Question

Masses of $1\text{kg}$ each are placed at $1\text{m},2\text{m},4\text{m},8\text{m},...$ from a point $P$. The gravitational field intensity at $P$ due to these masses is

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

Answer 1

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

Gravitational field due to a point mass $M$ at a distance $r$ from it is given by $E=\frac{GM}{r^2}$.
where $G$ is gravitational constant.

The gravitational field obeys the principle of superposition i.e. the net field at point $P$ will be equal to the vector sum of fields due to all the masses at that point.

Since the direction of fields will be away from point $P$ along the line on which point masses are placed i.e. in the same direction, we can directly add them up.


Gravitational field intensity at $P$ due to mass at $1\text{m}$ from $P$ is
$E_1=\frac{G\times 1}{1^2}=\frac{G}{1}$;
Gravitational field intensity at $P$ due to mass at $2\text{m}$ from $P$ is
$E_2=\frac{G\times 1}{2^2}=\frac{G}{2^2}$
Gravitational field intensity at $P$ due to mass at $4\text{m}$ from $P$ is
$E_3=\frac{G\times 1}{4^2}=\frac{G}{4^2}$ and so on.

So, the net electric field at $P$ is
$E_{\text{net}}=\frac{G}{1^2}+\frac{G}{2^2}+\frac{G}{4^2}.......\infty$
$\Rightarrow E_{\text{net}}=G[1+\frac{1}{4}+\frac{1}{16}+.....\infty ]$
This is an infinite geometric progression (G.P.) having common ratio, $r=\frac{1}{4}$.
Sum of infinite G.P. $=\frac{a}{1-r}$
$\therefore E_{net}=G\left[\frac{1}{1-\frac{1}{4}}\right]=\frac{4G}{3}$

Hence, option (d) is correct.

Why this question? To familiarize students with the concept of gravitational field, it’s value and the applicability of the principle of superposition. Key Concept: Principle of Superposition - The intensity of gravitational field at a point is the vector sum of gravitational field intensities due to all masses in the vicinity.