Question

Many identical point masses $m$ are placed along $x-$axis, at $x=0,1,2,4,....$. The magnitude of gravitational force on 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 due to masses $m$ at $x=1,2,4,8....$ respectively.

$\Rightarrow F_1=\frac{G{m}^2}{1^2}$

Similarly,

$F_2=\frac{G{m}^2}{2^2};F_4=\frac{G{m}^2}{4^2};F_8=\frac{G{m}^2}{8^2}$

So, net force $F_{net}$ acting on mass $m$ at origin is given as

$F_{net}=F_1+F_2+F_4+F_8...=G{m}^2(\frac{1}{1}+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}...)$

This is infinite G.P. with common ratio, $r=\frac{1}{4}$ and first term, $a=1$

For an infinite G.P,
$\text{sum}=\left(\frac{a}{1-r}\right)$

$\Rightarrow \text{Sum}=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$

So,
$\Rightarrow F_{net}=F_1+F_2+F_4+F_8....=\frac{4}{3}G{m}^2$

Hence, option $\left(b\right)$ is correct.

Key Concept: Principle of Superposition - The gravitational force between two masses is independent of other masses in the vicinity and the net force on a mass is equal to the vector sum of forces by all the masses.