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# 5.8.9: Solid Sphere

$$\text{FIGURE V.24A}$$

The potential outside a solid sphere is just the same as if all the mass were concentrated at a point in the centre. This is so, even if the density is not uniform, and long as it is spherically distributed. We are going to find the potential at a point $$\text{P}$$ inside a uniform sphere of radius $$a$$, mass $$M$$, density $$ρ$$, at a distance $$r$$ from the centre ($$r < a$$). We can do this in two parts. First, there is the potential from that part of the sphere “below” $$\text{P}$$. This is $$-GM_r/r$$, where $$M_r = \frac{r^3M}{a^3}$$ is the mass within radius $$r$$. Now we need to deal with the material “above” $$\text{P}$$. Consider a spherical shell of radii $$x$$, $$x + δx$$. Its mass is $$δM = \frac{4 \pi x^2 δx}{\frac{4}{3} \pi a^3} \cdot M = \frac{3Mx^2 δ x}{a^3}$$. The potential from this shell is $$-GδM/x = -\frac{3GMxδx}{a^3}$$. This is to be integrated from $$x = 0$$ to $$a$$, and we must then add the contribution from the material “below” $$\text{P}$$ . The final result is

$ψ = -\frac{GM}{2a^3} \left(3a^2 - r^2 \right) . \label{5.8.23} \tag{5.8.23}$

Figure $$\text{V.25}$$ shows the potential both inside and outside a uniform solid sphere. The potential is in units of $$−GM/r$$, and distance is in units of $$a$$, the radius of the sphere.

$$\text{FIGURE V.25}$$