Certainly! To find the mean distance of Jupiter from the center of the Sun, denoted as [tex]\( r \)[/tex], we'll use Kepler's Third Law, given by the formula:
[tex]\[ T^2 = \frac{4 \pi^2}{G M} r^3, \][/tex]
where:
- [tex]\( T \)[/tex] is the orbital period of Jupiter,
- [tex]\( G \)[/tex] is the gravitational constant,
- [tex]\( M \)[/tex] is the mass of the Sun,
- [tex]\( \pi \)[/tex] is a mathematical constant Pi,
- [tex]\( r \)[/tex] is the mean distance from the Sun.
Given values are:
- [tex]\( T = 3.79 \times 10^8 \)[/tex] seconds,
- [tex]\( M = 1.99 \times 10^{30} \)[/tex] kg,
- [tex]\( G = 6.67 \times 10^{-11} \, \frac{\text{Nm}^2}{\text{kg}^2} \)[/tex],
- [tex]\(\pi = 3.14\)[/tex].
1. Calculate [tex]\( T^2 \)[/tex]:
[tex]\[ T^2 = (3.79 \times 10^8)^2 = 1.43641 \times 10^{17} \, \text{seconds}^2 \][/tex]
2. Calculate the constant [tex]\(\frac{4 \pi^2}{G M}\)[/tex]:
[tex]\[
\frac{4 \pi^2}{G M} = \frac{4 \times (3.14)^2}{(6.67 \times 10^{-11}) \times (1.99 \times 10^{30})} \approx 2.971258089548191 \times 10^{-19} \, \text{m}^{-3} \text{s}^2
\][/tex]
3. Using Kepler's Third Law to find [tex]\( r^3 \)[/tex]:
[tex]\[
r^3 = T^2 \times \left( \frac{G M}{4 \pi^2} \right) = \frac{1.43641 \times 10^{17}}{2.971258089548191 \times 10^{-19}} = 4.834349479948476 \times 10^{35} \, \text{m}^3
\][/tex]
4. Take the cube root of [tex]\( r^3 \)[/tex] to find [tex]\( r \)[/tex]:
[tex]\[
r = (4.834349479948476 \times 10^{35})^{1/3} \approx 784836780537.7351 \, \text{meters}
\][/tex]
Thus, the mean distance [tex]\( r \)[/tex] from Jupiter to the center of the Sun is approximately [tex]\( 7.8 \times 10^{11} \)[/tex] meters.
Hence, the correct answer is:
E. [tex]\( 7.8 \times 10^{11} \)[/tex] meters
