What is the mutual gravitational potential energy of a point…

What is the mutual gravitational potential energy of a point mass $m$ and a thin homogeneous rod of mass $M$ and length $l$, lying on the same straight line, if the point is at distance $r_0$ from the near end of the rod?

Step 1: Idea

Potential energy is a scalar, so the contributions of different parts of the rod simply add, with no directions to worry about. We cut the rod into small pieces, each of which acts as a point mass, use the result of Problem 14 for each piece, and add them with an integral.

Step 2: One piece of the rod

Measure the distance $x$ from the point mass along the line. The rod occupies $r_0 \le x \le r_0 + l$. Being homogeneous, its mass per unit length is $M/l$, so a piece of length $dx$ has mass $dm = (M/l)\,dx$. Its energy with the point mass is

\[ dE_p = -\gamma\frac{m\,dm}{x} = -\gamma\frac{mM}{l}\frac{dx}{x}. \]

Step 3: Whole rod

\[ E_p = -\gamma\frac{mM}{l}\int_{r_0}^{r_0 + l}\frac{dx}{x} = -\gamma\frac{mM}{l}\Big[\ln x\Big]_{r_0}^{r_0 + l}. \]

Since $\ln(r_0 + l) - \ln r_0 = \ln\frac{r_0 + l}{r_0}$: $E_p = -\gamma\frac{mM}{l}\ln\left(1 + \frac{l}{r_0}\right)$.

Check

For $l \ll r_0$: $ \ln(1 + l/r_0) \approx l/r_0 $, so $ E_p \to -\gamma mM/r_0 $, the point-mass result $\checkmark$

The rod's centre is at $r_0 + l/2$; the exact energy is slightly more negative than $-\gamma mM/(r_0 + l/2)$, because the near half of the rod counts more $\checkmark$