Chapter 13, Problem 1Q

In Fig. 13-21, a central particle of mass M is surrounded by a square array of other particles, separated by either distance d or distance d/2 along 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?

Figure 13-21 Question 1.

To determine

To find:

The magnitude and direction of the net gravitational force acting on the central particle due to the other particles

Answer to Problem 1Q

Solution:

The net gravitational force acting on the central particle due to the other particles is

$\frac{3G{M}^2}{d^2}$

and is directed towards left.

Explanation of Solution

1) Concept:

Observing the figure, we can cancel out the equal and opposite forces. The sum of the remaining forces will be the net force acting on the central particle M.

2) Given:

Thefigure of system of particles is given.

3) Formulae:

The Gravitational force of attraction between two bodies of masses M and m separated by distance R is,

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

4) Calculations:

From the given figure, we can infer that the gravitation 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 3M. Particle 3M would exert force on M, but there is no force on the opposite side which can cancel this force.

So, the net force acting on the central particle M is due to the particle 3M which is

$F=\frac{GM\left(3M\right)}{d^2}$

$F=\frac{3G{M}^2 }{d^2}$

As we know the gravitational force is an attractive force, the force on particle M by particle 3M is pointing towards the left.

Conclusion:

Using the formula for gravitational force, we can find the net force acting on the particle due to the system of particles.