# 5.8.9: Solid Sphere

- Page ID
- 8153

\(\text{FIGURE V.24A}\)

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.

\(\text{FIGURE V.25}\)