Kvant Physics Problem 43

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Problem

Find the pressure at the center of a liquid planet of radius $R$, if the liquid is incompressible and has density $\rho$.

All-Union Physics Olympiad (1969)

Setup and Assumptions

A spherical planet of radius $R$ is filled with an incompressible жидкость of constant density $\rho$ in hydrostatic equilibrium under its own gravitational field. The unknown is the pressure at the center, denoted $P_c$, measured relative to the pressure at the surface. The gravitational constant is $G$. The pressure at the surface is taken as zero, corresponding to vacuum outside the planet.

The liquid is assumed to be perfectly static, with no rotation and no external gravitational fields. The density does not vary with pressure. The gravitational field is produced only by the liquid mass itself, and the configuration is spherically symmetric.

Physical Principles

Hydrostatic equilibrium in a self-gravitating fluid is described by the differential equation

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

where $g(r)$ is the magnitude of the gravitational acceleration at radius $r$.

The gravitational field inside a spherically symmetric mass distribution follows from Newton’s law,

$g(r) = \frac{G M(r)}{r^2},$

where $M(r)$ is the mass enclosed within radius $r$.

For constant density,

$M(r) = \frac{4}{3}\pi r^3 \rho.$

Derivation

Substituting the enclosed mass into the expression for the gravitational field gives

$g(r) = \frac{G \cdot \frac{4}{3}\pi r^3 \rho}{r^2} = \frac{4}{3}\pi G \rho r.$

The hydrostatic equilibrium equation becomes

$\frac{dP}{dr} = -\rho \cdot \frac{4}{3}\pi G \rho r = -\frac{4}{3}\pi G \rho^2 r.$

Integrating from the center to the surface, with $P(R)=0$, yields

$P(R) - P(0) = \int_{0}^{R} -\frac{4}{3}\pi G \rho^2 r , dr.$

Evaluating the integral gives

$-P_c = -\frac{4}{3}\pi G \rho^2 \int_{0}^{R} r , dr = -\frac{4}{3}\pi G \rho^2 \cdot \frac{R^2}{2}.$

Rearranging,

$P_c = \frac{2}{3}\pi G \rho^2 R^2.$

Result

$P_c = \frac{2}{3}\pi G \rho^2 R^2.$

The expression contains no additional numerical parameters in the problem statement, so it is already in final form.

The units follow from $[G] = \mathrm{m^3,kg^{-1},s^{-2}}$, giving

$[P_c] = \mathrm{m^3,kg^{-1},s^{-2}} \cdot \mathrm{kg^2,m^{-6}} \cdot \mathrm{m^2} = \mathrm{kg,m^{-1},s^{-2}} = \mathrm{Pa}.$

Sanity Checks

The dependence on $R^2$ matches the linear growth of gravitational acceleration inside a uniform sphere, since $g(r) \propto r$ leads to a pressure gradient proportional to $r$ and therefore a quadratic accumulation with radius.

The dependence on $\rho^2$ arises because one factor of $\rho$ determines the mass generating the gravitational field, while the second factor determines the weight of the fluid responding to that field.

Dimensional consistency reduces to $G\rho^2R^2$, which produces pressure units $\mathrm{Pa}$.

In the limit $R \to 0$, the pressure tends to zero because both the gravitational field and the volume of fluid vanish.