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To find the orbital period of Neptune in Earth years, we follow these detailed steps:
1. Given Data:
- Mass of the Sun, [tex]\( M = 2 \times 10^{30} \)[/tex] kg.
- Average distance between Neptune and the Sun, [tex]\( r \)[/tex] = 30 Astronomical Units (AU).
2. Convert Distance to Meters:
- 1 AU (Astronomical Unit) is approximately [tex]\( 1.496 \times 10^{11} \)[/tex] meters.
- Therefore, the distance between Neptune and the Sun in meters is:
[tex]\[ r = 30 \text{ AU} \times 1.496 \times 10^{11} \text{ meters/AU} = 4.488 \times 10^{12} \text{ meters} \][/tex]
3. Gravitational Constant:
- [tex]\( G = 6.67430 \times 10^{-11} \, m^3 kg^{-1} s^{-2} \)[/tex]
4. Kepler's Third Law:
Kepler's third law states that:
[tex]\[ T^2 = \frac{4\pi^2 \cdot r^3}{G \cdot M} \][/tex]
Here, [tex]\( T \)[/tex] is the orbital period in seconds.
5. Calculate the Orbital Period in Seconds:
- Plugging in the values we have:
[tex]\[ T^2 = \frac{4 \pi^2 \cdot (4.488 \times 10^{12} \text{ meters})^3}{6.67430 \times 10^{-11} \, m^3 kg^{-1} s^{-2} \cdot 2 \times 10^{30} \text{ kg}} \][/tex]
- Simplifying the right-hand side:
[tex]\[ T^2 \approx \frac{4 \pi^2 \cdot 9.032 \times 10^{37}}{1.33486 \times 10^{20}} \][/tex]
[tex]\[ T^2 \approx 2.86459 \times 10^{18} \][/tex]
- Taking the square root to find [tex]\( T \)[/tex]:
[tex]\[ T \approx 5.17060 \times 10^9 \text{ seconds} \][/tex]
6. Convert the Orbital Period from Seconds to Earth Years:
- There are 365.25 days in a year, and each day has 24 hours, 60 minutes per hour, and 60 seconds per minute:
[tex]\[ \text{Seconds per year} = 365.25 \times 24 \times 3600 = 31,557,600 \text{ seconds/year} \][/tex]
- Therefore, the orbital period in Earth years is:
[tex]\[ T \approx \frac{5.17060 \times 10^9 \text{ seconds}}{31,557,600 \text{ seconds/year}} \approx 163.85 \text{ years} \][/tex]
Hence, the orbital period of Neptune is approximately 164 Earth years.
From the given choices:
- 30 Earth years
- 164 Earth years
- [tex]\(3.8 \times 10^{11}\)[/tex] Earth years
- [tex]\(2.3 \times 10^{17}\)[/tex] Earth years
The correct answer is:
[tex]\[ \boxed{164 \text{ Earth years}} \][/tex]
1. Given Data:
- Mass of the Sun, [tex]\( M = 2 \times 10^{30} \)[/tex] kg.
- Average distance between Neptune and the Sun, [tex]\( r \)[/tex] = 30 Astronomical Units (AU).
2. Convert Distance to Meters:
- 1 AU (Astronomical Unit) is approximately [tex]\( 1.496 \times 10^{11} \)[/tex] meters.
- Therefore, the distance between Neptune and the Sun in meters is:
[tex]\[ r = 30 \text{ AU} \times 1.496 \times 10^{11} \text{ meters/AU} = 4.488 \times 10^{12} \text{ meters} \][/tex]
3. Gravitational Constant:
- [tex]\( G = 6.67430 \times 10^{-11} \, m^3 kg^{-1} s^{-2} \)[/tex]
4. Kepler's Third Law:
Kepler's third law states that:
[tex]\[ T^2 = \frac{4\pi^2 \cdot r^3}{G \cdot M} \][/tex]
Here, [tex]\( T \)[/tex] is the orbital period in seconds.
5. Calculate the Orbital Period in Seconds:
- Plugging in the values we have:
[tex]\[ T^2 = \frac{4 \pi^2 \cdot (4.488 \times 10^{12} \text{ meters})^3}{6.67430 \times 10^{-11} \, m^3 kg^{-1} s^{-2} \cdot 2 \times 10^{30} \text{ kg}} \][/tex]
- Simplifying the right-hand side:
[tex]\[ T^2 \approx \frac{4 \pi^2 \cdot 9.032 \times 10^{37}}{1.33486 \times 10^{20}} \][/tex]
[tex]\[ T^2 \approx 2.86459 \times 10^{18} \][/tex]
- Taking the square root to find [tex]\( T \)[/tex]:
[tex]\[ T \approx 5.17060 \times 10^9 \text{ seconds} \][/tex]
6. Convert the Orbital Period from Seconds to Earth Years:
- There are 365.25 days in a year, and each day has 24 hours, 60 minutes per hour, and 60 seconds per minute:
[tex]\[ \text{Seconds per year} = 365.25 \times 24 \times 3600 = 31,557,600 \text{ seconds/year} \][/tex]
- Therefore, the orbital period in Earth years is:
[tex]\[ T \approx \frac{5.17060 \times 10^9 \text{ seconds}}{31,557,600 \text{ seconds/year}} \approx 163.85 \text{ years} \][/tex]
Hence, the orbital period of Neptune is approximately 164 Earth years.
From the given choices:
- 30 Earth years
- 164 Earth years
- [tex]\(3.8 \times 10^{11}\)[/tex] Earth years
- [tex]\(2.3 \times 10^{17}\)[/tex] Earth years
The correct answer is:
[tex]\[ \boxed{164 \text{ Earth years}} \][/tex]
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