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(C) Let the third point mass $m_0$ be placed at a distance $r$ from the mass $m$. Then its distance from the mass $4m$ is $(d-r)$.The gravitational fo

Vedclass · Accepted Answer

$\frac{d}{3}$

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(C) Let the third point mass $m_0$ be placed at a distance $r$ from the mass $m$. Then its distance from the mass $4m$ is $(d-r)$.The gravitational force on $m_0$ due to mass $m$ is $F_1 = \frac{G m m_0}{r^2}$.The gravitational force on $m_0$ due to mass $4m$ is $F_2 = \frac{G (4m) m_0}{(d-r)^2}$.For the net gravitational force on $m_0$ to be zero,the magnitudes of these forces must be equal:$F_1 = F_2$$\frac{G m m_0}{r^2} = \frac{4 G m m_0}{(d-r)^2}$$\frac{1}{r^2} = \frac{4}{(d-r)^2}$Taking the square root on both sides:$\frac{1}{r} = \frac{2}{d-r}$$d - r = 2r$$d = 3r$$r = \frac{d}{3}$Therefore,the distance of the point from the mass $m$ is $\frac{d}{3}$.

Two point masses $m$ and $4m$ 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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