Problem 69 - University Physics with Modern Physics

Question

Kirkwood Gaps. Hundreds of thousands of asteroids orbit the sun within the asteroid belt, which extends from about \(3 \times 10^{8} \mathrm{km}\) to about \(5 \times 10^{8} \mathrm{km}\) from the sun. (a) Find the orbital period (in years) of (i) an asteroid at the inside of the belt and (ii) an asteroid at the outside of the belt. Assume circular orbits. (b) In 1867 the American astronomer Daniel Kirkwood pointed out that several gaps exist in the asteroid belt where relatively few asteroids are found. It is now understood that these Kirkwood gaps are caused by the gravitational attraction of Jupiter, the largest planet, which orbits the sun once every 11.86 years. As an example, if an asteroid has an orbital period half that of Jupiter, or 5.93 years, on every other orbit this asteroid would be at its closest to Jupiter and feel a strong attraction toward the planet. This attraction, acting over and over on successive orbits, could sweep asteroids out of the Kirkwood gap. Use this hypothesis to determine the orbital radius for this Kirkwood gap. (c) One of several other Kirkwood gaps appears at a distance from the sun where the orbital period is 0.400 that of Jupiter. Explain why this happens, and find the orbital radius for this Kirkwood gap.

Step-by-Step Solution

Answer

(a) 2.83 years at inside; 6.12 years at outside. (b) 3.275 AU for half period gap. (c) 2.708 AU for 0.400 period gap.

Step 1: Understanding Kepler's Third Law

Kepler's Third Law states that the square of the orbital period ( T ) of a planet is directly proportional to the cube of the semi-major axis ( a ) of its orbit, i.e., T^2 = k a^3 , where k is a constant. For the solar system, if the period is in years and the radius in astronomical units (AU), this relationship simplifies to T^2 = a^3 .

Step 2: Converting Distances to AU

The inside of the asteroid belt is at a distance of 3 imes 10^8 km and the outside at 5 imes 10^8 km. One AU (astronomical unit) is approximately 1.496 imes 10^8 km. Convert these distances to AU: 3 imes 10^8 km is 2.006 AU and 5 imes 10^8 km is 3.342 AU .

Step 3: Calculating Orbital Periods using Kepler's Law (Inside)

For an asteroid at the inside of the belt ( a = 2.006 AU ), find the orbital period using T^2 = a^3 . Calculate a^3 = (2.006)^3 and then find T by computing sqrt{a^3} . T approximately equals 2.83 years .

Step 4: Calculating Orbital Periods using Kepler's Law (Outside)

For an asteroid at the outside of the belt ( a = 3.342 AU ), find the orbital period using T^2 = a^3 . Calculate a^3 = (3.342)^3 and then find T by computing sqrt{a^3} . T approximately equals 6.12 years .

Step 5: Calculating Orbital Radius for Half Jupiter's Period

Jupiter's orbital period is 11.86 years . If an asteroid's period is half of this, then T_{asteroid} = 5.93 years . Using T^2 = a^3 , where T = 5.93 , solve for a to find the radius. (a^3 = (5.93)^2 ), which gives a approximately 3.275 AU .

Step 6: Explaining and Finding Orbital Radius for 0.400 Jupiter's Period

For an orbital period that is 0.400 of Jupiter's period, T_{asteroid} = 0.400 imes 11.86 = 4.744 years . Again use T^2 = a^3 , solve for a : (a^3 = (4.744)^2) gives a approximately 2.708 AU . This gap exists because of the strong periodic gravitational tug of Jupiter when the asteroid's period is a simple fraction of Jupiter's.

Key Concepts

Kepler's Third Law

Kepler's Third Law of planetary motion is an essential cornerstone in understanding orbital mechanics. This law provides a relationship between the time it takes for an object to orbit the sun (its orbital period) and the size of its orbit (the semi-major axis).
Specifically, Kepler's Third Law states that the square of the orbital period (\(T\)) is proportional to the cube of the semi-major axis (\(a\)) of its orbit. This can be mathematically expressed as: \[ T^2 = k \cdot a^3 \] where \(k\) is a constant that depends on the units used. For our solar system, if the period is measured in years and the semi-major axis in astronomical units (AU), \(k\) is effectively 1, simplifying the equation to: \[ T^2 = a^3 \] - This simplicity allows us to compute one of these variables easily when we know the other.
For instance, for asteroids within the asteroid belt, their distances from the sun can be converted to AU, and their orbital periods can be directly calculated using Kepler's Third Law. This relationship helps us understand the motion of not just asteroids but all celestial bodies orbiting larger masses.

Asteroid Belt

The asteroid belt is a fascinating region in our solar system located between the orbits of Mars and Jupiter. This belt contains millions of rocky bodies called asteroids.
The distance of the asteroid belt from the sun varies, with its inner boundary approximately 3 times \(10^8\) km and the outer edge extends to about 5 times \(10^8\) km. In terms of astronomical units (AU), this distance ranges from about 2.006 AU to 3.342 AU.
Understanding the distribution and movement of these asteroids is important in astronomy.

• Souces: Asteroids are remnants from the early solar system that never coalesced into a planet due to gravitational influences and other dynamic processes.
• Perturbations: They are in constant motion and experience perturbations due to gravitational pulls from nearby massive planets, particularly Jupiter.
• Kirkwood Gaps: Furthermore, the asteroid belt contains distinct gaps known as Kirkwood Gaps. These gaps are regions with fewer asteroids, further influenced by gravitational resonances from Jupiter, which affects their stable orbits.
  

Gravitational Resonance

Gravitational resonance occurs when two orbiting bodies exert a regular, periodic gravitational influence on each other. In the context of the asteroid belt, this is particularly important when considering the relationship between asteroids and Jupiter.

A gravitational resonance with Jupiter can occur when the orbital period of an asteroid is a simple fraction (like 1/2, 1/3, etc.) of Jupiter's orbital period. One example explained in the exercise is that if an asteroid's period is half that of Jupiter, every other orbit it comes close enough to Jupiter to experience a significant gravitational tug.

• Kirkwood Gaps: These repeated gravitational interactions over time can result in "Kirkwood Gaps," which are noticeable dips in the distribution of asteroids. They occur at orbital distances that correspond to these fractional resonances with Jupiter (e.g., 1/2, 1/3).
• Orbital Radius Impact: The gravitational resonance causes asteroids to be moved out of these gaps to orbits where they are less perturbed. For example, if their period resonates at 0.4 times Jupiter's period (\(4.744\) years), it can deflect asteroids to different paths, thus maintaining the gap.

These resonances showcase the dynamic and complex nature of celestial mechanics, emphasizing how massive bodies like Jupiter can influence the stability of smaller objects in nearby spatial zones.