 5\sqrt{5} \times 1 \text{ year} \]

8. **Final result**:
Thus, the time period of the planet is:
\[ T_2 = 5\sqrt{5} \text{ years} \]