Three uniform spheres each having a mass M and radius a are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any of the spheres due to the other two.

Explanation

To find the magnitude of the gravitational force on any of the spheres due to the other two, we can follow these steps:

### Step 1: Understand the Configuration
We have three uniform spheres, each of mass \( M \) and radius \( a \), arranged such that each sphere touches the other two. The centers of the spheres form an equilateral triangle with each side measuring \( 2a \).

### Step 2: Identify the Forces
Let’s denote the spheres as A, B, and C. We need to calculate the gravitational force on sphere C due to spheres A and B.

### Step 3: Calculate the Force Between Spheres
The gravitational force \( F \) between two masses \( M \) separated by a distance \( r \) is given by Newton's law of gravitation:
\[ F = \frac{G M^2}{r^2} \]
In our case, the distance \( r \) between the centers of any two spheres (e.g., A and B) is \( 2a \). Therefore, the force between any two spheres (say C due to A) is:
\[ F_{CA} = \frac{G M^2}{(2a)^2} = \frac{G M^2}{4a^2} \]

### Step 4: Resolve the Forces into Components
Since the spheres form an equilateral triangle, the angle between the lines connecting the centers of the spheres is \( 60^\circ \). We need to resolve the forces \( F_{CA} \) and \( F_{CB} \) into their x and y components.

- The force \( F_{CA} \) can be resolved into components:
- \( F_{CA,x} = F_{CA} \cos(60^\circ) = \frac{G M^2}{4a^2} \cdot \frac{1}{2} = \frac{G M^2}{8a^2} \)
- \( F_{CA,y} = F_{CA} \sin(60^\circ) = \frac{G M^2}{4a^2} \cdot \frac{\sqrt{3}}{2} = \frac{G M^2 \sqrt{3}}{8a^2} \)

- Similarly, for the force \( F_{CB} \):
- \( F_{CB,x} = -\frac{G M^2}{8a^2} \) (negative because it acts in the opposite direction)
- \( F_{CB,y} = \frac{G M^2 \sqrt{3}}{8a^2} \)

### Step 5: Calculate the Resultant Force
Now, we can find the resultant force on sphere C by adding the x and y components of the forces due to A and B.

- In the x-direction:
\[ F_{C,x} = F_{CA,x} + F_{CB,x} = \frac{G M^2}{8a^2} - \frac{G M^2}{8a^2} = 0 \]

- In the y-direction:
\[ F_{C,y} = F_{CA,y} + F_{CB,y} = \frac{G M^2 \sqrt{3}}{8a^2} + \frac{G M^2 \sqrt{3}}{8a^2} = \frac{G M^2 \sqrt{3}}{4a^2} \]

### Step 6: Magnitude of the Resultant Force
Since the x-component is zero, the magnitude of the gravitational force on sphere C due to the other two spheres is simply the y-component:
\[ F = F_{C,y} = \frac{G M^2 \sqrt{3}}{4a^2} \]

### Final Answer
The magnitude of the gravitational force on any of the spheres due to the other two is:
\[ F = \frac{G M^2 \sqrt{3}}{4a^2} \]

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