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 the particles A, B, C, and D, such that they remain at the vertices of a square while moving in a circular path due to mutual gravitational attraction. Here’s a step-by-step solution:

### Step 1: Understand the Geometry
The particles are at the vertices of a square with side length \( a \). The diagonal \( d \) of the square can be calculated using the Pythagorean theorem:
\[ d = \sqrt{a^2 + a^2} = a\sqrt{2} \]
The radius \( R \) of the circular path that the particles are moving along is half of the diagonal:
\[ R = \frac{d}{2} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}} \]

### Step 2: Calculate the Gravitational Force
Each particle experiences a gravitational force due to the other three particles. The gravitational force \( F \) between two particles of mass \( M \) separated by a distance \( r \) is given by:
\[ F = \frac{G M^2}{r^2} \]
For our configuration, the distance between adjacent particles is \( a \) and the distance between diagonal particles is \( a\sqrt{2} \).

### Step 3: Find the Net Gravitational Force
The net gravitational force acting on one particle can be calculated by considering the contributions from the other three particles:
- The force due to one adjacent particle is:
\[ F_1 = \frac{G M^2}{a^2} \]
- The force due to the diagonal particle is:
\[ F_3 = \frac{G M^2}{(a\sqrt{2})^2} = \frac{G M^2}{2a^2} \]
Since there are two adjacent particles contributing equally and one diagonal particle, the net force \( F_{\text{net}} \) on one particle is:
\[ F_{\text{net}} = 2F_1 + F_3 = 2\left(\frac{G M^2}{a^2}\right) + \frac{G M^2}{2a^2} = \frac{2G M^2}{a^2} + \frac{G M^2}{2a^2} = \frac{4G M^2 + G M^2}{2a^2} = \frac{5G M^2}{2a^2} \]

### Step 4: Centripetal Force Requirement
For the particles to move in a circle, the net gravitational force must provide the necessary centripetal force:
\[ F_{\text{centripetal}} = \frac{M v^2}{R} \]
Setting the gravitational force equal to the centripetal force gives:
\[ \frac{5G M^2}{2a^2} = \frac{M v^2}{R} \]

### Step 5: Substitute for \( R \)
Substituting \( R = \frac{a}{\sqrt{2}} \) into the equation:
\[ \frac{5G M^2}{2a^2} = \frac{M v^2}{\frac{a}{\sqrt{2}}} \]
This simplifies to:
\[ \frac{5G M^2}{2a^2} = \frac{M v^2 \sqrt{2}}{a} \]
Canceling \( M \) from both sides:
\[ \frac{5G}{2a^2} = \frac{v^2 \sqrt{2}}{a} \]

### Step 6: Solve for \( v^2 \)
Rearranging gives:
\[ v^2 = \frac{5G a \sqrt{2}}{2} \]

### Step 7: Final Expression for \( v \)
Taking the square root to find \( v \):
\[ v = \sqrt{\frac{5G a \sqrt{2}}{2}} \]

### Conclusion
Thus, the velocity \( v \) required for the particles to remain at the vertices of the square while moving in a circle is:
\[ v = \sqrt{\frac{5G a \sqrt{2}}{2}} \]