- Convert the orbital period from days to seconds: \(T = 365.26 \times 24 \times 3600 \approx 3.156 \times 10^7 \text{ s}\).
- Rearrange Kepler's Third Law to solve for the mass of the Sun: \(M = \frac{4\pi^2 r^3}{GT^2}\).
- Substitute the given values \(r = 1.496 \times 10^{11} \text{ m}\), \(T \approx 3.156 \times 10^7 \text{ s}\), and \(G \approx 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}\).
- The resulting mass of the Sun is approximately \(1.99 \times 10^{30} \text{ kg}\), so the correct answer is \(\boxed{1.99 \times 10^{30} \text{ kg}}\).
Explanation
- Problem Analysis and Physical Principles
We are given the orbital parameters of the Earth to determine the mass of the Sun. The average distance between the Earth and the Sun, which is the semi-major axis of the Earth's orbit, is given as \(r = 1.496 \times 10^{11} \text{ m}\). The orbital period of the Earth is \(T = 365.26 \text{ days}\). We also use the universal gravitational constant \(G = 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}\) and Kepler's Third Law, which relates these quantities by the equation:
\[T^2 = \frac{4\pi^2 r^3}{GM}\]
Our goal is to solve for the Sun's mass, \(M\).
- Unit Conversion and Formula Rearrangement
Before performing the calculation, we must ensure all units are in the SI system. The orbital period \(T\) is currently given in days, so we convert it into seconds:
\[T = 365.26 \text{ days} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} = 31,558,464 \text{ s}\]
Now, we rearrange Kepler's Third Law to isolate the mass of the Sun, \(M\):
\[M = \frac{4\pi^2 r^3}{GT^2}\]
This equation will allow us to compute the mass by substituting the values for \(r\), \(T\), and \(G\).
- Calculation and Comparison
Substituting the known values into the equation:
\[M = \frac{4 \times \pi^2 \times (1.496 \times 10^{11})^3}{(6.674 \times 10^{-11}) \times (31,558,464)^2}\]
Calculating the components:
- \(r^3 = (1.496 \times 10^{11})^3 \approx 3.348 \times 10^{33} \text{ m}^3\)
- \(T^2 = (31,558,464)^2 \approx 9.959 \times 10^{14} \text{ s}^2\)
- \(M = \frac{4 \times 9.8696 \times 3.348 \times 10^{33}}{6.674 \times 10^{-11} \times 9.959 \times 10^{14}} \approx \frac{1.321 \times 10^{35}}{6.646 \times 10^4} \approx 1.989 \times 10^{30} \text{ kg}\)
Comparing this value with the given options, we find it corresponds to \(1.99 \times 10^{30} \text{ kg}\).
Examples
Kepler's third law is fundamental in astrophysics. For instance, astronomers use it to determine the mass of distant stars or even supermassive black holes by observing the orbital period and distance of stars or gas clouds orbiting them. By measuring how fast an object orbits and how far it is from the center, we can weigh invisible massive objects in our galaxy, showcasing how physics bridges the gap between observation and deep understanding of the universe.