Taking that earth revolves round the sun ina circular orbit of radius `15xx10^10` m, with a time period 1 yr the time taken by another planet , which is at a distance of `540 xx 10^10` m, to revolve rounnd the sun in circular orbit once , will be

Options

A. 216 years
B. 144 years
C. 72 years
D. 36 years

Answer

To solve the problem of finding the time taken by another planet to revolve around the sun, we can use Kepler's Third Law of planetary motion. This law states that the square of the time period of a planet is directly proportional to the cube of the semi-major axis (radius) of its orbit.

### Step-by-Step Solution:

1. **Understand Kepler's Third Law**:
According to Kepler's Third Law, we have:
\[ \left( \frac{T_p}{T_e} \right)^2 = \left( \frac{R_p}{R_e} \right)^3 \]
where:
- \( T_p \) = time period of the planet
- \( T_e \) = time period of the Earth (1 year)
- \( R_p \) = radius of the planet's orbit
- \( R_e \) = radius of Earth's orbit

2. **Identify the Given Values**:
- \( R_e = 15 \times 10^{10} \) m (radius of Earth's orbit)
- \( R_p = 540 \times 10^{10} \) m (radius of the other planet's orbit)
- \( T_e = 1 \) year

3. **Substitute the Values into the Equation**:
We can substitute the known values into Kepler's Third Law:
\[ \left( \frac{T_p}{1 \text{ year}} \right)^2 = \left( \frac{540 \times 10^{10}}{15 \times 10^{10}} \right)^3 \]

4. **Simplify the Radius Ratio**:
Calculate the ratio of the radii:
\[ \frac{540 \times 10^{10}}{15 \times 10^{10}} = \frac{540}{15} = 36 \]
Thus, we have:
\[ \left( \frac{T_p}{1} \right)^2 = 36^3 \]

5. **Calculate \( 36^3 \)**:
Now, calculate \( 36^3 \):
\[ 36^3 = 36 \times 36 \times 36 = 1296 \times 36 = 46656 \]

6. **Find the Time Period of the Planet**:
Taking the square root to find \( T_p \):
\[ T_p = \sqrt{46656} = 216 \text{ years} \]

### Final Answer:
The time taken by the other planet to revolve around the sun in a circular orbit once is **216 years**.