Question

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 is

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

Answer 1

Answer: (A)

The force of attraction between sphere and shaded position $dF=GM\frac{\left(\frac{m}{l}dx\right)}{x^2}$

$F={\int }_r^{r+1}\frac{GMm}{l{x}^2}dx=\frac{GMm}{l}{\int }_r^{r+1}\frac{1}{x^2}dx$

$=\frac{GMm}{l}{\int }_r^{r+1}{x}^{-2}dx=\frac{GMm}{l}{\left[\frac{x^{-2+1}}{-2+1}\right]}_r^{r+1}$

$=-\frac{GMm}{l}{\left[x^{-1}\right]}_r^{r+1}=-\frac{GMm}{l}{\left[\frac{1}{x}\right]}_r^{r+1}=\frac{GMm}{r(r+1)}$