Find the potential energy of the gravitational interaction
(a) of two mass points of masses `m_1` and `m_2` located at a distance r from each other,
(b) 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.

Explanation

(a) The gravitational potential due to `m_1` at the point of location of `m_2`:
`V_2=underset(r)overset(oo)intvecG*dvecr=underset(r)overset(oo)int-(gammam_1)/(x^2)dx=-(gammam_1)/(r)`
So, `U_(21)=m_2V_2=-(gammam_1m_2)/(r)`
Similarly `U_(12)=-(gammam_1m_2)/(r)`
Hence `U_(12)=U_(21)=U=-(gammam_1m_2)/(r)`

(b) Choose the location of the point mass as the origin. Then the potential energy `dU` of an element of mass `dM=(M)/(l)dx` of the rod in the field of the point mass is
`dU=-gammam(M)/(l)dx(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(mM)/(l)underset(a)overset(a+l)int(dx)/(x)=-gammam(M)/(l)1n(1+l/a)`
The force of interaction is
`F=-(deltaU)/(deltaa)`
`=gamma(mM)/(l)xx(1)/(1+l/a)(-(l)/(a^2))=-(gammamM)/(a(a+l))`
Minus sign means attraction.