Question

Two particles of masses ‘$m$’ and ‘$9m$’ are separated by a distance ‘$r$’. At a point on the line joining them the gravitational field is zero. The gravitational potential at that point is ($G =$ Universal constant of gravitation)

• A. $- \frac{4Gm}{r}$
• B. $- \frac{8Gm}{r}$
• C. $- \frac{16Gm}{r}$
• D. $- \frac{32Gm}{r}$

Answer 1

Answer: (C)

Let electric field at point $P$ is zero which is situated at a distance $x$ from mass $m$.
$\therefore \frac{G m}{x^{2}}=\frac{G(9 m)}{(r-x)^{2}}$
Or $\frac{(r-x)^{2}}{x^{2}}=9$
Or $\frac{(r-x)}{x}=3$
$\Rightarrow x=\frac{r}{4}$
Potential at point $P V _{ p }=-\frac{ Gm }{ x }-\frac{ G (9 m )}{ r - x }$ Or $V_{p}=-\frac{G m}{r / 4}-\frac{G(9 m)}{3 r / 4}$
Or $V _{ p }=-\frac{ Gm }{ r }[4+12]=\frac{-16 Gm }{ r }$