Question

Find the potential energy of the gravitational interaction of a mass point of mass $m$ and a thin uniform rod of mass $M$ and length $l$, if they are located along a straight line at a distance $a$ from each other; also find the force of their interaction.

Answer 1

Choose the location of the point mass as the origin. Then the potential energy $dU$ of an element of mass $dM=\frac{M}{l}dx$ of the rod in the field of the point mass is
$dU=-\gamma m\frac{M}{l}dx\frac{1}{x}$
where $x$ is the distance between the element and the point. (Note that the rod and the point mass are on a straight line.) If then $a$ is the distance of the nearer end of the rod from the point mass.
$U=-\gamma \frac{mM}{l}{\int }_a^{a+l}\frac{dx}{x}=-\gamma m\frac{M}{l}\ln (1+\frac{l}{a})$
The force of interaction is
$F=\frac{\partial U}{\partial a}$
$=\gamma \frac{mM}{l}\times \frac{1}{1+\frac{l}{a}}(-\frac{l}{a^2})=-\frac{\gamma mM}{a(a+l)}$
Minus sign means attraction.