Kvant Physics Problem 43
A planet is modeled as a sphere of radius $R$ filled with an incompressible liquid of constant density $\rho$.
Verified: yes
Verdicts: PASS + PASS
Solve time: 5m48s
Source on kvant.digital
Problem
Find the pressure at the center of a liquid planet of radius $R$, assuming that the liquid is incompressible and has density $\rho$.
All-Union Physics Olympiad (1969)
Setup and Assumptions
A planet is modeled as a sphere of radius $R$ filled with an incompressible liquid of constant density $\rho$.
The required quantity is the pressure at the center of the planet, denoted by $p_c$.
The given quantities are the planet radius $R$ (m) and the liquid density $\rho$ (kg/m$^3$).
The liquid is assumed to be at rest in hydrostatic equilibrium. The density is uniform throughout the planet. The planet is perfectly spherical. Rotation, tidal forces, temperature effects, and compressibility are neglected. The pressure at the surface is taken to be zero, corresponding to a vacuum outside the planet. Pressure depends only on the distance from the center.
Physical Principles
For a fluid in hydrostatic equilibrium, the pressure gradient balances the gravitational force:
$$\frac{dp}{dr}=-\rho g(r),$$
where $r$ is the distance from the center and $g(r)$ is the gravitational acceleration at that radius.
The gravitational field inside a spherically symmetric body depends only on the mass enclosed within radius $r$:
$$g(r)=\frac{G M(r)}{r^2},$$
where $G$ is the gravitational constant.
For a sphere of constant density $\rho$, the enclosed mass is
$$M(r)=\frac{4}{3}\pi r^3\rho.$$
Derivation
Substituting the enclosed mass into the expression for the gravitational field gives
$$g(r)=\frac{G}{r^2}\left(\frac{4}{3}\pi r^3\rho\right) =\frac{4}{3}\pi G\rho, r.$$
The hydrostatic equation becomes
$$\frac{dp}{dr} = -\rho\left(\frac{4}{3}\pi G\rho, r\right) = -\frac{4}{3}\pi G\rho^2 r.$$
Integrating from the center to an arbitrary radius $r$,
$$p(r)-p_c = -\frac{4}{3}\pi G\rho^2 \int_0^r r',dr'.$$
Evaluating the integral,
$$p(r)-p_c = -\frac{4}{3}\pi G\rho^2 \left(\frac{r^2}{2}\right).$$
Hence
$$p(r) = p_c-\frac{2}{3}\pi G\rho^2 r^2.$$
At the surface, $r=R$, and the pressure is zero:
$$p(R)=0.$$
Substituting $r=R$,
$$0 = p_c-\frac{2}{3}\pi G\rho^2 R^2.$$
Solving for the central pressure,
$$p_c = \frac{2}{3}\pi G\rho^2 R^2.$$
Result
The pressure at the center of an incompressible liquid planet is
$$p_c=\frac{2}{3}\pi G\rho^2R^2.$$
Since the problem gives only the parameters $\rho$ and $R$, no further numerical substitution is possible.
The final answer is
$$\boxed{p_c=\frac{2}{3}\pi G\rho^2R^2}$$
with SI units
$$[\text{Pa}] = \left[\frac{\text{m}^3}{\text{kg},\text{s}^2}\right] \left[\frac{\text{kg}^2}{\text{m}^6}\right] [\text{m}^2] = \frac{\text{kg}}{\text{m},\text{s}^2}.$$
Thus the result has the correct unit of pressure, pascals.
Sanity Checks
The expression contains the factor $\rho^2$, so the central pressure vanishes when $\rho \to 0$. A body with no mass produces no self-gravity, and no pressure is required to support the liquid.
The expression contains the factor $R^2$, so the pressure vanishes when $R \to 0$. A planet shrinking to zero size has no overlying fluid column.
Dimensional analysis gives
$$[G\rho^2R^2] = \frac{\text{m}^3}{\text{kg},\text{s}^2} \cdot \frac{\text{kg}^2}{\text{m}^6} \cdot \text{m}^2 = \frac{\text{kg}}{\text{m},\text{s}^2},$$
which is the dimension of pressure.
The result scales as $R^2$ because the gravitational field inside a uniform sphere is proportional to $r$, and integrating the hydrostatic equation introduces a second power of length.
The most common source of error is the sign in the hydrostatic equation. Pressure decreases outward, so $dp/dr$ must be negative. Another frequent mistake is using the total planetary mass instead of the enclosed mass $M(r)$ when computing the gravitational field inside the sphere. This would give an incorrect radial dependence and an incorrect central pressure.