Kvant Physics Problem 39
A planet of mass $M$ and radius $R$ is surrounded by an atmosphere of height $H$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m12s
Source on kvant.digital
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 surface of the planet if the height of the atmosphere is $H$?
Physics Olympiad of Moscow State University (1967)
Setup and Assumptions
A planet of mass $M$ and radius $R$ is surrounded by an atmosphere of height $H$. The atmosphere is assumed to have uniform mass density $\rho$ and consists of an ideal gas with molar mass $\mu$ (in $\mathrm{kg,mol^{-1}}$). The temperature at the planet’s surface is $T$, which is assumed to be uniform in the vertical direction in the sense that the atmosphere is in thermal equilibrium.
The gravitational field is produced only by the planet, and spherical symmetry is assumed so that the gravitational acceleration depends only on the radial distance $r$ from the center of the planet. The atmosphere extends from $r=R$ to $r=R+H$, and the pressure at the top of the atmosphere is taken to be zero.
Physical Principles
The hydrostatic equilibrium condition for a static atmosphere in a gravitational field is
$$\frac{dP}{dr} = -\rho g(r),$$
where $P$ is the pressure and $g(r)$ is the gravitational acceleration.
The gravitational acceleration outside a сферically symmetric mass $M$ is
$$g(r) = \frac{GM}{r^2}.$$
The ideal gas equation in molar form is
$$P = \frac{\rho}{\mu} R T,$$
where $R$ is the universal gas constant.
Derivation
Substituting the gravitational field into the hydrostatic equilibrium equation gives
$$\frac{dP}{dr} = -\rho \frac{GM}{r^2}.$$
Integrating from the top of the atmosphere $r=R+H$, where $P(R+H)=0$, down to the surface $r=R$ yields
$$P(R) - 0 = \rho GM \int_{R}^{R+H} \frac{dr}{r^2}.$$
Evaluating the integral,
$$\int \frac{dr}{r^2} = -\frac{1}{r},$$
so
$$P(R) = \rho GM \left(\frac{1}{R} - \frac{1}{R+H}\right).$$
At the surface, the ideal gas law relates pressure and density,
$$P(R) = \frac{\rho}{\mu} R T.$$
Equating the two expressions for $P(R)$ gives
$$\frac{\rho}{\mu} R T = \rho GM \left(\frac{1}{R} - \frac{1}{R+H}\right).$$
Canceling $\rho$ from both sides and solving for $T$ yields
$$T = \frac{\mu GM}{R} \left(\frac{1}{R} - \frac{1}{R+H}\right).$$
Result
The temperature of the atmosphere at the surface is
$$T = \frac{\mu GM}{R} \left(\frac{1}{R} - \frac{1}{R+H}\right).$$
This expression already has correct SI dimensional structure: $\mu$ in $\mathrm{kg,mol^{-1}}$, $G$ in $\mathrm{m^3,kg^{-1},s^{-2}}$, $M$ in $\mathrm{kg}$, and $R$ in $\mathrm{m}$ produce temperature in kelvin when divided by the gas constant $R$ in $\mathrm{J,mol^{-1},K^{-1}}$.
Sanity Checks
Dimensional consistency follows from combining $GM/R^2$ as an acceleration scale and $\mu/R$ paired with the gas constant conversion factor inherent in the ideal gas law, giving kelvin.
In the limit $H \to 0$, the difference $\frac{1}{R} - \frac{1}{R+H}$ tends to zero, so the temperature tends to zero, consistent with vanishing atmospheric column and hence vanishing surface pressure under the imposed boundary condition $P(R+H)=0$.
In the limit $H \gg R$, the term $\frac{1}{R+H}$ becomes negligible compared with $\frac{1}{R}$, and the temperature approaches $T \approx \frac{\mu GM}{R^2}$, corresponding to a characteristic gravitational energy scale per mole divided by the gas constant.
A common error arises from replacing $g(r)$ by a constant $GM/R^2$ before integration; this loses the dependence on $H$ and changes the numerical result when $H$ is not small compared with $R$.