Orbits, Kepler's laws and resonances
Kepler's third law connects an asteroid's distance from the Sun to its orbital period. Combined with Jupiter's gravity, it explains the gaps in the asteroid belt and the delivery of meteorites.
- Derive Kepler's third law for circular orbits from Newton's law of gravitation.
- Locate a Kirkwood gap from a resonance with Jupiter.
- Explain how the Yarkovsky effect and resonances deliver meteoroids to Earth.
Links to: A-level physics, gravitational fields and circular motion: orbital speed and period, and T² ∝ r³.
Kepler's third law
For a circular orbit the gravitational force provides the centripetal force: GMm/r² = mω²r, with ω = 2π/T. Rearranging gives T² = (4π²/GM) r³, so T² ∝ r³ for all bodies orbiting the same central mass.
For the Sun, with r in astronomical units and T in years, the constant is 1, so T² = r³. Jupiter, at 5.20 AU, has a period of about 11.9 years. The law also holds for elliptical orbits if r is replaced by the semi-major axis, a.
Kirkwood gaps
In 1866 Daniel Kirkwood noticed gaps in the distribution of asteroid distances. They occur where an asteroid's period is a simple fraction of Jupiter's: 1/3, 2/5, 3/7 and 1/2. At such a mean-motion resonance, Jupiter's pull is repeated at the same point of the orbit and accumulates rather than averaging out, increasing the orbit's eccentricity until the asteroid crosses the orbits of Mars or Earth, or is thrown out. A secular resonance with Saturn, at the inner edge of the belt, does the same.
The conveyor belt
Asteroids do not sit in the gaps; something must feed them in. The Yarkovsky effect does so: a rotating body absorbs sunlight and re-emits it as infrared slightly later, from its afternoon side, giving a tiny thrust that changes its semi-major axis. Bodies metres to kilometres across drift slowly until they reach a resonance, which then pumps up their eccentricity within about a million years. Cosmic-ray exposure ages of stony meteorites, mostly a few to tens of millions of years, are consistent with this slow drift.
Finding the 3:1 Kirkwood gap
Jupiter orbits at 5.20 AU. Where is the 3:1 resonance, at which an asteroid completes three orbits for every one of Jupiter's?
- The asteroid's period is T = T_J/3.
- Kepler's third law: (a/a_J)³ = (T/T_J)², so a = a_J × (1/3)^(2/3).
- (1/3)^(2/3) = 3^(−2/3) = 1/2.080 ≈ 0.481.
- a = 5.20 × 0.481 ≈ 2.50 AU.
Answer: About 2.50 AU. The 3:1 gap is a major route by which fragments, including those that become ordinary chondrites, reach Earth-crossing orbits.
Calculate the semi-major axes for the 5:2 and 2:1 resonances with Jupiter. Then take a few asteroids from the asteroids section of this site, note their semi-major axes and orbital periods, and check how well T² = a³ holds for each.
Quiz
5 questions. Choose an answer to see whether you are right and why. Finish the quiz to complete the lesson.
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Question 1From GMm/r² = mω²r, which relationship follows?
Substituting ω = 2π/T and rearranging gives T² = 4π²r³/GM.
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Question 2An asteroid has a semi-major axis of 4.0 AU. What is its period?
T² = a³ = 64, so T = 8.0 years.
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Question 3Why are Kirkwood gaps nearly empty?
Resonant perturbations accumulate, driving asteroids onto planet-crossing orbits.
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Question 4What causes the Yarkovsky effect?
Heat absorbed on the day side is radiated later from the afternoon side, producing a tiny net force.
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Question 5What do cosmic-ray exposure ages of a few to tens of millions of years tell us?
Exposure ages record time spent as metre-scale bodies exposed to cosmic rays, matching slow delivery by Yarkovsky drift and resonances.