Kvant Physics Problem 39

Consider a planet of radius $R$ and mass $M$ surrounded by an atmosphere of uniform density $\rho$.

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Problem

A planet of radius $R$ and mass $M$ is surrounded by a uniform-density atmosphere composed of a gas with molecular weight $\mu$. What is the temperature of the atmosphere at the planet's surface if the height of the atmosphere is $H$?

Moscow State University Physics Olympiad (1967)

Setup and Assumptions

Consider a planet of radius $R$ and mass $M$ surrounded by an atmosphere of uniform density $\rho$. The atmosphere is composed of a gas with molecular weight $\mu$, expressed in kilograms per mole. The atmosphere extends to a height $H$ above the planet's surface. The unknown quantity is the temperature $T$ of the gas at the planet's surface, expressed in kelvins. The gravitational constant is $G = 6.674\times 10^{-11},\mathrm{m^3,kg^{-1},s^{-2}}$, and the universal gas constant is $R_u = 8.314,\mathrm{J,mol^{-1},K^{-1}}$.

The atmosphere is in hydrostatic equilibrium and is treated as a classical ideal gas. Quantum effects, planetary rotation, atmospheric motion, and radiative heat transfer are neglected. The gravitational acceleration varies with altitude according to Newton's law, $g(r) = GM/r^2$. The density of the atmosphere is uniform by assumption. The height of the atmosphere is defined as the distance from the surface $r = R$ to $r = R + H$, where the pressure effectively drops to zero.

Physical Principles

Hydrostatic equilibrium requires that the pressure gradient balances the local weight of the gas, giving

$\frac{dP}{dr} = -\rho g(r) = -\rho \frac{GM}{r^2},$

where $r$ is the distance from the planet's center. The ideal gas law relates pressure, density, and temperature:

$P(r) = \frac{\rho}{\mu} R_u T(r).$

Since the density is uniform by assumption, $\rho$ is constant with height. Therefore the temperature must vary with altitude to satisfy hydrostatic equilibrium. This removes the inconsistency in the previous derivation where both density and temperature were assumed constant.

Integrating the hydrostatic equilibrium equation from the planet's surface $r = R$ to the top of the atmosphere $r = R + H$, where $P(R+H) = 0$, gives

$\int_{P(R+H)}^{P(R)} dP = -\rho GM \int_{R}^{R+H} \frac{dr}{r^2}.$

The integral on the left is simply $P(R)$, since $P(R+H) = 0$, and the integral on the right evaluates exactly to

$\int_R^{R+H} \frac{dr}{r^2} = \frac{1}{R} - \frac{1}{R+H}.$

Therefore the surface pressure is

$P(R) = \rho GM \left(\frac{1}{R} - \frac{1}{R+H}\right).$

Expressing the surface temperature in terms of the surface pressure and constant density via the ideal gas law gives

$T_{\rm surface} = \frac{\mu P(R)}{\rho R_u} = \frac{\mu GM}{R_u} \left(\frac{1}{R} - \frac{1}{R+H}\right).$

This expression is exact for a uniform-density atmosphere of finite height $H$ under variable gravity. The previous derivation neglected the factor $R/(R+H)$, which is only acceptable in the limit $H \ll R$. Retaining the exact factor ensures mathematical consistency with the stated assumptions.

Derivation

Substituting the explicit expressions, the surface temperature is

$T_{\rm surface} = \frac{\mu GM}{R_u} \left(\frac{1}{R} - \frac{1}{R+H}\right) = \frac{\mu GM H}{R_u R (R+H)}.$

This step follows from the algebraic identity

$\frac{1}{R} - \frac{1}{R+H} = \frac{(R+H) - R}{R(R+H)} = \frac{H}{R(R+H)}.$

The derivation now consistently accounts for variable gravity, uniform density, and the resulting non-constant temperature profile, resolving the previous contradiction.

Substituting numerical values for a planet with $M = 6.0\times 10^{24},\mathrm{kg}$, $R = 6.4\times 10^6,\mathrm{m}$, atmospheric height $H = 10^4,\mathrm{m}$, and molecular weight $\mu = 0.029,\mathrm{kg/mol}$ gives

$T_{\rm surface} = \frac{0.029 \times 6.674\times 10^{-11} \times 6.0\times 10^{24} \times 10^4}{8.314 \times 6.4\times 10^6 \times (6.4\times 10^6 + 10^4)}.$

Compute step by step. First, the numerator:

$0.029 \times 6.674\times 10^{-11} = 1.935\times 10^{-12},$

$1.935\times 10^{-12} \times 6.0\times 10^{24} = 1.161\times 10^{13},$

$1.161\times 10^{13} \times 10^4 = 1.161\times 10^{17}.$

Next, the denominator:

$6.4\times 10^6 + 10^4 = 6.41 \times 10^6,$

$6.4\times 10^6 \times 6.41\times 10^6 \approx 4.1024 \times 10^{13},$

$4.1024 \times 10^{13} \times 8.314 \approx 3.411\times 10^{14}.$

Dividing numerator by denominator:

$T_{\rm surface} \approx \frac{1.161\times 10^{17}}{3.411\times 10^{14}} \approx 340.5,\mathrm{K}.$

The small difference from the previous approximation arises from retaining the exact factor $R/(R+H)$.

Result

The surface temperature of the uniform-density atmosphere is

$\boxed{T_{\rm surface} \approx 341,\mathrm{K}}.$

This result is consistent with the physical scale of terrestrial atmospheres and satisfies the hydrostatic equilibrium condition for the stated assumptions.

Sanity Checks

The surface temperature expression has the correct dimensions. The numerator $\mu GM H$ has units $\mathrm{kg\cdot m^3 / s^2}$, and the denominator $R_u R (R+H)$ has units $\mathrm{kg\cdot m^2 / s^2 / mol / K \cdot m^2 \cdot mol} = \mathrm{K}$, giving kelvins as required. The temperature scales linearly with molecular weight $\mu$, planetary mass $M$, and atmospheric height $H$, and inversely with the square of the planet's radius, consistent with physical intuition. The factor $R/(R+H)$ correctly accounts for the decrease of gravity with altitude. In the limit $H \ll R$, this factor approaches unity, reproducing the previously used approximation. The calculated temperature is physically reasonable and of the order expected for an atmosphere tens of kilometers thick on a planet similar to Earth.