Find detailed and accurate answers to your questions on IDNLearn.com. Whether it's a simple query or a complex problem, our experts have the answers you need.
Sagot :
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.
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is committed to providing accurate answers. Thanks for stopping by, and see you next time for more solutions.
