Four particles, each of mass M and equidistant from each other, move along a circle of radius R unde

Q: Why do we only consider the force towards the center?

Because for uniform circular motion, the net force must be purely centripetal (towards the center). The tangential components of the gravitational forces from the adjacent particles perfectly cancel each other out.

Q: How did we get the distance $R\sqrt{2}$ for adjacent particles?

The four particles form a square inscribed in a circle of radius $R$. The distance from the center to a corner is $R$. Using Pythagoras theorem on the right-angled triangle formed by two radii and a side, the side length is $\sqrt{R^2 + R^2} = R\sqrt{2}$.

Q: Why is the angle $45^\circ$?

The diagonals of a square intersect at $90^\circ$ and bisect the corner angles. The line joining a particle to the center is a diagonal, and the line joining it to an adjacent particle is a side. The angle between the diagonal and the side of a square is $45^\circ$.

Q: Can we use energy conservation instead of forces?

Energy conservation can tell us about the total energy of the system, but to find the orbital speed directly, equating the net force to the centripetal force is the most straightforward and standard method.

Q: What if the masses were different?

If the masses were different, the symmetry would be broken. The net force might not point exactly towards the geometric center, and the particles wouldn't necessarily move in a simple circular orbit with constant speed. The problem relies heavily on the symmetry of identical masses.

Question

Four particles, each of mass $M$ and equidistant from each other, move along a circle of radius $R$ under the action of their mutual gravitational attraction, the speed of each particle is

Options:

(A) $\sqrt{\frac{GM}{R}}$

(B) $\sqrt{2\sqrt{2}\frac{GM}{R}}$

(C) $\sqrt{\frac{GM}{R}(1+2\sqrt{2})}$

(D) $\frac{1}{2}\sqrt{\frac{GM}{R}(1+2\sqrt{2})}$

Sigma Percentile · Accepted Answer

### Visualizing the Cosmic Dance

Imagine a beautiful, symmetrical setup in the vastness of space: four identical particles, each with mass $M$, are positioned at the corners of a square. This square is perfectly inscribed within a circle of radius $R$.

Due to their mutual gravitational attraction, these particles don't just sit there; they pull on each other, causing the entire system to rotate. They move along the circumference of the circle with a constant speed $v$.

Our goal is to find this speed $v$. To do this, we need to understand the forces keeping them in this circular dance.

### The Master Equation: Centripetal Force

For any object to move in a circular path, it requires a net force directed exactly towards the center of the circle. This is known as the **centripetal force**, given by the formula:

$$F_c = \frac{Mv^2}{R}$$

In our system, there are no strings or invisible tracks. The only force acting on any particle is the gravitational pull from the other three particles. Therefore, the net gravitational force pointing towards the center must be exactly equal to the required centripetal force.

### Analyzing the Gravitational Pulls

Let's focus our attention on one specific particle, say the one on the far right. It experiences three distinct gravitational forces:
1. A pull from the particle diagonally opposite to it.
2. A pull from the particle directly above it.
3. A pull from the particle directly below it.

First, we need the distances. The diagonally opposite particle is at a distance equal to the diameter of the circle, which is $2R$.

The adjacent particles (above and below) form right-angled triangles with the center. Using the Pythagorean theorem, the distance to these adjacent particles is $\sqrt{R^2 + R^2} = R\sqrt{2}$.

Now, let's calculate the forces using Newton's Law of Universal Gravitation, $F = \frac{GM_1M_2}{r^2}$.

The force from the diagonal particle, let's call it $F_1$, points directly towards the center:

$$F_1 = \frac{GM^2}{(2R)^2} = \frac{GM^2}{4R^2}$$

The forces from the adjacent particles, let's call them $F_2$, act at an angle. Because the diagonals of a square bisect the corner angles, these forces act at $45^\circ$ to the line pointing towards the center.

$$F_2 = \frac{GM^2}{(R\sqrt{2})^2} = \frac{GM^2}{2R^2}$$

### Resolving Components and Finding the Net Force

To find the total force directed towards the center, we must resolve the $F_2$ forces. The components perpendicular to the radius cancel each other out perfectly. The components along the radius add up.

The net inward force is the sum of $F_1$ and the inward components of the two $F_2$ forces:

$$F_{net} = F_1 + 2F_2 \cos(45^\circ)$$

Substituting our calculated values and knowing that $\cos(45^\circ) = \frac{1}{\sqrt{2}}$:

$$F_{net} = \frac{GM^2}{4R^2} + 2 \left( \frac{GM^2}{2R^2} \right) \frac{1}{\sqrt{2}}$$

Let's simplify this expression. The $2$ in the numerator and denominator of the second term cancel out:

$$F_{net} = \frac{GM^2}{R^2} \left( \frac{1}{4} + \frac{1}{\sqrt{2}} \right)$$

To make it neater, we can find a common denominator. Since $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} = \frac{2\sqrt{2}}{4}$, we can write:

$$F_{net} = \frac{GM^2}{R^2} \left( \frac{1 + 2\sqrt{2}}{4} \right)$$

### The Final Calculation

Now, we bring back our master equation. We equate the net gravitational force to the centripetal force:

$$\frac{Mv^2}{R} = \frac{GM^2}{R^2} \left( \frac{1 + 2\sqrt{2}}{4} \right)$$

We can cancel one mass $M$ and one radius $R$ from both sides:

$$v^2 = \frac{GM}{R} \left( \frac{1 + 2\sqrt{2}}{4} \right)$$

Finally, taking the square root of both sides gives us the orbital speed $v$. The square root of $4$ in the denominator is $2$:

$$v = \frac{1}{2}\sqrt{\frac{GM}{R}(1+2\sqrt{2})}$$

This elegant result shows how the speed depends on the mass of the particles and the radius of their orbit, perfectly matching option (d).

Animated Solution for Physics - Gravitation: Four particles, each of mass $M$ and equidistant from each other, move along a circle of radius $R$ under the action of their mutual gravitational attraction, the speed of each particle is

Select Answer:

Visualized Solution

Visualizing the Setup

Centripetal Force Requirement

Identifying Distances and Forces

Substituting Gravitational Law

Simplifying the Expression

Taking Common Denominator

Equating to Centripetal Force

Final Velocity Calculation

The Way Forward

The Sigma Insight: Gravitational Force

Visualizing the Cosmic Dance

The Master Equation

Analyzing the Gravitational Pulls

Resolving Components and Finding the Net Force

The Final Calculation

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