A, B, C and D are four particles with each of mass M lying on the vertices of a square of side a. They always move along a common circle with velocity v under mutual gravitational force. Find v so that they always remain on the vertices of the square

Options

A. `sqrt((GM(2sqrt2+1))/(2sqrt2a))`
B. `sqrt((GM(sqrt2+1))/(2sqrt2a))`
C. `sqrt((GMsqrt2(2+1))/(2sqrt2a))`
D. none

Answer

To solve the problem, we need to find the velocity \( v \) of four particles, each of mass \( M \), positioned at the vertices of a square of side \( a \) such that they move along a common circle under mutual gravitational attraction. Here’s a step-by-step solution:

### Step 1: Determine the Geometry
The particles are positioned at the vertices of a square. The diagonal of the square can be calculated using the formula:
\[ d = a\sqrt{2} \]
This diagonal will serve as the diameter of the circle along which the particles are moving.

### Step 2: Calculate the Radius of the Circle
The radius \( r \) of the circle is half the diagonal:
\[ r = \frac{d}{2} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}} \]

### Step 3: Calculate the Gravitational Forces
Each particle experiences gravitational forces due to the other three particles. We will denote the gravitational force between two masses \( M \) separated by a distance \( r \) as:
\[ F = \frac{GM^2}{r^2} \]
For the forces acting on one particle located at vertex A, we have:
- The distance to the particles at vertices B and D is \( a \).
- The distance to the particle at vertex C (the opposite vertex) is \( \sqrt{2}a \).

Calculating the forces:
1. Force due to particle B:
\[ F_1 = \frac{GM^2}{a^2} \]
2. Force due to particle D:
\[ F_2 = \frac{GM^2}{a^2} \]
3. Force due to particle C:
\[ F_3 = \frac{GM^2}{( \sqrt{2}a )^2} = \frac{GM^2}{2a^2} \]

### Step 4: Resultant Force Calculation
The resultant force \( F_R \) acting on the particle at vertex A due to particles B and D can be calculated using vector addition. Since \( F_1 \) and \( F_2 \) are equal and directed at right angles, we can find the resultant:
\[ F_R = \sqrt{F_1^2 + F_2^2} = \sqrt{2F_1^2} = \sqrt{2}F_1 = \sqrt{2} \cdot \frac{GM^2}{a^2} \]

### Step 5: Total Force Acting on the Particle
The total force acting on the particle at A is:
\[ F_{total} = F_R + F_3 = \sqrt{2} \cdot \frac{GM^2}{a^2} + \frac{GM^2}{2a^2} \]
To combine these forces, we need a common denominator:
\[ F_{total} = \frac{2\sqrt{2}GM^2 + GM^2/2}{a^2} = \frac{(2\sqrt{2} + 0.5)GM^2}{a^2} \]

### Step 6: Centripetal Force Requirement
For the particle to move in a circle, the gravitational force must provide the necessary centripetal force:
\[ F_{centripetal} = \frac{Mv^2}{r} \]
Setting the total gravitational force equal to the centripetal force:
\[ \frac{(2\sqrt{2} + 0.5)GM^2}{a^2} = \frac{Mv^2}{\frac{a}{\sqrt{2}}} \]

### Step 7: Solve for Velocity \( v \)
Rearranging gives:
\[ v^2 = \frac{(2\sqrt{2} + 0.5)GM^2 \cdot \sqrt{2}}{Ma} \]
\[ v^2 = \frac{(2\sqrt{2} + 0.5)GM\sqrt{2}}{a} \]
Taking the square root:
\[ v = \sqrt{\frac{(2\sqrt{2} + 0.5)GM\sqrt{2}}{a}} \]

### Final Result
Thus, the velocity \( v \) required for the particles to remain at the vertices of the square while moving along a common circle is:
\[ v = \sqrt{\frac{GM(1 + 2\sqrt{2})}{2a}} \]