Question

Two point masses m and 4 m are separated by a distance d on a line. A third point mass $m_0$ is to be placed at a point on the line such that the net gravitational force on it is zero.
The distance of that point from the m mass is

Answer 1

Let the mass $Z$ be keptata distance of $y$ units from $m,$

hence the distance from $4m$ will be $(x-z)$ units

now since the attractive force from both the masses is equal and opposite so

$Gmz/\left(y\right)=G4mz/\left(x-y\right)2$

$or,1/y=4/\left(x-y\right)2$

$or,1/y=2/\left(x-y\right)$

$or,2y=x-y$

$or,3y=x$

$or,y=x/3$

Hence the mass should be kept at distance of $x/3$ units from $m$ particle$.$