Show that the gravitational potential $\mathit{V}\mathbf{=}\mathbf{-}\frac{\mathbf{G}\mathbf{m}}{\mathbf{r}}$satisfies Laplace's equation, that is, show that ${\mathbf{\nabla }}^{\mathbf{2}}\mathbf{\left(}\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{r}$}\right.\mathbf{\right)}\mathbf{=}\mathbf{0}$where${\mathit{r}}^{\mathbf{2}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{,}\mathit{r}\mathbf{\ne }\mathbf{0}$.

The given equations are as follows.

$\begin{array}{l}{r}^2=x^2+y^2+z^2\\ {\nabla }^2=\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2}\end{array}$

The total of the second-order partial derivatives of R , the unknown function, with respect to the Cartesian coordinates, according to Laplace's equation, equals zero.

Put given values in equation${\nabla }^2\left(\frac{1}{r}\right)=0$.

${\nabla }^2\left(\frac{1}{r}\right)=(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2})\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)$ …….(1)

Find the derivative to prove the equation zero.

$\begin{array}{c}\frac{{\partial }^2}{\partial x^2}\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)=\frac{\partial }{\partial x}[-\frac{1}{2}\frac{2x}{{(x^2+y^2+z^2)}^{3/2}}]\\ =-\frac{\partial }{\partial x}\left[\frac{x}{{(x^2+y^2+z^2)}^{3/2}}\right]\end{array}$

Similarly, the other two derivatives are as follows.

$\begin{array}{c}\frac{{\partial }^2}{\partial x^2}\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)=-\frac{1}{{(x^2+y^2+z^2)}^{3/2}}+\frac{3}{2}\frac{x\left(2x\right)}{{(x^2+y^2+z^2)}^{5/2}}\\ =-\frac{1}{{(x^2+y^2+z^2)}^{3/2}}+3\frac{x^2}{{(x^2+y^2+z^2)}^{5/2}}\end{array}$

Put the derivative value in equation(1).

$\begin{array}{c}{\nabla }^2\left(\frac{1}{r}\right)=-3\frac{1}{{(x^2+y^2+z^2)}^{3/2}}+3\frac{x^2+y^2+z^2}{{(x^2+y^2+z^2)}^{5/2}}\\ =-3\frac{1}{{(x^2+y^2+z^2)}^{3/2}}+3\frac{1}{{(x^2+y^2+z^2)}^{3/2}}\end{array}$

${\nabla }^2\left(\frac{1}{r}\right)=0$

Hence, the gravitational potential $V=-\frac{Gm}{r}$satisfies the Laplace equation is proved.