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Q. Two bodies of masses $ m $ and $ 4\,m $ are placed at a distance $ r $ . The gravitational potential at a point on the line joining them where the gravitational field is zero is

AMUAMU 2017

Solution:

Let gravitational field at the point $A$ in zero as
shown in figure below
image
So, $\frac{Gm}{x^2} = \frac{G(4m)}{(r -x)^2} $ or $(r - x)^2 = (2x)^2$
or $ r - x = 2x$
$\Rightarrow 3x = r$(neglecting negative values)
$\therefore x = \frac{r}{3}$
Hence, gravitational potential at point $A$,
$V = \frac{-GM}{x} - \frac{G(4m)}{(r - x)}$
$ = - \frac{3Gm}{r} - \frac{3G(4m)}{2r} $
$= \frac{-9Gm}{r}$