To solve for the distance of the planet from the Sun in Astronomical Units (AU), we can use Kepler's Third Law of Planetary Motion, which states that
\[ T^2 \propto r^3 \]
We can express this as:
\[ T^2 = k \cdot r^3 \]
where \( k \) is a constant that can be determined based on Earth’s orbit, where \( T = 1 \) year and \( r = 1 \) AU. For Earth:
\[ 1^2 = k \cdot 1^3 \] \[ k = 1 \]
Now we can use this relationship to determine \( r \) for the given orbital period \( T = 29.46 \) years:
- Calculate \( T^2 \):
\[ T^2 = (29.46)^2 = 869.0516 \]
- Set that equal to \( r^3 \):
\[ r^3 = 869.0516 \]
- Now solve for \( r \) by taking the cube root:
\[ r = \sqrt[3]{869.0516} \]
Calculating the cube root:
\[ r \approx 9.54 , \text{AU} \]
Thus, the planet is approximately 9.54 AU away from the Sun.
Therefore, the correct response is:
9.54 AU