Let\,
\, be a point bearing a mass\,
\, and\,
\, a variable point. If the distance of these points is
, we can define the potential of\,
\, in\,
\, as
The relevance of this concept appears from the fact that its partial derivatives
are the components of the gravitational force with which the material point\,
\, acts on one mass unit in the point\,
\, (provided that the measure units are chosen suitably).
The potential of a set of points\,
\, is the sum of the potentials of individual points, i.e. it may lead to an integral.\\
We determine the potential of all points\,
\, of a hollow ball, where the matter is located between two concentric spheres with radii
and
. Here the density of mass is assumed to be presented by a continuous function \,
\, at the distance
from the centre
. Let
be the distance from
of the point
, where the potential is to be determined. We chose
the origin and the ray
the positive
-axis.
For obtaining the potential in
we must integrate over the ball shell where
. We use the spherical coordinates
,
and
which are tied to the Cartesian coordinates via
for attaining all points we set
The cosines law implies that\,
. 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a5d43b9c709ab863e022d32d23e88907f10b6e2)
where the factor\,
\, is the coefficient for the coordinate changing
We get from the latter integral
Failed to parse (unknown function "\sijoitus"): {\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:
.\, The point
is outwards the hollow ball, i.e.
.\, Then we have\,
\, for all\,
.\, The value of the integral (2) is
, and (1) gets the form
where
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
.\, The point
is in the cavity of the hollow ball, i.e.
.\, Then\,
\, on the interval of integration of (2). The value of (2) is equal to 2, and (1) yields
which is independent on
. 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 .