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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 
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Solution

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$$.$$

Physics

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