Question

Four point masses each of mass ${}^{\prime }{m}^{\prime }$ are placed at four vertices $\text{A, B, C}$ and $\text{D}$ of a regular hexagon of side ${}^{\prime }{a}^{\prime }$ as shown in figure. Find the gravitational field strength at the centre $\text{O}$ of the hexagon.

• A. $\frac{\sqrt{3}Gm}{a^2}$
• B. $\frac{\sqrt{2}Gm}{a^2}$
• C. $\frac{\sqrt{3}Gm}{2a^2}$
• D. $\frac{Gm}{a^2}$

Answer 1

The correct option is A $\frac{\sqrt{3}Gm}{a^2}$

Gravitational field strength is a vector quantity. So, it is a vector sum of four gravitational field vectors of equal magnitudes act at the center.

Let, $E_A,E_B,E_C$ and $E_D$ are the gravitational field strength acting at the centre of the hexagon due to masses placed at the vertices $\text{A, B, C}$ and $\text{D}$ respectively.


From the figure,
$E_A=E_B=E_C=E_D=\frac{Gm}{a^2}$

$E_A$ and $E_D$ are equal in magnitude but opposite in diection, so both are cancelled.

So, net gravitational field strength is a vector sum $E_B$ and $E_C$ at angle ${60}^o$

$E_{net}=\sqrt{{\left(E\right)}^2+{\left(E\right)}^2+2\left(E\right)\left(E\right)\cos {60}^o}$

$\Rightarrow E_{net}=\sqrt{3}E$

$\therefore E=\frac{\sqrt{3}Gm}{a^2}$

Hence, option (a) is the correct answer.