Question

An infinite number of masses, each of one kg are placed on the +ve X axis at 1 m, 2 m, 4 m,
_______ from the origin . The magnitude of the gravitational field at origin due to this distribution of masses is :

• A. $2G$
• B. $\frac{4G}{3}$
• C. $\frac{G}{4}$
• D. $\infty$

Answer 1

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

Each mass is of $m=1$ kg placed on the +ve X axis.

$\because 1^{st}$ mass is at ($2^{1-1}=2^0=1$ m), $2^{nd}$ mass is at ($2^{2-1}=2^1=2$ m), $3^{rd}$ mass is at ($2^{3-1}=2^2=4$ m) from the origin,

and so on.

Thus, we can see that the $n^{th}$ mass is placed at a distance of $r_n=2^{n-1}$m from the origin on the +ve X axis.

Thus, the gravitational field intensity at origin for the $n^{th}$ particle will be given as

$g_n=\frac{Gm}{r_n^2}$

$\therefore g_n=\frac{G}{(2^{n-1}{)}^2}=\frac{G}{4^{n-1}}$

Thus, the net magnitude of the gravitational field at the origin due to this distribution will be a sum of this individual $g_n$

$\therefore g_{net}=\sum _{n=1}^{\infty }\frac{G}{4^{n-1}}$

$\therefore g_{net}=G\sum _{n=1}^{\infty }\frac{1}{4^{n-1}}$

$\because \sum _{n=1}^{\infty }\frac{1}{4^{n-1}}$ is the sum of an infinite Geometrical Progression with ratio = $1/4$,

$\therefore \sum _{n=1}^{\infty }\frac{1}{4^{n-1}}=\frac{1}{1-\left(1/4\right)}=\frac{1}{3/4}=\frac{4}{3}$

substituting

$\therefore g_{net}=\frac{4G}{3}$

Magnitude at origin for the given distribution of masses is $\frac{4G}{3}$.