Which statements regarding Kepler’s third law, p2 = a3 are true? (select all that apply) Group of answer choices If Jupiter were as close to the Sun as Earth, it would orbit faster than Mercury. If Earth were replaced by with a small asteroid, the asteroid would take 1 year to orbit. A planet orbiting in a circular orbit at 30 AU will have the same orbital period as a comet in a very eccentric orbit at 30 AU. An asteroid orbiting the Sun at 3 AU will take less than 4 years to orbit the Sun.
The Correct Answer and Explanation is:
Kepler’s Third Law of planetary motion states that the square of the orbital period (p²) of a planet is directly proportional to the cube of the semi-major axis (a³) of its orbit. Mathematically, this is expressed as:
p2=a3p^2 = a^3p2=a3
Where:
- ppp is the orbital period in Earth years,
- aaa is the semi-major axis (average distance from the Sun) in astronomical units (AU).
Now, let’s evaluate each statement:
- If Jupiter were as close to the Sun as Earth, it would orbit faster than Mercury. False. According to Kepler’s Third Law, the closer a planet is to the Sun (smaller semi-major axis, aaa), the faster its orbital period is. Jupiter, being much farther from the Sun than Earth, has a much longer orbital period. If it were at Earth’s distance, it would have a similar orbital period to Earth, not faster than Mercury, which is much closer to the Sun.
- If Earth were replaced by a small asteroid, the asteroid would take 1 year to orbit. True. According to Kepler’s Third Law, any object orbiting the Sun at the same distance as Earth (1 AU) will have an orbital period of 1 year, regardless of its size, as the orbital period is determined only by the semi-major axis of the orbit. Whether it’s a planet or an asteroid, the orbital period will remain the same if the distance from the Sun is the same.
- A planet orbiting in a circular orbit at 30 AU will have the same orbital period as a comet in a very eccentric orbit at 30 AU. True. The orbital period of a planet or any object depends solely on the semi-major axis of the orbit, not the shape of the orbit (whether circular or elliptical). If both the planet and the comet are at 30 AU from the Sun, their orbital periods will be the same, as Kepler’s Law only considers the average distance from the Sun (semi-major axis).
- An asteroid orbiting the Sun at 3 AU will take less than 4 years to orbit the Sun. True. Using Kepler’s Third Law, we can calculate the orbital period for an object at 3 AU. From the formula p2=a3p^2 = a^3p2=a3, if a=3a = 3a=3, then p2=33=27p^2 = 3^3 = 27p2=33=27. Thus, p=27≈5.2p = \sqrt{27} \approx 5.2p=27≈5.2 years. However, this seems to contradict the statement, as it should take about 5.2 years, not less than 4 years. If you reconsider the math, the correct interpretation here is that the period is around 5.2 years, not less than 4.
Summary of the True Statements:
- The asteroid orbiting at 3 AU will take more than 4 years (this part is actually incorrect).
- A planet and a comet at the same distance (30 AU) will have the same orbital period.
