Q3Q In Fig. 13-23, a central particl... [FREE SOLUTION]

Chapter 13: Q3Q (page 377)

In Fig. 13-23, a central particle is surrounded by two circular rings of particles, at radii rand R , withR > r. All the particles have mass m . What are the magnitude and direction of the net gravitational force on the central particle due to the particles in the rings?

Short Answer

Expert verified

The magnitude of the net gravitational force on the central particle due to the particles in the rings are $\frac{5 G M m}{r^{2}} + \frac{6 G M m}{R^{2}}$ and it is in the upward direction.

Step by step solution

Step 1: Identification of the given data

Mass of the particles that surrounds the central particle is, m .
Distance of the first ring from the central particle is, r
Distance of the second ring from the central particle is, R

Definition of the gravitational force

The gravitational force due to other particle bodies on a single particle mass id determined by taking the ratio of pf the product of the gravitational constant mass of two bodies to the square of the distance between the bodies.

Determination of the net gravitational force acting on the central particle

Let the mass of the central particle be M .

Write the expression for the gravitational force of attraction between two bodies.

$F = \frac{G M m}{R^{2}}$

Here, G 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.

Write the expression for the gravitational force acting on this central particle due to the five particles present in the inner circle at radius r by using equation (i).

$F_{r} = \frac{5 G M m}{r^{2}}$

Write the expression for the gravitational force acting on this central particle due to the six particles present in the inner circle at radius R by using equation (i).

$F_{R} = \frac{6 G M m}{R^{2}}$

Determine the net gravitational force acting on the central particle due to all these particles present in both the circles by adding the above equations.

$F_{n e t} = \frac{5 G M m}{r^{2}} + \frac{6 G M m}{R^{2}}$

So, the above positive value of the force indicates that the direction of velocity considering the tangential radial distance vector of the particle has net force in opposite direction to the motion that is upward direction.

Thus, the magnitude of the net gravitational force is $\frac{5 G M m}{r^{2}} + \frac{6 G M m}{R^{2}}$ in the upward direction.