Four point masses each of mass m are placed on vertices of a regular tetrahedron. Distance between any two masses is `r`.

Options

A. Gravitation field at centre at zero
B. Gravitation potential at centre is `(-4 Gm)/(r)`
C. Gravitation potential energy of system in `(-6 Gm^(2))/(r)`
D. Gravitation force on one of the point mass is `(sqrt(6) Gm^(2))/(r^(2))`

Explanation

To solve the problem of finding the gravitational potential energy and the net force acting on the point masses at the vertices of a regular tetrahedron, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
We have four point masses, each of mass \( m \), placed at the vertices of a regular tetrahedron. The distance between any two masses is given as \( r \).

2. **Gravitational Potential Energy Calculation**:
The gravitational potential energy \( U \) of a system of point masses is given by the formula:
\[ U = -\frac{G m_1 m_2}{r_{12}} - \frac{G m_1 m_3}{r_{13}} - \frac{G m_1 m_4}{r_{14}} - \frac{G m_2 m_3}{r_{23}} - \frac{G m_2 m_4}{r_{24}} - \frac{G m_3 m_4}{r_{34}} \]
In our case, since all masses are equal and the distance between any two masses is \( r \), we can simplify this:
\[ U = -6 \frac{G m^2}{r} \]
Here, there are 6 pairs of masses, each contributing \(-\frac{G m^2}{r}\) to the potential energy.

3. **Calculating the Gravitational Force Between Two Masses**:
The gravitational force \( F \) between any two point masses is given by:
\[ F = \frac{G m^2}{r^2} \]

4. **Finding the Net Force on One Mass**:
Each mass experiences gravitational attraction from the other three masses. The angle between the forces exerted by any two masses on a third mass is \( 60^\circ \) due to the geometry of the tetrahedron.

The net force \( \vec{F}_{net} \) can be calculated using vector addition. The magnitude of the net force can be found using the formula:
\[ F_{net}^2 = F_1^2 + F_2^2 + F_3^2 + 2F_1F_2\cos(60^\circ) + 2F_2F_3\cos(60^\circ) + 2F_3F_1\cos(60^\circ) \]
Since \( F_1 = F_2 = F_3 = F \):
\[ F_{net}^2 = 3F^2 + 3F^2 \cdot \frac{1}{2} = 3F^2 + \frac{3}{2}F^2 = \frac{6}{2}F^2 = 3F^2 \]
Thus,
\[ F_{net} = \sqrt{3}F = \sqrt{3}\left(\frac{G m^2}{r^2}\right) \]

### Final Results:
- The gravitational potential energy of the system is:
\[ U = -\frac{6G m^2}{r} \]
- The net force acting on one mass due to the other three is:
\[ F_{net} = \sqrt{3} \frac{G m^2}{r^2} \]