Kepler's third law states that the square of the period of a planet (\( T \)) is proportional to the cube of the semi-major axis of its orbit (\( a \)) around the Sun. This can be expressed mathematically as:
\[ \frac{T_1^2}{a_1^3} = \frac{T_2^2}{a_2^3} \]
Where:
- \( T_1 \) and \( a_1 \) are the period and semi-major axis of Earth.
- \( T_2 \) and \( a_2 \) are the period and semi-major axis of Mars.
Given:
- \( a_1 = 149.6 \times 10^6 \text{ km} \)
- \( T_1 = 1 \text{ year} \)
- \( a_2 = 227.9 \times 10^6 \text{ km} \)
We need to find \( T_2 \).
First, we can plug in the values into Kepler's third law:
\[ \frac{1^2}{(149.6 \times 10^6)^3} = \frac{T_2^2}{(227.9 \times 10^6)^3} \]
Rearranging the equation to solve for \( T_2^2 \):
\[ T_2^2 = \left( \frac{(227.9 \times 10^6)^3}{(149.6 \times 10^6)^3} \right) \times 1^2 \]
Calculating the ratio of the distances:
\[ \frac{(227.9)^3}{(149.6)^3} = \left( \frac{227.9}{149.6} \right)^3 \]
Calculating the ratio:
\[ \frac{227.9}{149.6} \approx 1.524 \]
Now calculating \( (1.524)^3 \):
\[ (1.524)^3 \approx 3.531 \]
So now we have:
\[ T_2^2 \approx 3.531 \]
Taking the square root to find \( T_2 \):
\[ T_2 \approx \sqrt{3.531} \approx 1.88 \text{ years} \]
Thus, the period of Mars is approximately 1.9 years.
The answer is D. 1.9 years.