Question

The gravitational force of attraction between a uniform sphere of mass $M$ and a uniform rod of length / and mass $m$ oriented as shown is

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

Answer 1

Answer: (A)

The gravitational field of $M$ at $x=\frac{G M}{x^{2}}$
$d F=\frac{G M}{x^{2}} d m$
where $d m=\frac{m}{l} d x$
$ \Rightarrow d F=\frac{G M m d x}{l \cdot x^{2}}$
$F=\frac{G M m}{l} \int_{r}^{r+l} \frac{1}{x^{2}} d x=\frac{G M m}{l}\left[-\frac{1}{x}\right]_{r}^{r+l}=\frac{G M m}{l}\left[\frac{1}{r}-\frac{1}{r+l}\right]$
$F=\frac{G M m l}{l \cdot r(r+l)}=\frac{G M \cdot m}{r(r+l)}$