To determine Earth's distance from the Sun using Kepler's third law, we start by understanding the given data and the relationship between the orbital periods and distances of Mars and Earth.
Given data:
- Mars's orbital period ([tex]\(T_{mars}\)[/tex]): 687 days
- Mars's distance from the Sun ([tex]\(R_{mars}\)[/tex]): [tex]\(2.279 \times 10^{11}\)[/tex] meters
Possible distances for Earth:
1. [tex]\(1.49 \times 10^{11}\)[/tex] meters
2. [tex]\(1.49 \times 10^{33}\)[/tex] meters
3. [tex]\(3.34 \times 10^{11}\)[/tex] meters
4. [tex]\(3.34 \times 10^{33}\)[/tex] meters
Earth's orbital period ([tex]\(T_{earth}\)[/tex]): 365.25 days
According to Kepler's third law, the ratio of the squares of the periods ([tex]\(T\)[/tex]) of two planets is equal to the ratio of the cubes of their semi-major axes ([tex]\(R\)[/tex]):
[tex]\[
\left(\frac{T_{earth}}{T_{mars}}\right)^2 = \left(\frac{R_{earth}}{R_{mars}}\right)^3
\][/tex]
Rearranging this equation to solve for [tex]\(R_{earth}\)[/tex], we get:
[tex]\[
R_{earth} = R_{mars} \left(\frac{T_{earth}}{T_{mars}}\right)^{\frac{2}{3}}
\][/tex]
Now, plug in the given values:
- [tex]\(R_{mars} = 2.279 \times 10^{11}\)[/tex] meters
- [tex]\(T_{mars} = 687\)[/tex] days
- [tex]\(T_{earth} = 365.25\)[/tex] days
Substitute these values into the equation:
[tex]\[
R_{earth} = 2.279 \times 10^{11} \left(\frac{365.25}{687}\right)^{\frac{2}{3}}
\][/tex]
Calculate the ratio [tex]\(\frac{365.25}{687}\)[/tex]:
[tex]\[
\frac{365.25}{687} \approx 0.53151
\][/tex]
Now raise this ratio to the power of [tex]\(\frac{2}{3}\)[/tex]:
[tex]\[
0.53151^{\frac{2}{3}} \approx 0.655939
\][/tex]
Finally, multiply this result by [tex]\(R_{mars}\)[/tex]:
[tex]\[
R_{earth} = 2.279 \times 10^{11} \times 0.655939 \approx 1.495661507937e11
\][/tex]
[tex]\[
R_{earth} \approx 1.496 \times 10^{11} \text{ meters}
\][/tex]
Based on this calculation, the distance of Earth from the Sun is approximately [tex]\(1.496 \times 10^{11}\)[/tex] meters.
Among the provided options, the correct one is:
[tex]\[ 1.49 \times 10^{11} \text{ meters} \][/tex]
This result is in close agreement with the first option, confirming that [tex]\(1.49 \times 10^{11}\)[/tex] meters is the correct distance of Earth from the Sun according to Kepler's third law and the given data for Mars.
