To determine the mean distance of Jupiter from the center of the Sun, we start with Kepler's third law of planetary motion, given by the equation:
[tex]\[
T^2 = \frac{4 \pi^2}{G M} r^3
\][/tex]
Here:
- [tex]\( T \)[/tex] is the period of revolution, which is [tex]\( 3.79 \times 10^8 \)[/tex] seconds.
- [tex]\( M \)[/tex] is the mass of the Sun, which is [tex]\( 1.99 \times 10^{30} \)[/tex] kilograms.
- [tex]\( G \)[/tex] is the gravitational constant, which is [tex]\( 6.67 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)[/tex].
- [tex]\( \pi \approx 3.14 \)[/tex].
- [tex]\( r \)[/tex] is the mean distance from the center of the Sun, which we need to find.
First, we rearrange the equation to solve for [tex]\( r^3 \)[/tex]:
[tex]\[
r^3 = \frac{T^2 G M}{4 \pi^2}
\][/tex]
Next, we substitute the known values into the equation:
[tex]\[
r^3 = \frac{(3.79 \times 10^8)^2 \times 6.67 \times 10^{-11} \times 1.99 \times 10^{30}}{4 \times (3.14)^2}
\][/tex]
We calculate the numerator and the denominator separately:
1. Calculate [tex]\( T^2 \)[/tex]:
[tex]\[
T^2 = (3.79 \times 10^8)^2 = 1.43641 \times 10^{17}
\][/tex]
2. Multiply by [tex]\( G \)[/tex] and [tex]\( M \)[/tex]:
[tex]\[
1.43641 \times 10^{17} \times 6.67 \times 10^{-11} \times 1.99 \times 10^{30} = 1.90979499 \times 10^{37}
\][/tex]
3. Calculate [tex]\( 4 \pi^2 \)[/tex]:
[tex]\[
4 \times 3.14^2 = 4 \times 9.8596 = 39.4384
\][/tex]
Finally, divide the results to find [tex]\( r^3 \)[/tex]:
[tex]\[
r^3 = \frac{1.90979499 \times 10^{37}}{39.4384} \approx 4.84 \times 10^{35}
\][/tex]
To find [tex]\( r \)[/tex], take the cube root of [tex]\( r^3 \)[/tex]:
[tex]\[
r = \left( 4.84 \times 10^{35} \right)^{1/3}
\][/tex]
Approximating the cube root:
[tex]\[
r \approx 7.8 \times 10^{11} \, \text{meters}
\][/tex]
Thus, the mean distance of Jupiter from the center of the Sun is approximately [tex]\( 7.8 \times 10^{11} \)[/tex] meters. The correct answer is:
E. [tex]\( 7.8 \times 10^{11} \)[/tex] meters
