Atmospheric entry and fragmentation
A meteoroid is decelerated, heated and often shattered by the air. Simple models of drag and ram pressure explain where it breaks up and how much survives.
- Model drag deceleration using momentum transfer to the air.
- Calculate ram pressure and compare it with material strength.
- Explain why meteorites are stronger than the bodies they came from.
Links to: A-level physics, mechanics and materials: momentum change, force as rate of change of momentum, pressure and breaking stress.
Drag as momentum transfer
A meteoroid of cross-sectional area A moving at speed v through air of density ρ sweeps up a mass of air ρAv each second. If that air is pushed to roughly the meteoroid's speed, the rate of change of momentum, and so the drag force, is of order ρAv². Including a drag coefficient C_D of order 1 gives F = ½C_DρAv².
The meteoroid decelerates significantly once it has swept up a mass of air comparable to its own mass. Small bodies are therefore slowed high up, while large bodies keep their speed much deeper into the atmosphere.
Ram pressure and break-up
The pressure on the front face is approximately the ram pressure, P ≈ ρv². Air density increases roughly exponentially with decreasing altitude, so ram pressure rises steeply as a meteoroid descends. When P exceeds the body's strength, it fragments. The pieces spread sideways, present a larger total area, decelerate and release their energy rapidly, producing a flare or an airburst.
Strength depends on size
Laboratory samples of ordinary chondrite have compressive strengths of tens to a few hundred megapascals, yet fireball observations show that metre-scale meteoroids typically break up at ram pressures of only about 0.1 to 10 MPa. Large bodies contain cracks and weak zones, and many asteroids are loosely bound rubble piles. Break-up continues until the surviving pieces are small enough, and crack-free enough, to withstand the load. Meteorites in our collections are therefore the strongest survivors of much weaker parent bodies.
Ram pressure at 30 km
At 30 km altitude the air density is about 0.018 kg m⁻³. Calculate the ram pressure on a meteoroid travelling at 19 km s⁻¹ and compare it with a typical bulk strength of 1 to 5 MPa.
- P ≈ ρv² = 0.018 × (1.9 × 10⁴)².
- (1.9 × 10⁴)² = 3.61 × 10⁸ m² s⁻².
- P ≈ 0.018 × 3.61 × 10⁸ ≈ 6.5 × 10⁶ Pa = 6.5 MPa.
Answer: About 6.5 MPa, above the bulk strength, so a body like Chelyabinsk's was expected to disintegrate at around this altitude, as observed.
Air density is roughly ρ = 1.2 exp(−h/7,000) kg m⁻³, with h in metres. Use a spreadsheet to calculate ram pressure against altitude from 80 km to 10 km for entry speeds of 15, 20 and 30 km s⁻¹ (ignoring deceleration). For each, find the altitude where pressure first exceeds 1 MPa. Discuss what the simplification of constant speed overlooks.
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 1Which expression approximates the ram pressure on a meteoroid?
Ram pressure is the momentum flux of the oncoming air, of order ρv².
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Question 2Why does ram pressure rise steeply as a meteoroid descends?
Even as it slows, the rapid rise in air density increases ρv².
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Question 3When does a meteoroid begin to decelerate strongly?
Momentum transfer to the air becomes significant once the intercepted air mass approaches the body's mass.
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Question 4Why do metre-scale meteoroids break up at pressures far below the strength of meteorite samples?
Bigger bodies are more likely to contain flaws, so their bulk strength is lower.
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Question 5Air at 40 km has ρ ≈ 0.004 kg m⁻³. What is the ram pressure at 20 km s⁻¹?
P = 0.004 × (2 × 10⁴)² = 0.004 × 4 × 10⁸ = 1.6 × 10⁶ Pa.