Three point masses 'm' each are placed at the three vertices of an equilateral traingle of side 'a'. Find net gravitational force on any point mass.

Explanation

To find the net gravitational force on any point mass placed at one vertex of an equilateral triangle formed by three point masses, we can follow these steps:

### Step 1: Identify the Configuration
We have three point masses, each of mass 'm', placed at the vertices of an equilateral triangle with side length 'a'. Let's label the vertices as A, B, and C.

### Step 2: Calculate the Gravitational Force Between the Masses
The gravitational force between any two point masses is given by Newton's law of gravitation:

\[ F = \frac{G m_1 m_2}{r^2} \]

In our case, the force between any two masses (e.g., mass at A and mass at C) is:

\[ F_{AC} = \frac{G m^2}{a^2} \]

Similarly, the force between mass at B and mass at C is:

\[ F_{BC} = \frac{G m^2}{a^2} \]

### Step 3: Determine the Directions of the Forces
The forces \( F_{AC} \) and \( F_{BC} \) act along the lines connecting the masses. The angle between these two forces is 60 degrees because the triangle is equilateral.

### Step 4: Resolve the Forces into Components
To find the net gravitational force acting on the mass at vertex C, we can resolve the forces \( F_{AC} \) and \( F_{BC} \) into their x and y components.

- The x-component of \( F_{AC} \) (acting towards A) is:
\[ F_{ACx} = F_{AC} \cdot \cos(60^\circ) = \frac{G m^2}{a^2} \cdot \frac{1}{2} = \frac{G m^2}{2a^2} \]

- The y-component of \( F_{AC} \) is:
\[ F_{ACy} = F_{AC} \cdot \sin(60^\circ) = \frac{G m^2}{a^2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3} G m^2}{2a^2} \]

- The x-component of \( F_{BC} \) (acting towards B) is:
\[ F_{BCx} = F_{BC} \cdot \cos(60^\circ) = \frac{G m^2}{a^2} \cdot \frac{1}{2} = \frac{G m^2}{2a^2} \]

- The y-component of \( F_{BC} \) is:
\[ F_{BCy} = F_{BC} \cdot \sin(60^\circ) = \frac{G m^2}{a^2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3} G m^2}{2a^2} \]

### Step 5: Sum the Components
Now, we sum the x and y components to find the net force on mass C.

- The total x-component of the net force \( F_{net,x} \):
\[ F_{net,x} = F_{ACx} + F_{BCx} = \frac{G m^2}{2a^2} + \frac{G m^2}{2a^2} = \frac{G m^2}{a^2} \]

- The total y-component of the net force \( F_{net,y} \):
\[ F_{net,y} = F_{ACy} + F_{BCy} = \frac{\sqrt{3} G m^2}{2a^2} + \frac{\sqrt{3} G m^2}{2a^2} = \sqrt{3} \frac{G m^2}{a^2} \]

### Step 6: Calculate the Magnitude of the Net Force
The magnitude of the net gravitational force \( F_{net} \) can be calculated using the Pythagorean theorem:

\[ F_{net} = \sqrt{(F_{net,x})^2 + (F_{net,y})^2} \]

Substituting the values we found:

\[ F_{net} = \sqrt{\left(\frac{G m^2}{a^2}\right)^2 + \left(\sqrt{3} \frac{G m^2}{a^2}\right)^2} \]

\[ = \sqrt{\frac{G^2 m^4}{a^4} + \frac{3 G^2 m^4}{a^4}} = \sqrt{\frac{4 G^2 m^4}{a^4}} = \frac{2 G m^2}{a^2} \]

### Final Answer
Thus, the net gravitational force on any point mass placed at one vertex of the triangle is:

\[ F_{net} = \frac{2 G m^2}{a^2} \]