Find the gravitational force of attraction between a uniform sphere of mass M and a uniform rod of length l and mass m oriented as shown in the figure.

 

Since the sphere is uniform, its entire mass may be considered to be concentrated at its centre. The force on the elementary mass dm on the rod is
 $$\begin{gathered} force of attractions between two masses= F=\frac{GMm}{r^2} here universal gravitation cons\tan t=G; mass=m; dis\tan ce=r; \\ divide rod into elements of equal width dx having mass dm, then force between sphere and element dm is dF given by \\ dF=\frac{GMdm}{x^2}---\left(1\right) \\ here mass per unit length is cons\tan t \\ hence, \\ \frac{total mass of rod\left(m\right)}{total length of rod\left(l\right)}=\frac{element mass\left(dm\right)}{element length\left(dx\right)} \\ \Rightarrow dm=\frac{mdx}{l}---\left(2\right) \\ substitute eqn\left(2\right) in eqn\left(1\right) \\ dF=\frac{GMmdx}{x^{2}l} \\ integrating, \\ \int dF=\int \frac{GMmdx}{x^{2}l} \end{gathered}$$
 
Gravitational force of attraction between the uniform sphere and the uniform rod is
$$\begin{gathered} \Rightarrow F=\underset{r}{\overset{r+l}{\int }}\frac{GMm}{l{x}^2}dx \\ F=-\frac{GMm}{l}{\left[\frac{1}{x}\right]}_r^{r+l} \\ F=-\frac{GMm}{l}\left[\frac{1}{r+l}-\frac{1}{r}\right] \end{gathered}$$

$$\begin{gathered} F=\frac{GmM}{l}\frac{l}{r\left(r+l\right)} \\ F=\frac{GMm}{r\left(r+l\right)} \end{gathered}$$