Europa, a satellite of Jupiter, has an orbital diameter of $1.34 \times 10 ^ { 9 } \mathrm {~m}$ and a period of 3.55 days. What is the mass of Jupiter? $\left( G = 6.67 \times 10 ^ { - 11 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { kg } ^ { 2 } \right)$
A) $1.89 \times 10 ^ { 27 } \mathrm {~kg}$
B) $1.07 \times 10^{27} \mathrm {~kg}$
C) $1.53 \times 10 ^ { 27 } \mathrm {~kg}$
D) $3.08 \times 10 ^ { 27 } \mathrm {~kg}$
E) $1.65 \times 10^{27} \mathrm {~kg}$

Question

Europa, a satellite of Jupiter, has an orbital diameter of $1.34 \times 10 ^ { 9 } \mathrm {~m}$  and a period of  3.55  days. What is the mass of Jupiter? $\left( G = 6.67 \times 10 ^ { - 11 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { kg } ^ { 2 } \right)$
A) $1.89 \times 10 ^ { 27 } \mathrm {~kg}$
B) $1.07 \times 10^{27} \mathrm {~kg}$
C) $1.53 \times 10 ^ { 27 } \mathrm {~kg}$
D) $3.08 \times 10 ^ { 27 } \mathrm {~kg}$
E) $1.65 \times 10^{27} \mathrm {~kg}$

Quizplus · Accepted Answer

To find the mass of Jupiter, we can use the formula derived from Kepler's third law and Newton's law of universal gravitation: $T^2 = \frac{4\pi^2}{G M}r^3$, where $T$ is the orbital period, $G$ is the gravitational constant, $M$ is the mass of Jupiter, and $r$ is the semi-major axis of the orbit (half the orbital diameter). Rearranging for $M$, we get $M = \frac{4\pi^2r^3}{G T^2}$.Given:- Orbital diameter = $1.34 \times 10^9 \, \text{m}$, so $r = 0.5 \times 1.34 \times 10^9 \, \text{m} = 6.7 \times 10^8 \, \text{m}$- Period ($T$) = 3.55 days = $3.55 \times 24 \times 3600 \, \text{s}$- Gravitational constant ($G$) = $6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2$Plugging these values into the formula:$$M = \frac{4\pi^2(6.7 \times 10^8)^3}{6.67 \times 10^{-11} \times (3.55 \times 24 \times 3600)^2}$$$$M \approx 1.89 \times 10^{27} \, \text{kg}$$Therefore, the mass of Jupiter is approximately $1.89 \times 10^{27} \, \text{kg}$.

Europa, a Satellite of Jupiter, Has an Orbital Diameter Of 1.34