Question

The point masses having mass $m$ and $2m$ are placed at distance $d$ from each other. The point on the line joining the point masses, where gravitational field intensity is zero will be at distance

• A. $\frac{2d}{\sqrt{3}+1}$ from point mass $2m$.
• B. $\frac{2d}{\sqrt{3}-1}$ from point mass $2m$.
• C. $\frac{d}{1+\sqrt{2}}$from point mass $m$.
• D. $\frac{d}{1-\sqrt{2}}$from point mass $m$.

Answer 1

The correct option is C $\frac{d}{1+\sqrt{2}}$from point mass $m$.

Let $r$ be the distance of the point from $m$ where the gravitational field is zero.
The gravitational field due to mass $M$ at a distance $r$ from it is given by
$E=\frac{GM}{r^2}$
where $G$ is gravitation constant.

The gravitational field intensity will be zero at a point where the field due to mass $m$ will be cancelled by the field due to mass $2m$.


The gravitational field intensity due to mass $m$ at $r$ $=\frac{Gm}{r^2}$
The gravitational field intensity due to mass $2m$ at $d-r$ from $2m=\frac{G\left(2m\right)}{(d-r{)}^2}$

According to question,
$\frac{Gm}{r^2}=\frac{2Gm}{(d-r{)}^2}$
$\Rightarrow \frac{1}{r}=\frac{\sqrt{2}}{d-r}$
$\Rightarrow d-r=\sqrt{2}r$
$\Rightarrow r(1+\sqrt{2})=d$
$\Rightarrow r=\frac{d}{1+\sqrt{2}}$

Hence, option (c) is correct.

Why this question? To make students familiar with the concept of gravitational field intensity and that it is a vector quantity. Key Concept: Between two masses, there will be a point where the gravitational field of one mass will be equal and opposite to the gravitational field of the other mass.