Mountain Top On A Faraway Planet

Classical Mechanics Level 5

Consider a planet with an atmosphere of ideal gas at constant temperature. The following is known:

  • Pressure at the surface: P 0 = 100 kPa ; P_0 = 100\ \text{kPa};
  • Pressure gradient at the surface: − d P / d y ∣ 0 = 12 Pa/m ; -dP/dy|_0 = 12\ \text{Pa/m};
  • Height of highest mountain: y ⋆ = 9000 m . y^\star = 9000\ \text{m}.

Determine the atmospheric pressure near the top of the highest mountain. Give your answer in kilopascals, rounded to the nearest integer.


The answer is 34.

Arjen Vreugdenhil by Arjen Vreugdenhil , 44, USA

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1 solution

The density of an ideal gas is directly proportional to the pressure: ρ ∝ P \rho \propto P .

The pressure gradient is d P d y = − ρ g = − κ P , \frac{dP}{dy} = -\rho\:g = -\kappa P, where κ \kappa is a constant depending on the gravitational constant, temperature, and molar mass of the gas.

Solving this differential equation we find P = P 0 e − κ y . P = P_0\ e^{-\kappa y}. It follows that γ : = − d P d y ∣ 0 = κ P 0 ∴ κ = γ P 0 , \gamma := -\left.\frac{dP}{dy}\right|_0 = \kappa P_0\ \ \ \ \ \therefore\ \ \ \ \ \kappa = \frac\gamma{P_0}, so that P = P 0 e − γ y / P 0 . P = P_0\ e^{-\gamma y/P_0}. Substitute the given values: P = 1 0 5 ⋅ e − 12 ⋅ 9000 / 1 0 5 = 1 0 5 ⋅ e − 10.8 = 33960 Pa , P = 10^5\cdot e^{-12\cdot 9000/10^5} = 10^5\cdot e^{-10.8} = 33960\ \text{Pa}, which we convert to 34 \boxed{34} kPa.

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Did you mean universal gas constant instead of gravitational constant?

Vilakshan Gupta - 1 year, 1 month ago

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