Answer :
To determine Earth's distance from the Sun using Mars's orbital information and Kepler's third law, follow these steps:
### Step-by-Step Solution:
1. Identify the 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
2. Write Kepler's Third Law in ratio form:
Kepler's Third Law can be represented as:
[tex]\[ \left(\frac{T_{1}}{T_{2}}\right)^2 = \left(\frac{R_{1}}{R_{2}}\right)^3 \][/tex]
Where [tex]\(T_{1}\)[/tex] and [tex]\(R_{1}\)[/tex] are the orbital period and distance for Mars, and [tex]\(T_{2}\)[/tex] and [tex]\(R_{2}\)[/tex] are the orbital period and distance for Earth respectively.
3. Given [tex]\(T_{2}\)[/tex] (Earth's orbital period):
[tex]\(T_{Earth} = 365.25\)[/tex] days
4. Set up the ratio equation for Mars and Earth:
[tex]\[ \left(\frac{T_{Earth}}{T_{Mars}}\right)^2 = \left(\frac{R_{Earth}}{R_{Mars}}\right)^3 \][/tex]
Substituting in the known values:
[tex]\[ \left(\frac{365.25}{687}\right)^2 = \left(\frac{R_{Earth}}{2.279 \times 10^{11}}\right)^3 \][/tex]
5. Solve for [tex]\(R_{Earth}\)[/tex]:
First, compute the ratio of the periods squared:
[tex]\[ \left(\frac{365.25}{687}\right)^2 \][/tex]
Simplify this to get a numerical value:
[tex]\[ \left(\frac{365.25}{687}\right)^2 \approx 0.282 \][/tex]
Now set this equal to the cubed ratio of the distances:
[tex]\[ 0.282 = \left(\frac{R_{Earth}}{2.279 \times 10^{11}}\right)^3 \][/tex]
6. Isolate [tex]\(R_{Earth}\)[/tex] by taking the cube root of both sides:
[tex]\[ \left(0.282\right)^{1/3} = \frac{R_{Earth}}{2.279 \times 10^{11}} \][/tex]
Calculate the cube root of 0.282:
[tex]\[ \left(0.282\right)^{1/3} \approx 0.657 \][/tex]
So,
[tex]\[ 0.657 = \frac{R_{Earth}}{2.279 \times 10^{11}} \][/tex]
7. Solve for [tex]\(R_{Earth}\)[/tex]:
[tex]\[ R_{Earth} = 0.657 \times 2.279 \times 10^{11} \][/tex]
Compute the multiplication:
[tex]\[ R_{Earth} \approx 1.498 \times 10^{11} \, \text{meters} \][/tex]
### Conclusion:
The Earth's distance from the Sun is approximately [tex]\(1.495 \times 10^{11}\)[/tex] meters, which is closest to the value [tex]\(1.49 \times 10^{11} \)[/tex] meters among the given options. Therefore, the correct answer is:
[tex]\[1.49 \times 10^{11} \, \text{meters}\][/tex]
### Step-by-Step Solution:
1. Identify the 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
2. Write Kepler's Third Law in ratio form:
Kepler's Third Law can be represented as:
[tex]\[ \left(\frac{T_{1}}{T_{2}}\right)^2 = \left(\frac{R_{1}}{R_{2}}\right)^3 \][/tex]
Where [tex]\(T_{1}\)[/tex] and [tex]\(R_{1}\)[/tex] are the orbital period and distance for Mars, and [tex]\(T_{2}\)[/tex] and [tex]\(R_{2}\)[/tex] are the orbital period and distance for Earth respectively.
3. Given [tex]\(T_{2}\)[/tex] (Earth's orbital period):
[tex]\(T_{Earth} = 365.25\)[/tex] days
4. Set up the ratio equation for Mars and Earth:
[tex]\[ \left(\frac{T_{Earth}}{T_{Mars}}\right)^2 = \left(\frac{R_{Earth}}{R_{Mars}}\right)^3 \][/tex]
Substituting in the known values:
[tex]\[ \left(\frac{365.25}{687}\right)^2 = \left(\frac{R_{Earth}}{2.279 \times 10^{11}}\right)^3 \][/tex]
5. Solve for [tex]\(R_{Earth}\)[/tex]:
First, compute the ratio of the periods squared:
[tex]\[ \left(\frac{365.25}{687}\right)^2 \][/tex]
Simplify this to get a numerical value:
[tex]\[ \left(\frac{365.25}{687}\right)^2 \approx 0.282 \][/tex]
Now set this equal to the cubed ratio of the distances:
[tex]\[ 0.282 = \left(\frac{R_{Earth}}{2.279 \times 10^{11}}\right)^3 \][/tex]
6. Isolate [tex]\(R_{Earth}\)[/tex] by taking the cube root of both sides:
[tex]\[ \left(0.282\right)^{1/3} = \frac{R_{Earth}}{2.279 \times 10^{11}} \][/tex]
Calculate the cube root of 0.282:
[tex]\[ \left(0.282\right)^{1/3} \approx 0.657 \][/tex]
So,
[tex]\[ 0.657 = \frac{R_{Earth}}{2.279 \times 10^{11}} \][/tex]
7. Solve for [tex]\(R_{Earth}\)[/tex]:
[tex]\[ R_{Earth} = 0.657 \times 2.279 \times 10^{11} \][/tex]
Compute the multiplication:
[tex]\[ R_{Earth} \approx 1.498 \times 10^{11} \, \text{meters} \][/tex]
### Conclusion:
The Earth's distance from the Sun is approximately [tex]\(1.495 \times 10^{11}\)[/tex] meters, which is closest to the value [tex]\(1.49 \times 10^{11} \)[/tex] meters among the given options. Therefore, the correct answer is:
[tex]\[1.49 \times 10^{11} \, \text{meters}\][/tex]
