Temperature at the centre of the earth

[Michael Smith]

Assume a straight hole is drilled in the Earth from one side to the other, and is filled with an ideal gas which comes to equilibrium. I know the temperature at the surface, and this is proportional to the mean kinetic energy per gas molecule. If the molecule is allowed to fall to the centre it gains velocity and energy that can be calculated. Therefore the temperature at the centre of the Earth is the subject of an armchair calculation.(??)

 

[Bill_K]

Interesting question! But the idea of dropping a molecule won't get you the right answer. It's a standard problem in thermo courses to calculate the 'lapse rate' of the atmosphere, i.e. how rapidly temperature falls off with altitude. And the argument used there can easily be extended. The key assumption they make is that the temperature and pressure changes are adiabatic. I won't go through the steps but they wind up with

dT/dz = - (γ - 1)/γ (μg/R)

where γ is the ratio of specific heats c_P/c_V, μ is the molecular weight of the gas (grams per mole), g is the acceleration due to gravity, and R is the ideal gas constant. What's different about the problem you suggest is that g varies on the way down. In fact as you fall down the hole it diminishes linearly: g(r) = g r/R_E, where R_E is the Earth's radius. So using that, we can say that

dT = - (γ - 1)/γ (μg/R) (r/R_E) dr

or integrating from r = 0 to R_E: (just ∫r dr)

T = T₀ - (γ - 1)/γ (μg/R) R_E/2

What a mess! But everything in there is a known constant. g = 980 cm/sec², R = 8.3 x 10⁷ ergs/deg/mole, and R_E = 6.4 x 10⁸ cm. For air, γ = 1.4 and μ = 28 g/mole. Multiply out and you get (I think!) T - T₀ ≈ 30,000 degrees K.