Join IDNLearn.com to access a wealth of knowledge and get your questions answered by experts. Discover comprehensive answers from knowledgeable members of our community, covering a wide range of topics to meet all your informational needs.
Sagot :
To determine the mass of the Sun using Kepler's third law and the given data about Earth's orbit, follow these detailed steps:
### Given Information:
1. Average distance between Earth and the Sun ([tex]\(r\)[/tex]): [tex]\(1.496 \times 10^{11}\)[/tex] meters.
2. Earth's orbital period around the Sun ([tex]\(T\)[/tex]): 365.26 days.
### Convert Orbital Period into Seconds:
First, you need to convert the orbital period from days to seconds because the gravitational constant ([tex]\(G\)[/tex]) is in SI units (meters, kilograms, seconds).
1. There are 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute.
2. Thus, there are 24 60 60 = 86400 seconds in a day.
3. Now convert days to seconds:
[tex]\[ T = 365.26 \times 86400 = 31,558,464 \text{ seconds} \][/tex]
### Kepler's Third Law:
Kepler's third law is given by the formula:
[tex]\[ T^2 = \frac{4 \pi^2}{G M} r^3 \][/tex]
Where:
- [tex]\(T\)[/tex] is the orbital period,
- [tex]\(r\)[/tex] is the average distance between Earth and the Sun,
- [tex]\(G\)[/tex] is the gravitational constant ([tex]\(G = 6.67430 \times 10^{-11} \, m^3 kg^{-1} s^{-1}\)[/tex]),
- [tex]\(M\)[/tex] is the mass of the Sun.
### Rearrange the Formula to Solve for [tex]\(M\)[/tex]:
[tex]\[ M = \frac{4 \pi^2 r^3}{G T^2} \][/tex]
### Substitute the Known Values into the Formula:
[tex]\[ M = \frac{4 \pi^2 \left(1.496 \times 10^{11} \, m\right)^3}{6.67430 \times 10^{-11} \, \left(31,558,464 \, s\right)^2} \][/tex]
### Calculate the Numerical Values:
1. [tex]\(\pi\)[/tex] is approximately 3.141592653589793.
2. Calculate [tex]\( \left(1.496 \times 10^{11} \, m\right)^3 \)[/tex]:
[tex]\[ \left(1.496 \times 10^{11}\right)^3 = 3.35281216 \times 10^{33} \, m^3 \][/tex]
3. Calculate [tex]\( T^2 \)[/tex]:
[tex]\[ \left(31,558,464 \, s\right)^2 = 9.9623133 \times 10^{15} \, s^2 \][/tex]
Now substitute these into the formula:
[tex]\[ M = \frac{4 \times (3.141592653589793)^2 \times 3.35281216 \times 10^{33}}{6.67430 \times 10^{-11} \times 9.9623133 \times 10^{15}} \][/tex]
### Perform the Final Calculation:
[tex]\[ M \approx \frac{1.3271244 \times 10^{34}}{6.6627598 \times 10^{5}} \approx 1.9884610073711198 \times 10^{30} \, kg \][/tex]
### Match with Given Options:
Comparing the calculated mass to the given options:
[tex]\[ 1.99 \times 10^{30} \quad kg \][/tex]
Therefore, the mass of the Sun is approximately [tex]\(1.99 \times 10^{30} \, kg\)[/tex].
### Given Information:
1. Average distance between Earth and the Sun ([tex]\(r\)[/tex]): [tex]\(1.496 \times 10^{11}\)[/tex] meters.
2. Earth's orbital period around the Sun ([tex]\(T\)[/tex]): 365.26 days.
### Convert Orbital Period into Seconds:
First, you need to convert the orbital period from days to seconds because the gravitational constant ([tex]\(G\)[/tex]) is in SI units (meters, kilograms, seconds).
1. There are 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute.
2. Thus, there are 24 60 60 = 86400 seconds in a day.
3. Now convert days to seconds:
[tex]\[ T = 365.26 \times 86400 = 31,558,464 \text{ seconds} \][/tex]
### Kepler's Third Law:
Kepler's third law is given by the formula:
[tex]\[ T^2 = \frac{4 \pi^2}{G M} r^3 \][/tex]
Where:
- [tex]\(T\)[/tex] is the orbital period,
- [tex]\(r\)[/tex] is the average distance between Earth and the Sun,
- [tex]\(G\)[/tex] is the gravitational constant ([tex]\(G = 6.67430 \times 10^{-11} \, m^3 kg^{-1} s^{-1}\)[/tex]),
- [tex]\(M\)[/tex] is the mass of the Sun.
### Rearrange the Formula to Solve for [tex]\(M\)[/tex]:
[tex]\[ M = \frac{4 \pi^2 r^3}{G T^2} \][/tex]
### Substitute the Known Values into the Formula:
[tex]\[ M = \frac{4 \pi^2 \left(1.496 \times 10^{11} \, m\right)^3}{6.67430 \times 10^{-11} \, \left(31,558,464 \, s\right)^2} \][/tex]
### Calculate the Numerical Values:
1. [tex]\(\pi\)[/tex] is approximately 3.141592653589793.
2. Calculate [tex]\( \left(1.496 \times 10^{11} \, m\right)^3 \)[/tex]:
[tex]\[ \left(1.496 \times 10^{11}\right)^3 = 3.35281216 \times 10^{33} \, m^3 \][/tex]
3. Calculate [tex]\( T^2 \)[/tex]:
[tex]\[ \left(31,558,464 \, s\right)^2 = 9.9623133 \times 10^{15} \, s^2 \][/tex]
Now substitute these into the formula:
[tex]\[ M = \frac{4 \times (3.141592653589793)^2 \times 3.35281216 \times 10^{33}}{6.67430 \times 10^{-11} \times 9.9623133 \times 10^{15}} \][/tex]
### Perform the Final Calculation:
[tex]\[ M \approx \frac{1.3271244 \times 10^{34}}{6.6627598 \times 10^{5}} \approx 1.9884610073711198 \times 10^{30} \, kg \][/tex]
### Match with Given Options:
Comparing the calculated mass to the given options:
[tex]\[ 1.99 \times 10^{30} \quad kg \][/tex]
Therefore, the mass of the Sun is approximately [tex]\(1.99 \times 10^{30} \, kg\)[/tex].
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. Thank you for visiting IDNLearn.com. For reliable answers to all your questions, please visit us again soon.
