Three equal masses m are placed at the three vertices of an equilateral triangle of side a. The gravitational force exerted by this system on another particle of mass m placed at the centroid of triangle

Options

A. `(Gm^2)/(3a^2)`
B. `(Gm^2)/a^2`
C. `(Gm^2)/a^2`
D. 0

Answer

To find the gravitational force exerted by three equal masses \( m \) placed at the vertices of an equilateral triangle of side \( a \) on another mass \( m \) placed at the centroid of the triangle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Geometry**:
- We have three masses \( m \) located at the vertices \( A \), \( B \), and \( C \) of an equilateral triangle.
- The centroid \( O \) of the triangle is the point where we will calculate the gravitational force.

2. **Calculate the Distance from the Centroid to the Vertices**:
- The distance from the centroid \( O \) to any vertex (say \( A \)) of the triangle can be calculated using the formula:
\[ d = \frac{a}{\sqrt{3}} \]
- This is derived from the properties of an equilateral triangle.

3. **Calculate the Gravitational Force from One Mass**:
- The gravitational force \( F \) exerted by one mass \( m \) at vertex \( A \) on the mass at the centroid \( O \) is given by Newton's law of gravitation:
\[ F_A = \frac{G m^2}{d^2} = \frac{G m^2}{\left(\frac{a}{\sqrt{3}}\right)^2} = \frac{G m^2 \cdot 3}{a^2} = \frac{3G m^2}{a^2} \]

4. **Calculate the Gravitational Forces from All Three Masses**:
- Since the triangle is symmetric, the gravitational forces from masses at vertices \( B \) and \( C \) will have the same magnitude as \( F_A \):
\[ F_B = F_C = \frac{3G m^2}{a^2} \]

5. **Determine the Resultant Gravitational Force**:
- The forces \( F_A \), \( F_B \), and \( F_C \) act along the lines connecting the centroid to each vertex. Due to symmetry, the horizontal components of \( F_B \) and \( F_C \) will cancel each other out, while the vertical components will add up.
- The net gravitational force \( F_{net} \) acting on mass \( m \) at centroid \( O \) is:
\[ F_{net} = F_A + F_B + F_C = 3 \cdot \frac{3G m^2}{a^2} = \frac{9G m^2}{a^2} \]

### Final Answer:
The total gravitational force exerted by the three masses on the mass at the centroid is:
\[ F_{net} = \frac{9G m^2}{a^2} \]