Find the potential energy of the gravitational interaction of a point mass `m` and a rod of mass `m` and length `l`, if they are along a straight line. Point mass is at a distance of a from the end of the rod.

Find the potential energy of the gravitational

Options

A. `U=(-Gm^(2))/l ln((2a+l)/(2a-l))`
B. `U=(-Gm^(2))/(l^(2)) ln((2a+l)/(2a-l))`
C. `U=(-Gm^(2))/(l^(2)) ln((2a-l)/(2a+l))`
D. `U=(-Gm^(2))/(l^(2)) ln((2a-l)/(2a+l))^(2)`

Answer

A

Explanation

`dm=m/l.dx`
`dU=-(Gm.(dm))/((a-x))=(-Gm^(2)dx)/(l(a-x))`
`U=int_(-1//2)^(1//2)dU=-(Gm^(2))/l int_(-l//2)^(l//2)(dx)/(a-x)`
`=-(Gm^(2))/l ln((a+l//2)/(a-l//2))=-(Gm^(2))/lln((2a+l)/(2a-l))`