# PlanetPhysics/Potential of Spherical Shell

Let\, ${\displaystyle (\xi ,\,\eta ,\,\zeta )}$\, be a point bearing a mass\, ${\displaystyle m}$\, and\, ${\displaystyle (x,\,y,\,z)}$\, a variable point. If the distance of these points is ${\displaystyle r}$, we can define the potential of\, ${\displaystyle (\xi ,\,\eta ,\,\zeta )}$\, in\, ${\displaystyle (x,\,y,\,z)}$\, as ${\displaystyle {\frac {m}{r}}={\frac {m}{\sqrt {(x-\xi )^{2}+(y-\eta )^{2}+(z-\zeta )^{2}}}}.}$ The relevance of this concept appears from the fact that its partial derivatives ${\displaystyle {\frac {\partial }{\partial x}}\!\left({\frac {m}{r}}\right)=-{\frac {m(x-\xi )}{r^{3}}},\quad {\frac {\partial }{\partial y}}\!\left({\frac {m}{r}}\right)=-{\frac {m(y-\eta )}{r^{3}}},\quad {\frac {\partial }{\partial z}}\!\left({\frac {m}{r}}\right)=-{\frac {m(z-\zeta )}{r^{3}}}}$ are the components of the gravitational force with which the material point\, ${\displaystyle (\xi ,\,\eta ,\,\zeta )}$\, acts on one mass unit in the point\, ${\displaystyle (x,\,y,\,z)}$\, (provided that the measure units are chosen suitably).

The potential of a set of points\, ${\displaystyle (\xi ,\,\eta ,\,\zeta )}$\, is the sum of the potentials of individual points, i.e. it may lead to an integral.\\

We determine the potential of all points\, ${\displaystyle (\xi ,\,\eta ,\,\zeta )}$\, of a hollow ball, where the matter is located between two concentric spheres with radii ${\displaystyle R_{0}}$ and ${\displaystyle R\,(>R_{0})}$. Here the density of mass is assumed to be presented by a continuous function \, ${\displaystyle \varrho =\varrho (r)}$\, at the distance ${\displaystyle r}$ from the centre ${\displaystyle O}$. Let ${\displaystyle a}$ be the distance from ${\displaystyle O}$ of the point ${\displaystyle A}$, where the potential is to be determined. We chose ${\displaystyle O}$ the origin and the ray ${\displaystyle OA}$ the positive ${\displaystyle z}$-axis.

For obtaining the potential in ${\displaystyle A}$ we must integrate over the ball shell where ${\displaystyle R_{0}\leq r\leq R}$. We use the spherical coordinates ${\displaystyle r}$, ${\displaystyle \varphi }$ and ${\displaystyle \psi }$ which are tied to the Cartesian coordinates via ${\displaystyle x=r\cos \varphi \cos \psi ,\quad y=r\cos \varphi \sin \psi ,\quad z=r\sin \varphi ;}$ for attaining all points we set ${\displaystyle R_{0}\leq r\leq R,\quad -{\frac {\pi }{2}}\leq \varphi \leq {\frac {\pi }{2}},\quad 0\leq \psi <2\pi .}$ The cosines law implies that\, ${\displaystyle PA={\sqrt {r^{2}-2ar\sin \varphi +a^{2}}}}$. Thus the potential is the triple integral

${\displaystyle {\begin{matrix}V(a)=\int _{R_{0}}^{R}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\int _{0}^{2\pi }\!\!{\frac {\varrho (r)\,r^{2}\cos \varphi }{\sqrt {r^{2}-2ar\sin \varphi +a^{2}}}}\,dr\,d\varphi \,d\psi =2\pi \int _{R_{0}}^{R}\varrho (r)\,r\,dr\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}{\frac {r\cos \varphi \,d\varphi }{\sqrt {r^{2}-2ar\sin \varphi +a^{2}}}},\end{matrix}}}$

where the factor\, ${\displaystyle r^{2}\cos \varphi }$\, is the coefficient for the coordinate changing ${\displaystyle \left|{\frac {\partial (x,\,y,\,z)}{\partial (r,\,\varphi ,\,\psi )}}\right|=\!\mod \!\left|{\begin{matrix}\cos \varphi \cos \psi &\cos \varphi \sin \psi &\sin \varphi \\-r\sin \varphi \cos \psi &-r\sin \varphi \sin \psi &r\cos \varphi \\-r\cos \varphi \sin \psi &r\cos \varphi \cos \psi &0\end{matrix}}\right|.}$

We get from the latter integral

$\displaystyle \begin{matrix} \int_{-\frac{\pi}{2}}^\frac{\pi}{2} \frac{r\cos\varphi\,d\varphi}{\sqrt{r^2-2ar\sin\varphi+a^2}} = -\frac{1}{a}\sijoitus{\varphi=-\frac{\pi}{2}}{\quad\frac{\pi}{2}}\sqrt{r^2-2ar\sin\varphi+a^2} = \frac{1}{a}[(r+a)-|r-a|]. \end{matrix}$

Accordingly we have the two cases:

${\displaystyle 1^{\circ }}$.\, The point ${\displaystyle A}$ is outwards the hollow ball, i.e. ${\displaystyle a>R}$.\, Then we have\, ${\displaystyle |r-a|=a-r}$\, for all\, ${\displaystyle r\in [R_{0},\,R]}$.\, The value of the integral (2) is ${\displaystyle {\frac {2r}{a}}}$, and (1) gets the form ${\displaystyle V(a)={\frac {4\pi }{a}}\int _{R_{0}}^{R}\varrho (r)\,r^{2}\,dr={\frac {M}{a}},}$ where ${\displaystyle M}$ is the mass of the hollow ball. Thus the potential outwards the hollow ball is exactly the same as in the case that all mass were concentrated to the centre . A correspondent statement concerns the attractive force ${\displaystyle V'(a)=-{\frac {M}{a^{2}}}.}$

${\displaystyle 2^{\circ }}$.\, The point ${\displaystyle A}$ is in the cavity of the hollow ball, i.e. ${\displaystyle a .\, Then\, ${\displaystyle |r-a|=r-a}$\, on the interval of integration of (2). The value of (2) is equal to 2, and (1) yields ${\displaystyle V(a)=4\pi \int _{R_{0}}^{R}\varrho (r)\,r\,dr,}$ which is independent on ${\displaystyle a}$. That is, the potential of the hollow ball, when the density of mass depends only on the distance from the centre, has in the cavity a constant value, and the hollow ball influences in no way on a mass inside it .

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## References

1. {\sc Ernst Lindel\"of}: {\em Differentiali- ja integralilasku ja sen sovellutukset II}.\, Mercatorin Kirjapaino Osakeyhti\"o, Helsinki (1932).