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To determine the mean distance from the center of the Sun to Jupiter, we can use Kepler's third law, which is expressed as:
[tex]\[ T^2 = \frac{4 \pi^2}{GM} r^3 \][/tex]
Where:
- [tex]\( T \)[/tex] is the orbital period,
- [tex]\( G \)[/tex] is the gravitational constant,
- [tex]\( M \)[/tex] is the mass of the Sun,
- [tex]\( r \)[/tex] is the mean distance from the center of the Sun.
Given data:
- [tex]\( G = 6.67 \times 10^{-11} \, \frac{\text{Nm}^2}{\text{kg}^2} \)[/tex]
- [tex]\( T = 3.79 \times 10^8 \, \text{seconds} \)[/tex]
- [tex]\( M = 1.99 \times 10^{30} \, \text{kg} \)[/tex]
- [tex]\[ \pi \approx 3.14 \][/tex]
Rearrange the formula to solve for [tex]\( r \)[/tex]:
[tex]\[ r^3 = \frac{G M T^2}{4 \pi^2} \][/tex]
First, calculate the numerator:
[tex]\[ G M = (6.67 \times 10^{-11}) \times (1.99 \times 10^{30}) = 1.32733 \times 10^{20} \][/tex]
Next, calculate the denominator:
[tex]\[ 4 \pi^2 \approx 4 \times (3.14)^2 \approx 4 \times 9.8596 = 39.4384 \][/tex]
Now, substitute these values into the equation:
[tex]\[ r^3 = \frac{(1.32733 \times 10^{20}) \times (3.79 \times 10^8)^2}{39.4384} \][/tex]
Calculate [tex]\( T^2 \)[/tex]:
[tex]\[ T^2 = (3.79 \times 10^8)^2 = 1.43641 \times 10^{17} \][/tex]
Now substitute [tex]\( T^2 \)[/tex] into the formula for [tex]\( r^3 \)[/tex]:
[tex]\[ r^3 = \frac{(1.32733 \times 10^{20}) \times (1.43641 \times 10^{17})}{39.4384} \][/tex]
[tex]\[ r^3 \approx \frac{1.90565 \times 10^{37}}{39.4384} \][/tex]
[tex]\[ r^3 \approx 4.83434948 \times 10^{35} \][/tex]
Finally, take the cube root to find [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt[3]{4.83434948 \times 10^{35}} \][/tex]
[tex]\[ r \approx 784836780537.7351 \, \text{meters} \][/tex]
[tex]\[ r \approx 7.8 \times 10^{11} \, \text{meters} \][/tex]
Thus, the mean distance from the center of the Sun to Jupiter is approximately [tex]\( 7.8 \times 10^{11} \)[/tex] meters. The correct answer is:
E. [tex]\( 7.8 \times 10^{11} \)[/tex] meters
[tex]\[ T^2 = \frac{4 \pi^2}{GM} r^3 \][/tex]
Where:
- [tex]\( T \)[/tex] is the orbital period,
- [tex]\( G \)[/tex] is the gravitational constant,
- [tex]\( M \)[/tex] is the mass of the Sun,
- [tex]\( r \)[/tex] is the mean distance from the center of the Sun.
Given data:
- [tex]\( G = 6.67 \times 10^{-11} \, \frac{\text{Nm}^2}{\text{kg}^2} \)[/tex]
- [tex]\( T = 3.79 \times 10^8 \, \text{seconds} \)[/tex]
- [tex]\( M = 1.99 \times 10^{30} \, \text{kg} \)[/tex]
- [tex]\[ \pi \approx 3.14 \][/tex]
Rearrange the formula to solve for [tex]\( r \)[/tex]:
[tex]\[ r^3 = \frac{G M T^2}{4 \pi^2} \][/tex]
First, calculate the numerator:
[tex]\[ G M = (6.67 \times 10^{-11}) \times (1.99 \times 10^{30}) = 1.32733 \times 10^{20} \][/tex]
Next, calculate the denominator:
[tex]\[ 4 \pi^2 \approx 4 \times (3.14)^2 \approx 4 \times 9.8596 = 39.4384 \][/tex]
Now, substitute these values into the equation:
[tex]\[ r^3 = \frac{(1.32733 \times 10^{20}) \times (3.79 \times 10^8)^2}{39.4384} \][/tex]
Calculate [tex]\( T^2 \)[/tex]:
[tex]\[ T^2 = (3.79 \times 10^8)^2 = 1.43641 \times 10^{17} \][/tex]
Now substitute [tex]\( T^2 \)[/tex] into the formula for [tex]\( r^3 \)[/tex]:
[tex]\[ r^3 = \frac{(1.32733 \times 10^{20}) \times (1.43641 \times 10^{17})}{39.4384} \][/tex]
[tex]\[ r^3 \approx \frac{1.90565 \times 10^{37}}{39.4384} \][/tex]
[tex]\[ r^3 \approx 4.83434948 \times 10^{35} \][/tex]
Finally, take the cube root to find [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt[3]{4.83434948 \times 10^{35}} \][/tex]
[tex]\[ r \approx 784836780537.7351 \, \text{meters} \][/tex]
[tex]\[ r \approx 7.8 \times 10^{11} \, \text{meters} \][/tex]
Thus, the mean distance from the center of the Sun to Jupiter is approximately [tex]\( 7.8 \times 10^{11} \)[/tex] meters. The correct answer is:
E. [tex]\( 7.8 \times 10^{11} \)[/tex] meters
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