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4 The rule (as in the book)

4 The rule (as in the book)

Point 19 of 21 — Module 9 — Components and Layers of the AtmosphereQuestion bank: 1 questions (0 with pictures)

19 5.4 The rule (as in the book)

  • If vrms < ve, the molecules cannot escape and the planet retains that gas — it keeps its atmosphere.
  • If vrms ≥ ve, the molecules can escape and drift into space; the gas becomes rare or absent on the surface.

Notes.

  • In reality the molecules in a gas have a range of speeds (a few move much faster than the average), so a planet loses a gas gradually when vrms is more than about one-sixth of ve over billions of years.
  • That is why Earth has lost most of its hydrogen and helium (very light) but keeps nitrogen, oxygen and carbon dioxide (heavy), and why the Moon and Mercury have little atmosphere at all, whereas giant planets keep hydrogen.

Problem. Calculate vrms for (a) hydrogen (M = 0.002 kg/mol) and (b) nitrogen (M = 0.028 kg/mol) at 300 K. Compare with Earth's escape velocity 11 200 m/s. Then calculate the vrms of oxygen (M = 0.032) at 700 K on Mercury's sunlit side and compare with Mercury's ve = 4250 m/s.

Solution. (a) H₂: √(3 × 8.314 × 300 / 0.002) = √(3.74 × 10⁶) = 1934 m/s (using M = 0.002; with 0.002016, 1924 m/s). (b) N₂: √(3 × 8.314 × 300 / 0.028) = √(2.67 × 10⁵) = 517 m/s. Both are far less than 11 200 m/s; Earth keeps nitrogen easily and (in the lower atmosphere) hydrogen; hydrogen at the hot top of the atmosphere (T ≈ 1000 K) has vrms ≈ 3500 m/s ≈ 31 % of ve, high enough for the fastest molecules to escape, so hydrogen slowly leaks away. (c) O₂ at 700 K: √(3 × 8.314 × 700/0.032) = √(5.46 × 10⁵) = 739 m/s — much less than 4250 m/s. So the simplified rule does not entirely explain why Mercury lacks an atmosphere: the answer is the combination of the weak gravity, the intense heating and solar wind that continuously strip gases, and the fact that fast molecules in the high-speed tail escape. The book's simplified rule is correct as a first approximation and is the one to use in answers.

Problem. Calculate the escape velocity from the Moon (M = 7.35 × 10²² kg, R = 1.737 × 10⁶ m).

Solution. 2GM/R = 2 × 6.67 × 10⁻¹¹ × 7.35 × 10²² / 1.737 × 10⁶ = 9.80 × 10¹² / 1.737 × 10⁶ = 5.64 × 10⁶ m²/s²; ve = 2375 m/s ≈ 2.4 km/s.

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