In free space, a rod of mass M and length $l$ has another point mass M on its axial line at a distance
d from one end as shown in figure. At a point P in between the rod and the mass, at a distance ‘a’
from one end of the rod, the gravitational field is zero. Then

Considering a small element dx at x from P.

${\mathrm{E}}_{\mathrm{R}}={\int }_{\mathrm{a}}^{(\mathrm{a}+l)}-\mathrm{G}\frac{\mathrm{M}}{l}\cdot \mathrm{dx}\cdot \frac{1}{{\mathrm{x}}^2}=-\frac{\mathrm{GM}}{l}{[-\frac{1}{\mathrm{x}}]}_{\mathrm{a}}^{l+\mathrm{a}}\}$

$=-\frac{\mathrm{GM}}{\mathrm{l}}(\frac{1}{\mathrm{a}}-\frac{1}{(\mathrm{n}+\mathrm{a})})=-\frac{\mathrm{GM}}{l}\left(\frac{l+\mathrm{a}-\mathrm{a}}{\mathrm{a}(l+\mathrm{a})}\right)$

$=-\frac{\mathrm{GM}}{\mathrm{a}(l+\mathrm{a})}={\mathrm{E}}_{\mathrm{M}}=-\frac{\mathrm{GM}}{{\mathrm{r}}^2}\Rightarrow \mathrm{r}=\sqrt{\mathrm{a}(l+\mathrm{a})}$

$\therefore \mathrm{d}=\mathrm{a}+\sqrt{\mathrm{a}(l+\mathrm{a})}$

At P, potential is  ${{\int }_{\mathrm{a}}^{\mathrm{a}+l}-\frac{\mathrm{GM}}{l}\frac{\mathrm{dx}}{\mathrm{x}}=-\frac{\mathrm{GM}}{l}\ln \mathrm{x}]}_{\mathrm{a}}^{l+\mathrm{a}}$

$=-\frac{\mathrm{GM}}{l}\ln \left(\frac{l+\mathrm{a}}{\mathrm{a}}\right)$

Hence the work done in bringing M from  infinity to P is $\mathrm{W}=-\frac{{\mathrm{GM}}^2}{l}\ln (1+\frac{l}{\mathrm{a}})$

which is the P.E. of the system. Obviously, PE of the point mass M is not zero.