Answer :
To determine the mean distance from Jupiter to the center of the Sun, we use Kepler’s third law of planetary motion, which can be expressed as:
[tex]\[ T^2 = \frac{4 \pi^2}{G M} r^3 \][/tex]
where:
- [tex]\( T \)[/tex] is the orbital period of the planet,
- [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 planet to the Sun (semi-major axis of the orbit),
- [tex]\( \pi \)[/tex] is the mathematical constant pi.
Given data:
- Orbital period [tex]\( T = 3.79 \times 10^8 \)[/tex] seconds,
- Mass of the Sun [tex]\( M = 1.99 \times 10^{30} \)[/tex] kilograms,
- Gravitational constant [tex]\( G = 6.67 \times 10^{-11} \frac{Nm^2}{kg^2} \)[/tex],
- [tex]\( \pi = 3.14 \)[/tex].
Let's proceed with the solution step-by-step:
1. Calculate [tex]\( T^2 \)[/tex]:
[tex]\[ T^2 = (3.79 \times 10^8)^2 = 1.43641 \times 10^{17} \][/tex]
2. Calculate the numerator [tex]\( G \cdot M \cdot T^2 \)[/tex]:
[tex]\[ G \cdot M \cdot T^2 = 6.67 \times 10^{-11} \cdot 1.99 \times 10^{30} \cdot 1.43641 \times 10^{17} \][/tex]
[tex]\[ = 1.9065900853 \times 10^{37} \][/tex]
3. Calculate the denominator [tex]\( 4 \pi^2 \)[/tex]:
[tex]\[ 4 \pi^2 = 4 \times (3.14)^2 = 4 \times 9.8596 = 39.4384 \][/tex]
4. Calculate [tex]\( r^3 \)[/tex]:
[tex]\[ r^3 = \frac{G \cdot M \cdot T^2}{4 \pi^2} = \frac{1.9065900853 \times 10^{37}}{39.4384} \][/tex]
[tex]\[ = 4.834349479948476 \times 10^{35} \][/tex]
5. Calculate [tex]\( r \)[/tex] by taking the cubic root of [tex]\( r^3 \)[/tex]:
[tex]\[ r = \left( 4.834349479948476 \times 10^{35} \right)^{\frac{1}{3}} = 7.848367805377351 \times 10^{11} \text{ meters} \][/tex]
Comparing this calculated value with the given options, we find that the closest option to our calculated mean distance is:
[tex]\[ E. \ 7.8 \times 10^{11} \text{ meters} \][/tex]
So, the correct answer is [tex]\( \boxed{7.8 \times 10^{11}} \text{ meters} \)[/tex].
[tex]\[ T^2 = \frac{4 \pi^2}{G M} r^3 \][/tex]
where:
- [tex]\( T \)[/tex] is the orbital period of the planet,
- [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 planet to the Sun (semi-major axis of the orbit),
- [tex]\( \pi \)[/tex] is the mathematical constant pi.
Given data:
- Orbital period [tex]\( T = 3.79 \times 10^8 \)[/tex] seconds,
- Mass of the Sun [tex]\( M = 1.99 \times 10^{30} \)[/tex] kilograms,
- Gravitational constant [tex]\( G = 6.67 \times 10^{-11} \frac{Nm^2}{kg^2} \)[/tex],
- [tex]\( \pi = 3.14 \)[/tex].
Let's proceed with the solution step-by-step:
1. Calculate [tex]\( T^2 \)[/tex]:
[tex]\[ T^2 = (3.79 \times 10^8)^2 = 1.43641 \times 10^{17} \][/tex]
2. Calculate the numerator [tex]\( G \cdot M \cdot T^2 \)[/tex]:
[tex]\[ G \cdot M \cdot T^2 = 6.67 \times 10^{-11} \cdot 1.99 \times 10^{30} \cdot 1.43641 \times 10^{17} \][/tex]
[tex]\[ = 1.9065900853 \times 10^{37} \][/tex]
3. Calculate the denominator [tex]\( 4 \pi^2 \)[/tex]:
[tex]\[ 4 \pi^2 = 4 \times (3.14)^2 = 4 \times 9.8596 = 39.4384 \][/tex]
4. Calculate [tex]\( r^3 \)[/tex]:
[tex]\[ r^3 = \frac{G \cdot M \cdot T^2}{4 \pi^2} = \frac{1.9065900853 \times 10^{37}}{39.4384} \][/tex]
[tex]\[ = 4.834349479948476 \times 10^{35} \][/tex]
5. Calculate [tex]\( r \)[/tex] by taking the cubic root of [tex]\( r^3 \)[/tex]:
[tex]\[ r = \left( 4.834349479948476 \times 10^{35} \right)^{\frac{1}{3}} = 7.848367805377351 \times 10^{11} \text{ meters} \][/tex]
Comparing this calculated value with the given options, we find that the closest option to our calculated mean distance is:
[tex]\[ E. \ 7.8 \times 10^{11} \text{ meters} \][/tex]
So, the correct answer is [tex]\( \boxed{7.8 \times 10^{11}} \text{ meters} \)[/tex].
