Answer :
To determine the orbital period of Neptune around the Sun, we will use Kepler's Third Law, which relates the orbital period [tex]\( T \)[/tex] to the semi-major axis of the orbit [tex]\( r \)[/tex] for an object orbiting the Sun. The formula is given by:
[tex]\[ T^2 = \frac{4 \pi^2 r^3}{GM} \][/tex]
where:
- [tex]\( T \)[/tex] is the orbital period,
- [tex]\( r \)[/tex] is the average distance between the Sun and Neptune,
- [tex]\( G \)[/tex] is the gravitational constant,
- [tex]\( M \)[/tex] is the mass of the Sun.
### Step-by-Step Solution
1. Given Data:
[tex]\[ M = 2 \times 10^{30} \ \text{kg} \][/tex]
[tex]\[ r = 30 \ \text{AU} \][/tex]
2. Constants:
[tex]\[ \text{1 AU} = 1.496 \times 10^{11} \ \text{meters} \][/tex]
[tex]\[ G = 6.67430 \times 10^{-11} \ \text{m}^3 \ \text{kg}^{-1} \ \text{s}^{-2} \][/tex]
[tex]\[ \text{1 year} = 3.154 \times 10^7 \ \text{seconds} \][/tex]
3. Convert the distance from AU to meters:
[tex]\[ r = 30 \ \text{AU} \times 1.496 \times 10^{11} \ \text{meters/AU} \][/tex]
[tex]\[ r = 4.488 \times 10^{12} \ \text{meters} \][/tex]
4. Apply Kepler's Third Law to find the period in seconds:
[tex]\[ T^2 = \frac{4 \pi^2 r^3}{GM} \][/tex]
[tex]\[ T^2 = \frac{4 \pi^2 \left(4.488 \times 10^{12} \ \text{meters}\right)^3}{6.67430 \times 10^{-11} \ \text{m}^3 \ \text{kg}^{-1} \ \text{s}^{-2} \times 2 \times 10^{30} \ \text{kg}} \][/tex]
[tex]\[ T^2 = \frac{4 \pi^2 \left(9.036 \times 10^{37} \ \text{m}^3\right)}{1.33486 \times 10^{20} \ \text{m}^3 \ \text{s}^{-2}} \][/tex]
[tex]\[ T^2 = \frac{3.55407 \times 10^{39}}{1.33486 \times 10^{20}} \][/tex]
[tex]\[ T^2 = 2.663 \times 10^{19} \ \text{s}^2 \][/tex]
[tex]\[ T = \sqrt{2.663 \times 10^{19}} \][/tex]
[tex]\[ T \approx 5.1706 \times 10^9 \ \text{seconds} \][/tex]
5. Convert the period from seconds to years:
[tex]\[ T = \frac{5.1706 \times 10^9 \ \text{seconds}}{3.154 \times 10^7 \ \text{seconds/year}} \][/tex]
[tex]\[ T \approx 163.938 \ \text{years} \][/tex]
Therefore, the orbital period of Neptune around the Sun is approximately 164 Earth years.
[tex]\[ T^2 = \frac{4 \pi^2 r^3}{GM} \][/tex]
where:
- [tex]\( T \)[/tex] is the orbital period,
- [tex]\( r \)[/tex] is the average distance between the Sun and Neptune,
- [tex]\( G \)[/tex] is the gravitational constant,
- [tex]\( M \)[/tex] is the mass of the Sun.
### Step-by-Step Solution
1. Given Data:
[tex]\[ M = 2 \times 10^{30} \ \text{kg} \][/tex]
[tex]\[ r = 30 \ \text{AU} \][/tex]
2. Constants:
[tex]\[ \text{1 AU} = 1.496 \times 10^{11} \ \text{meters} \][/tex]
[tex]\[ G = 6.67430 \times 10^{-11} \ \text{m}^3 \ \text{kg}^{-1} \ \text{s}^{-2} \][/tex]
[tex]\[ \text{1 year} = 3.154 \times 10^7 \ \text{seconds} \][/tex]
3. Convert the distance from AU to meters:
[tex]\[ r = 30 \ \text{AU} \times 1.496 \times 10^{11} \ \text{meters/AU} \][/tex]
[tex]\[ r = 4.488 \times 10^{12} \ \text{meters} \][/tex]
4. Apply Kepler's Third Law to find the period in seconds:
[tex]\[ T^2 = \frac{4 \pi^2 r^3}{GM} \][/tex]
[tex]\[ T^2 = \frac{4 \pi^2 \left(4.488 \times 10^{12} \ \text{meters}\right)^3}{6.67430 \times 10^{-11} \ \text{m}^3 \ \text{kg}^{-1} \ \text{s}^{-2} \times 2 \times 10^{30} \ \text{kg}} \][/tex]
[tex]\[ T^2 = \frac{4 \pi^2 \left(9.036 \times 10^{37} \ \text{m}^3\right)}{1.33486 \times 10^{20} \ \text{m}^3 \ \text{s}^{-2}} \][/tex]
[tex]\[ T^2 = \frac{3.55407 \times 10^{39}}{1.33486 \times 10^{20}} \][/tex]
[tex]\[ T^2 = 2.663 \times 10^{19} \ \text{s}^2 \][/tex]
[tex]\[ T = \sqrt{2.663 \times 10^{19}} \][/tex]
[tex]\[ T \approx 5.1706 \times 10^9 \ \text{seconds} \][/tex]
5. Convert the period from seconds to years:
[tex]\[ T = \frac{5.1706 \times 10^9 \ \text{seconds}}{3.154 \times 10^7 \ \text{seconds/year}} \][/tex]
[tex]\[ T \approx 163.938 \ \text{years} \][/tex]
Therefore, the orbital period of Neptune around the Sun is approximately 164 Earth years.