5.13: Pressure at the Centre of a Uniform Sphere

What is the pressure at the centre of a sphere of radius $a$ and of uniform density $ρ$?

(Preliminary thought: Show by dimensional analysis that it must be something times $Gρ^2 a^2$.)

$\text{FIGURE V.27}$

Consider a portion of the sphere between radii $r$ and $r + δr$ and cross-sectional area $A$. Its volume is $Aδr$ and its mass is $ρAδr$. (Were the density not uniform throughout the sphere, we would here have to write $ρ(r)Aδr$. ) Its weight is $ρgAδr$, where $g = GM_r / r^2 = \frac{4}{3} \pi G ρ r$. We suppose that the pressure at radius $r$ is $P$ and the pressure at radius $r + δr$ is $P + δP$. ($δP$ is negative.) Equating the downward forces to the upward force, we have

$$A(P + δP) + \frac{4}{3} \pi A G ρ^2 rδr = AP. \label{5.13.1} \tag{5.13.1}$$

That is: $$δP = - \frac{4}{3} \pi G ρ^2 r δr. \label{5.13.2} \tag{5.13.2}$$

Integrate from the centre to the surface:

$$\int_{P_0}^0 dP = -\frac{4}{3} \pi G ρ^2 \int_0^a r dr. \label{5.13.3} \tag{5.13.3}$$

Thus: $$P = \frac{2}{3} \pi G ρ^2 a^2 . \label{5.13.4} \tag{5.13.4}$$