Question

A large number of identical point masses each of mass $m$ are placed along $\text{x-axis}$, at $x=0,1,2,4,...$ The magnitude of gravitational force on the mass at origin $(x=0)$, will be

• A. $G{m}^2$
• B. $\frac{4}{3}G{m}^2$
• C. $\frac{2}{3}G{m}^2$
• D. $\frac{5}{4}G{m}^2$

Answer 1

The correct option is B $\frac{4}{3}G{m}^2$

Let, $F_1,F_2,F_4,F_8,....$ be the forces of gravitation on mass $m$ placed at $x=0$ due masses placed at $x=1,$$2,4,8...$ respectively.


The gravitational force between any two point masses is given by $$F=\frac{G{m}_1{m}_2}{r^2}$$ According to principle of superposition, the net gravitational force on mass $m$ placed at $x=0$ can be written as

$\overrightarrow{F}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_4+{\overrightarrow{F}}_8...$

$\Rightarrow F=\frac{G{m}^2}{1^2}+\frac{G{m}^2}{2^2}+\frac{G{m}^2}{4^2}+\frac{G{m}^2}{8^2}+...$

$\Rightarrow F=G{m}^2(\frac{1}{1}+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}...)...\left(1\right)$

The above expression consists a infinite G.P with common ratio, $r=\frac{1}{4}$

So, sum of the G.P, $S=\frac{a}{1-r}$

where $a$ is the first term $r$ is the common ratio.

$\Rightarrow S=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$

From equation $\left(1\right)$,

$\Rightarrow F=F_1+F_2+F_4+F_8......=G{m}^{2}S$

$\therefore F=\frac{4}{3}G{m}^2$

Hence, option (b) is the correct answer.

Why this question: To make students understand that gravitational force follows the principle of superposition " The net force on a particle is equal to the vector sum of all the forces acting on that particle".