Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, it can be expressed 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 average distance from the sun of the first planet (Earth, in this case).
- \( T_2 \) and \( a_2 \) are the period and average distance from the sun of the second planet (Pluto).
Given:
- \( a_1 = 149.6 \times 10^6 \) km (Earth's average distance from the sun)
- \( T_1 = 1.0 \) year (Earth's period)
- \( a_2 = 1427 \times 10^6 \) km (Pluto's average distance from the sun)
We need to find \( T_2 \), the period of Pluto.
First, calculate \( a_1^3 \) and \( a_2^3 \):
\[ a_1^3 = (149.6 \times 10^6)^3 \] \[ a_2^3 = (1427 \times 10^6)^3 \]
Using the formula from Kepler's third law:
\[ \frac{T_1^2}{a_1^3} = \frac{T_2^2}{a_2^3} \]
Rearranging for \( T_2 \):
\[ T_2^2 = T_1^2 \cdot \frac{a_2^3}{a_1^3} \] \[ T_2 = T_1 \cdot \sqrt{\frac{a_2^3}{a_1^3}} \]
Now substituting the values:
- Compute \( \frac{a_2^3}{a_1^3} \):
\[ \frac{a_2^3}{a_1^3} = \left( \frac{a_2}{a_1} \right)^3 = \left( \frac{1427 \times 10^6}{149.6 \times 10^6} \right)^3 \]
This simplifies to:
\[ \frac{1427}{149.6} \approx 9.527 \] Now, we can raise this to the power of 3:
\[ 9.527^3 \approx 860 \]
- Now plug this into the period equation:
\[ T_2 = 1 \cdot \sqrt{860} \approx 29.32 \text{ years} \]
So, rounding off, the approximate period of Pluto is:
B. 29 years
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