In Fig. 13-24, two particles of masses, mand 2m, are fixed in place on an axis. (a) Where on the axis can a third particle of mass 3m be placed (other than at infinity) so that the net gravitational force on it from the first two particles is zero: to the left of the first two particles, to their right, between them but closer to the more massive particle, or between them but closer to the less massive particle? (b) Does the answer change if the third particle has, instead, a mass of 16m ? (c) Is there a point off the axis (other than infinity) at which the net force on the third particle would be zero?

The masses of the particles are m, 2m and 3m.

The force of attraction between any two bodies is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them, according to Newton's universal law of gravitation. Using the concept of gravitational force, we can observe in the figure which of the forces gets canceled out. Thus, the remaining force will be the net force acting on the central particle M.

Formula:

The Gravitational force of attraction between two bodies of masses M and m separated by distance R is, $F=\frac{GMm}{R^2}$ …(i)

If the third particle of mass 3m is placed on the left side of both particles of mass m and 2m, the forces acting on it due to these two masses would be towards right and the net force will be the addition of these two forces, and they will not get canceled with each other to give the net force equal to zero.

If the particle is placed at the right side of the two masses m and 2m , the net force will be towards the left and will be non-zero.

So, the particle of mass 3m should be placed in between the particles of masses and as the gravitational force is directly proportional to the masses and inversely proportional to the square of the distance between them. The forces due to m and 2m will be equal and opposite when the third particle is placed near the less massive particle ( m ) to balance the lower mass in the numerator with the lower value of $r^2$in the denominator.

The location of the third particle will be the same if its mass is 16 m instead of 3m as it will be the same in the numerator of the gravitational force acting on it due to the other two particles. So, the gravitational force will depend only on the masses of the other two particles and r.

For, we can write the force equation using equation (i):

$\frac{G\times m\times 3m}{r^2}=\frac{G\times 2m\times 3m}{{(d-r)}^2}$

In the above equation, d is the distance between m and 2m. The distance between m and 3m is r.

For 16m, we can write the force equation using equation (i):

$\frac{G\times m\times 16m}{r^2}=\frac{G\times 2m\times 16m}{{(d-r)}^2}$

From the above two equations, we can see that 3m is on both sides of the first equation and 16m is on both sides of the second equation. So, the mass of the 3^rd particle will be canceled in both equations.

Hence, the position where the 3^rd particle is placed would be independent of the mass of the 3^rd particle.

If the third particle of mass 3m is placed off the axis, the component of the force by particle m on particle 3m and the component of the force by particle 2m on particle 3m would be in the same direction and hence they would be added. So, the net force acting on it would not be zero.