Irodov 1.216: A Sphere Crushed by Its Gravity – Amit & Deepali Sinha Foundation

Problem Irodov 1.216

A solid sphere of uniform density has total mass $M$ and radius $R$ , held together only by its own gravitational attraction. Find the pressure $p(r)$ that this self-gravitational compression produces at a distance $r$ from the center, for $0\le r\le R$ . Then evaluate $p$ at the center ( $r=0$ ), treating the Earth itself as such a uniform sphere.

A uniform sphere of radius R; the dashed inner sphere of radius r is the surface at which the pressure is evaluated. The shell outside radius r presses inward on the material within it. — A uniform sphere of radius $R$ ; the dashed inner sphere of radius $r$ is the surface at which the pressure is evaluated. The shell outside radius $r$ presses inward on the material within it.

Given

Find

(a)

p(r)

for

0\le r\le R

, (b)

p(0)

for the Earth

Solution

Step 1Balance a thin column against its own weight

Consider a thin radial column of cross-section $dA$ between radii $r'$ and $r'+dr'$ . The net outward force from the pressure difference across it must support the inward-pulling weight of the column, so

\begin{align*}\bigl[p(r')-p(r'+dr')\bigr]\,dA &= \rho(r')\,g(r')\,dA\,dr' \\ -\,dp &= \rho\,g(r')\,dr'\end{align*}

\frac{dp}{dr'} = -\rho\,g(r') \tag{1}

where $\rho$ is the (uniform) density and $g(r')$ is the magnitude of the gravitational field at radius $r'$ .

Step 2The field inside a uniform sphere

Only the mass $M(r')$ enclosed within radius $r'$ contributes to the field there (gravitationally, the shell outside pulls in all directions and cancels, but its weight is still transmitted inward as pressure, which is the balance (1)). With uniform density $\rho = 3M/(4\pi R^3)$ ,

\begin{align*}M(r') &= \rho\cdot\frac{4}{3}\pi r'^3 \\ g(r') &= \frac{\gamma M(r')}{r'^2} = \frac{\gamma\cdot\frac{4}{3}\pi\rho r'^3}{r'^2}\end{align*}

g(r') = \frac{4}{3}\pi\gamma\rho\,r' \tag{2}

Step 3Integrate outward from the free surface

The outer surface carries no pressure, $p(R)=0$ , so integrating (1) with the field (2) from a general radius $r$ out to $R$ fixes $p(r)$ .

\begin{align*}\int_{p(r)}^{0} dp &= -\int_r^R \rho\cdot\frac{4}{3}\pi\gamma\rho\,r'\,dr' \\ -p(r) &= -\frac{4}{3}\pi\gamma\rho^2\left.\frac{r'^2}{2}\right|_r^R \\ -p(r) &= -\frac{2}{3}\pi\gamma\rho^2\bigl(R^2-r^2\bigr)\end{align*}

p(r) = \frac{2}{3}\pi\gamma\rho^2\bigl(R^2-r^2\bigr)

(3)

Step 4Express the pressure through

M

and

R

Substituting $\rho = 3M/(4\pi R^3)$ into (3) and simplifying gives the pressure in terms of the given quantities.

\begin{align*}p(r) &= \frac{2}{3}\pi\gamma\left(\frac{3M}{4\pi R^3}\right)^{2}\bigl(R^2-r^2\bigr) \\ &= \frac{2}{3}\pi\gamma\cdot\frac{9M^2}{16\pi^2R^6}\bigl(R^2-r^2\bigr) \\ &= \frac{3}{8}\cdot\frac{\gamma M^2}{\pi R^6}\bigl(R^2-r^2\bigr)\end{align*}

p(r) = \frac{3}{8}\left(1-\frac{r^2}{R^2}\right)\frac{\gamma M^2}{\pi R^4}

(4)

Step 5Evaluate at the center of the Earth

Setting $r=0$ in the boxed answer gives the center value; using the Earth’s mass and radius,

\begin{align*}p(0) &= \frac{3}{8}\cdot\frac{\gamma M^2}{\pi R^4} = \frac{3}{8}\cdot\frac{(6.67\times10^{-11})(5.96\times10^{24})^2}{\pi(6.37\times10^6)^4} \\ &= \frac{3}{8}\cdot\frac{2.37\times10^{39}}{5.17\times10^{27}} \\ &= 1.72\times10^{11}\ \text{Pa}\end{align*}

Dividing by one standard atmosphere converts this to the usual unit for such pressures.

p(0) = \frac{1.72\times10^{11}\ \text{Pa}}{1.013\times10^5\ \text{Pa/atm}} \approx 1.7\times10^6\ \text{atm}

p(0) \approx 1.7\times10^{6}\ \text{atm}

(5)

Check

$p(R)=0$ by construction (the free surface), and $p$ is largest at $r=0$ , where the whole overlying sphere presses down; both are exactly what a self-compressed body must do. Dimensionally, $[\gamma M^2/R^4] = (\mathrm{N\,m^2/kg^2})(\mathrm{kg^2})/\mathrm{m^4} = \mathrm{N/m^2}$ , a pressure. The center value, $1.72\times10^{11}\ \text{Pa} \approx 1.7\times10^6$ standard atmospheres, sits below the book’s printed $1.8\times10^6$ ; this is not rounding of $M$ and $R$ (matching $1.8\times10^6$ standard atm would need $M$ or $R$ shifted well outside their three-significant-figure precision), but the atmosphere convention. Expressed in the technical atmosphere, $1\ \text{at} = 9.81\times10^4\ \text{Pa}$ , common in the Russian literature, $1.72\times10^{11}/(9.81\times10^4) \approx 1.75\times10^6\ \text{at}$ , which rounds to the book’s $1.8\times10^6$ .

Remark

In an ordinary pool under uniform gravity, pressure grows linearly with depth because $g$ itself does not change. Here $g(r')$ is not uniform: it grows linearly with $r'$ inside the sphere (Step 2), so integrating it once more, as Step 3 does, produces a pressure that is quadratic in radius. The same self-gravitating balance, $dp/dr=-\gamma\rho\, m(r)/r^2$ , is the starting point for the interior structure of stars and planets, where only the equation of state relating $p$ to $\rho$ changes.