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. $\frac{G M m}{r(r+l)}$
• B. $\frac{G M}{r^{2}}$
• C. $Mmr ^{2}$
• D. $\left(r^{2}+l\right) m M$

Answer 1

Answer: (A)

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

Let $d F$ be the gravitational force of attraction between this
element and sphere. Then, $d F=G M \frac{\left(\frac{m}{l} d x\right)}{x^{2}}$
$\therefore F=\int_{r}^{r+l} \frac{G M m}{l x^{2}}$
$d x=\frac{G M m}{l} \int_{r}^{r+l} \frac{1}{x^{2}} d x$
$=\frac{G M m}{l} \int_{r}^{r+l} x^{-2} d x=\frac{G M m}{l}\left[\frac{x^{-2+1}}{-2+1}\right]_{r}^{r+l}$
$=\frac{G M m}{r(r+l)}$
If $r >> l,$ then $F=\frac{G m M}{r^{2}}$