The mean distance of mars from sun is 1.5 times that of earth from sun. What is approximately the number of years required by mars to make one revolution about sun ?

Options

A. 2.35 years
B. 1.84 years
C. 3.65 years
D. 2.75 years

Explanation

To solve the problem of determining the number of years required by Mars to make one revolution around the Sun, we can use Kepler's Third Law of planetary motion. This law states that the square of the period of revolution (T) of a planet is directly proportional to the cube of the semi-major axis of its orbit (r), which can be expressed mathematically as:

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

### Step-by-Step Solution:

1. **Identify the given information**:
- The mean distance of Mars from the Sun (r_M) is 1.5 times the mean distance of Earth from the Sun (r_E).
- Therefore, we can express this as:
\[ r_M = 1.5 \cdot r_E \]

2. **Apply Kepler's Third Law**:
- According to Kepler's Third Law, we can write:
\[ \frac{T_M^2}{T_E^2} = \frac{r_M^3}{r_E^3} \]
where:
- \( T_M \) is the orbital period of Mars,
- \( T_E \) is the orbital period of Earth (which is 1 year).

3. **Substitute the value of r_M**:
- Substitute \( r_M = 1.5 \cdot r_E \) into the equation:
\[ \frac{T_M^2}{T_E^2} = \frac{(1.5 \cdot r_E)^3}{r_E^3} \]

4. **Simplify the equation**:
- This simplifies to:
\[ \frac{T_M^2}{T_E^2} = \frac{1.5^3 \cdot r_E^3}{r_E^3} \]
- The \( r_E^3 \) cancels out:
\[ \frac{T_M^2}{T_E^2} = 1.5^3 \]

5. **Calculate \( 1.5^3 \)**:
- Calculate \( 1.5^3 \):
\[ 1.5^3 = 3.375 \]
- Therefore:
\[ \frac{T_M^2}{T_E^2} = 3.375 \]

6. **Find \( T_M \)**:
- Taking the square root of both sides gives:
\[ \frac{T_M}{T_E} = \sqrt{3.375} \]
- Calculate \( \sqrt{3.375} \):
\[ T_M \approx 1.84 \cdot T_E \]

7. **Substitute \( T_E \)**:
- Since \( T_E = 1 \) year:
\[ T_M \approx 1.84 \cdot 1 \text{ year} \]
- Thus:
\[ T_M \approx 1.84 \text{ years} \]

### Final Answer:
Mars takes approximately **1.84 years** to make one revolution around the Sun.