Question

Two bodies of masses $m$ and $9\, m$ are placed at a distance $r$. The gravitational potential at a point on the line joining them, where gravitational field is zero, is ( $G$ is universal gravitational constant)

• A. $\frac{-14\, Gm }{r}$
• B. $\frac{-16 \, Gm }{r}$
• C. $\frac{-12 \, Gm }{r}$
• D. $\frac{-8 \, Gm }{r}$

Answer 1

Answer: (B)

Mass of first body $\left(m_{1}\right)=m$
Mass of second body $\left(m_{2}\right)=9\, m$
Distance between the bodies $(d)=r$
Now, distance of the point from the body of mass $m$,
where gravitational field is zero is given by
$x=\frac{d}{\sqrt{\frac{m_{2}}{m_{1}}+1}}$
or, $x=\frac{r}{\sqrt{\frac{9 m}{m}}+1}$ or, $x=\frac{r}{4}$
Now, $(r-x)=r-\frac{r}{4}=\frac{3 r}{4}$
$\therefore $ Potential, $v=v_{1}+v_{2}$
$=-\frac{G m_{1}}{x}-\frac{G m_{2}}{(r-x)} $
$=-\frac{G m}{r / 4}-\frac{G(9 m)}{3 r / 4} $
$=-\frac{4 G m}{r}-\frac{12 G m}{r}$
$=-\frac{16 G m}{r}$