# 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: \[P = \frac{2}{3} \pi G ρ^2 a^2 . \label{5.13.4} \tag{5.13.4}\]