Europa, a moon of Jupiter, has an orbital diameter of 1.34 x 109 m, and a period of 3.55 days. What is the mass of Jupiter?

Question

BlackTom AI · Accepted Answer

We start by restating what we know and plan to compute the mass of Jupiter using the orbital parameters of Europa.
 Europa orbits Jupiter with an orbital diameter of 1.34 x 10^9 m, so the orbital radius a is half of that: a = 0.5 × 1.34 x 10^9 m = 6.7 x 10^8 m.
 The orbital period is 3.55 days. Convert this to seconds for SI units: T = 3.55 days × 86400 s/day = 306,720 s.
 Kepler's third law for a body in circular orbit around a much more massive planet gives M = (4π^2 a^3) / (G T^2), where G is the gravitational constant (G ≈ 6.674 × 10^-11 N·m^2/kg^2).
 Now compute step by step:
 - a^3 = (6.7 × 10^8 m)^3 = 6.7^3 × 10^(24) ≈ 300.763 × 10^24 ≈ 3.00763 × 10^26 m^3.
 - T^2 = (3.0672 × 10^5 s)^2 ≈ 9.405 × 10^10 s^2.
 - 4π^2 ≈ 39.478.
 - Numerator = (4π^2) × a^3 ≈ 39.478 × 3.00763 × 10^26 ≈ 1.187 × 10^28.
 - Denominator = G × T^2 ≈ (6.674 × 10^-11) × (9.405 × 10^10) ≈ 6.278 × 10^0 ≈ 6.278.
 - Therefore M ≈ (1.187 × 10^28) / 6.278 ≈ 1.89 × 10^27 kg.

Now evaluate the given answer options by comparison:
A. 1.87 × 10^27 kg — This value is very close to the computed 1.89 × 10^27 kg; it would be within typical rounding variation, but we assess all options individually.
B. 1.85 × 10^27 kg — Slightly lower than the calculated value; while plausible, it is a bit farther from the precise result given the inputs.
C. 1.89 × 10^27 kg — Matches the computed result from the orbital calculations using the provided a and T; this aligns with the derived formula and numbers.
D. 1.83 × 10^27 kg — Another lower value, further from the computed result than option B, indicating a larger deviation from the calculation.

In summary, the mass derived from applying the orbital relation with the given a and T yields about 1.89 × 10^27 kg, with nearby values differing by a few percent due to rounding in intermediate steps. The method hinges on using the correct radius a, convert period to seconds, apply Kepler’s law in the form M = (4π^2 a^3)/(G T^2), and carefully carry units through the calculation.

Europa, a moon of Jupiter, has an orbital diameter of 1.34 x 109 m, and a period of 3.55 days. What is the mass of Jupiter?单项选择题

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Europa, a moon of Jupiter, has an orbital diameter of 1.34 x 109 m, and a period of 3.55 days. What is the mass of Jupiter?单项选择题

类似问题

Why can Kepler’s third law be applied to estimate the Galaxy’s mass using the Sun’s orbit?

It takes the planet Jupiter 12 years to orbit the Sun once. What is the average distance from Jupiter to the Sun? (The distance from the Earth to the Sun is 1.5 × 1011 m.)

It takes the planet Jupiter 12 years to orbit the Sun once. What is the average distance from Jupiter to the Sun? (The distance from the Earth to the Sun is 1.5 × 1011 m.)

According to Kepler's 2nd law, during what month is Earth moving the fastest in its orbit?

Kepler's second law states that the line joining the Sun to a planet

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更多留学生实用工具

希望你的学习变得更简单

Kepler's First Law of Planetary Motion states that a planet will orbit on a(n) _ with the Sun located at _.