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# 5.13: Pressure at the Centre of a Uniform Sphere

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

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

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