PlanetPhysics/Potential of Spherical Shell

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

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

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

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

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

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

We get from the latter integral

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikiversity.org/v1/":): {\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:

$1^{\circ }$.\, The point $A$ is outwards the hollow ball, i.e. $a>R$.\, Then we have\, $|r-a|=a-r$\, for all\, $r\in [R_{0},\,R]$.\, The value of the integral (2) is ${\frac {2r}{a}}$, and (1) gets the form $V(a)={\frac {4\pi }{a}}\int _{R_{0}}^{R}\varrho (r)\,r^{2}\,dr={\frac {M}{a}},$ where $M$ 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 $V'(a)=-{\frac {M}{a^{2}}}.$

$2^{\circ }$.\, The point $A$ is in the cavity of the hollow ball, i.e. $a<R_{0}$ .\, Then\, $|r-a|=r-a$\, on the interval of integration of (2). The value of (2) is equal to 2, and (1) yields $V(a)=4\pi \int _{R_{0}}^{R}\varrho (r)\,r\,dr,$ which is independent on $a$. 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 .