Six particles each of mass`m` are placed at the corners of a regular hexagon of edge length `a`. If a point mass `m_0` is placed at the centre of the hexagon, then the net gravitational force on the point mass is

Options

A. `(6Gm^(2))/(a^(2))`
B. `(6Gmm_(0))/(a^(2))`
C. zero
D. `(6Gm)/(a^(4))`

Answer

To find the net gravitational force on the point mass \( m_0 \) placed at the center of a regular hexagon with six particles of mass \( m \) at its corners, we can follow these steps:

### Step 1: Understand the Configuration
We have a regular hexagon with six corners, each occupied by a particle of mass \( m \). The point mass \( m_0 \) is located at the center of the hexagon.

### Step 2: Determine the Distance from the Center to Each Corner
In a regular hexagon, the distance from the center to any corner (vertex) can be calculated using the relationship:
\[ d = \frac{a}{\sqrt{3}} \quad \text{(where \( a \) is the edge length)} \]
However, for gravitational force calculations, we can directly use the edge length \( a \) and the geometry of the hexagon.

### Step 3: Calculate the Gravitational Force from One Corner
The gravitational force \( F \) exerted by one corner mass \( m \) on the mass \( m_0 \) at the center is given by Newton's law of gravitation:
\[ F = \frac{G m m_0}{d^2} \]
where \( G \) is the gravitational constant and \( d \) is the distance from the center to a corner.

### Step 4: Analyze the Forces
Since the hexagon is symmetric, the forces exerted by the six corner masses on the central mass \( m_0 \) will have equal magnitudes but different directions.

### Step 5: Resolve the Forces into Components
Each force can be resolved into x and y components. Due to the symmetry of the hexagon:
- The forces acting in opposite directions will cancel out.
- For example, the force from mass at corner 1 will cancel with the force from mass at corner 4, and so on.

### Step 6: Calculate the Net Force
Since all forces from the corner masses cancel each other out due to symmetry, the net gravitational force \( F_{\text{net}} \) on the mass \( m_0 \) is:
\[ F_{\text{net}} = 0 \]

### Conclusion
Thus, the net gravitational force on the point mass \( m_0 \) at the center of the hexagon is:
\[ \text{Net Gravitational Force} = 0 \]