Question

Two point masses $m$ and $4m$ are separated by a distance $d$ on a straight line as shown in figure. A third point mass $m_o$ is to be placed at a point on the line such that the net gravitational force on it is zero.


The distance of the $m_0$ from the $m$ is

• A. $\frac{d}{2}$
• B. $\frac{d}{4}$
• C. $\frac{d}{3}$
• D. $\frac{d}{5}$

Answer 1

The correct option is C $\frac{d}{3}$

Let, the mass $m_0$ be placed at distance $r$ from mass $m$.

Force of gravitation on $m_o$ due to $m$ is

$F_1=\frac{Gm{m}_0}{r^2}$

Force of gravitation on $m_o$ due to $4m$ is

$F_2=\frac{G\left(4m\right)m_0}{(d-r{)}^2}$

Since, the net force acting on $m_0$ is to be zero, we have

$F_1=F_2$

$\Rightarrow \frac{Gm{m}_0}{r^2}=\frac{G\left(4m\right)m_0}{(d-r{)}^2}$

$\Rightarrow (d-r{)}^2=4r^2$

$\Rightarrow (d-r{)}^2=(2r{)}^2$

$\Rightarrow d-r=\pm 2r$

$\Rightarrow d=+3r\text{or}-r$

The null point must lie in between the two stationary masses,

$\therefore r=\frac{d}{3}$

Hence, option (c) is the correct answer.

Why this question: To make students familiar with the concept of equilibrium of masses under action of gravitational force.