Question

The gravitational force of attraction between a uniform sphere of mass $M$ and a uniform rod of length $l$ and mass $m$ as shown in figure is

• A. $\frac{GMm}{r(r+l)}$
• B. $\frac{GM}{r^2}$
• C. $Mm{r}^2$
• D. $\left(r^2+l\right)mM$

Answer 1

Answer: (A)

Mass per unit length of the rod $=\frac{m}{l}$ Mass of element of length, $dx=\frac{m}{l}dx$

Let $dF$ be the gravitational force of attraction between this
element and sphere. Then, $dF=GM\frac{\left(\frac{m}{l}dx\right)}{x^2}$
$\therefore F={\int }_r^{r+l}\frac{GMm}{l{x}^2}$
$dx=\frac{GMm}{l}{\int }_r^{r+l}\frac{1}{x^2}dx$
$=\frac{GMm}{l}{\int }_r^{r+l}{x}^{-2}dx=\frac{GMm}{l}{\left[\frac{x^{-2+1}}{-2+1}\right]}_r^{r+l}$
$=\frac{GMm}{r(r+l)}$
If $r>>l,$ then $F=\frac{GmM}{r^2}$