To determine the orbital period of Neptune, we can use Kepler's third law, which relates the orbital period of a planet to its semi-major axis (the distance from the Sun in this context):
[tex]\[ T^2 = \frac{4\pi^2 r^3}{GM_{\text{sun}}} \][/tex]
where:
- [tex]\( T \)[/tex] is the orbital period of the planet,
- [tex]\( r \)[/tex] is the average distance from the planet to the Sun (in meters),
- [tex]\( G \)[/tex] is the gravitational constant [tex]\((6.67430 \times 10^{-11} \, m^3 kg^{-1} s^{-2})\)[/tex],
- [tex]\( M_{\text{sun}} \)[/tex] is the mass of the Sun [tex]\((2 \times 10^{30} \, kg)\)[/tex].
Firstly, we need to convert the distance from Astronomical Units (AU) to meters:
[tex]\[ 1 \, \text{AU} = 1.496 \times 10^{11} \, \text{m} \][/tex]
[tex]\[ \text{distance}_{\text{Neptune-Sun}} = 30 \, \text{AU} \times 1.496 \times 10^{11} \, \text{m/AU} = 4.488 \times 10^{12} \, \text{m} \][/tex]
Next, we plug in the values into Kepler's third law to find [tex]\( T \)[/tex]:
[tex]\[ T^2 = \frac{4 \pi^2 (4.488 \times 10^{12} \, \text{m})^3}{(6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2})(2 \times 10^{30} \,\text{kg})} \][/tex]
Calculating the numerator:
[tex]\[ 4 \pi^2 (4.488 \times 10^{12} \, \text{m})^3 \approx 2.67077 \times 10^{38} \, \text{m}^3 \][/tex]
Calculating the denominator:
[tex]\[ (6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2})(2 \times 10^{30} \, \text{kg}) = 1.33486 \times 10^{20} \, \text{m}^3 \text{s}^{-2} \][/tex]
Combining these:
[tex]\[ T^2 = \frac{2.67077 \times 10^{38}}{1.33486 \times 10^{20}} \approx 2.6735 \times 10^{19} \][/tex]
[tex]\[ T \approx \sqrt{2.6735 \times 10^{19}} \approx 5.1706 \times 10^{9} \, \text{s} \][/tex]
Finally, we convert the orbital period from seconds to Earth years:
[tex]\[ 1 \, \text{year} = 365.25 \, \text{days} \times 24 \, \text{hours/day} \times 60 \, \text{minutes/hour} \times 60 \, \text{seconds/minute} = 3.15576 \times 10^{7} \, \text{s/year} \][/tex]
[tex]\[ T_{\text{years}} = \frac{5.1706 \times 10^{9} \, \text{s}}{3.15576 \times 10^{7} \, \text{s/year}} \approx 163.85 \, \text{years} \][/tex]
Therefore, the orbital period of Neptune is approximately:
[tex]\[ 164 \, \text{Earth years} \][/tex]
So the correct answer is:
[tex]\[ \boxed{164 \, \text{Earth years}} \][/tex]
