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].
