Three uniform spheres of mass M and radius R each are kept in such a way that each touches the other two. The gravitational force of attraction on any of the spheres due to the oth... Three uniform spheres of mass M and radius R each are kept in such a way that each touches the other two. The gravitational force of attraction on any of the spheres due to the other two is: Read more

Steps to Solve

1. Identify the positions of the spheres

Consider the three spheres positioned at the vertices of an equilateral triangle. The distance between the centers of any two spheres is equal to the sum of their radii, which is $2R$.

2. Apply the formula for gravitational force

The gravitational force between two point masses is given by the formula: $$ F = \frac{G m_1 m_2}{r^2} $$

where:

• $F$ is the gravitational force
• $G$ is the gravitational constant
• $m_1$ and $m_2$ are the masses
• $r$ is the distance between the centers.

3. Calculate the gravitational force between two spheres

For two spheres with mass $M$ and distance $2R$, the gravitational force ($F_{12}$) on one sphere from another is: $$ F_{12} = \frac{G M^2}{(2R)^2} = \frac{G M^2}{4R^2} $$

4. Consider the resultant force from both spheres

Since both spheres attract the sphere in the middle, the forces will act at an angle of $120^\circ$ to each other (because of the geometry of the equilateral triangle). The resultant force can be found using the formula for the vector addition of two forces: $$ F_{R} = \sqrt{F_{12}^2 + F_{12}^2 + 2F_{12}F_{12}\cos(120^\circ)} $$

Substituting $\cos(120^\circ) = -\frac{1}{2}$: $$ F_{R} = \sqrt{F_{12}^2 + F_{12}^2 - F_{12}^2} = \sqrt{F_{12}^2} = \sqrt{2} F_{12} $$

5. Substituting the force value

Substituting the value of $F_{12}$: $$ F_{R} = \sqrt{2} \cdot \frac{G M^2}{4R^2} = \frac{\sqrt{2} G M^2}{4R^2} $$

This represents the gravitational force of attraction on one sphere due to the other two.