In Fig. 13-21, a central particle of mass Mis surrounded by a square array of other particles, separated by either distance dor distance d /2along the perimeter of the square. What are the magnitude and direction of the net gravitational force on the central particle due to the other particles?

The given data is listed as follows,

• The mass of the central particle is M
• The distance by which the surrounded square array of particles is separated along its perimeter is, d or d/ 2

The expression for the gravitational force of attraction between two bodies is as follows,

$\mathbf{F}\mathbf{=}\frac{\mathbf{GMm}}{{\mathbf{R}}^{\mathbf{2}}}$

Here, is the gravitational constant, M is the mass of the heavy body, m is the mass of the light body, and R is the distance between two bodies.

From the given figure, it can be observed that the gravitational force due to each particle on one of the sides of the square is cancelled by the gravitational force due to one of the particles on the opposite side of the square with the same mass except particle with mass 3M. This particle would exert force on the central particle having mass M, because there is no force on the opposite side which can cancel this force.

So, the net force acting on the central particle is due to the particle with mass 3M. It is expressed by usingthe expression for the gravitational force of attraction between two bodies.

$$\begin{gathered} \mathrm{Substitute}3M\mathrm{for}m\mathrm{in}F=\frac{GMm}{R^2}. \\ F=\frac{GM\left(3M\right)}{d^2} \\ =\frac{3G{M}^2}{d^2} \end{gathered}$$

It is known that the gravitational force is an attractive force so, the force on particle having mass M by the particle having mass, 3M is pointing towards the left.

Thus, the magnitude of the force is $\frac{3G{M}^2}{d^2}$and is directed towards left side.