Two particles of equal mass m go around a circle of radius R under the action of their mutual gravit

Q: Why is the distance $2R$ and not $R$ in the gravitational force formula?

The gravitational force acts between the two masses. Since they are on opposite sides of the circle, the distance between them is the diameter, which is $2R$.

Q: Why do we use $R$ in the centripetal force formula?

Centripetal force depends on the radius of the circular path the particle is taking. Each particle is moving in a circle of radius $R$ around the center of mass.

Q: What if the masses were different?

If the masses were different, the center of mass would not be at the geometric center of the distance between them. They would revolve in circles of different radii, but with the same angular velocity.

Q: Is the speed constant?

Yes, since the orbit is perfectly circular and the masses are equal, the gravitational force is constant in magnitude, leading to a uniform circular motion with constant speed.

Q: Why is the speed with respect to the center of mass?

The center of mass of this isolated system is at rest (or moving with constant velocity). The problem asks for the speed in this center of mass frame, which is simply the orbital speed $v$ we calculated.

Question

Two particles of equal mass $m$ go around a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is

Options:

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

(B) $\sqrt{\frac{Gm}{4R}}$

(C) $\sqrt{\frac{Gm}{3R}}$

(D) $\sqrt{\frac{Gm}{2R}}$

Sigma Percentile · Accepted Answer

Imagine you are floating in deep space, observing a beautiful cosmic dance. Two identical stars, each of mass $m$, are gracefully revolving around their common center of mass in a perfect circle of radius $R$.

This is a classic two-body problem, and it beautifully illustrates the harmony between Newton's Law of Gravitation and the dynamics of circular motion. Let's break down the physics behind this celestial waltz.

### Analyzing the Setup

Since both particles have the exact same mass $m$, their center of mass must lie exactly at the midpoint of the line joining them. This means they will always be diametrically opposite to each other on the circular orbit.

If the radius of the circular orbit is $R$, the distance between the two particles at any given instant is the diameter of the circle, which is $2R$. This is a crucial detail that often trips students up!

### The Master Equation

For any object to move in a circle, it requires a centripetal force directed towards the center. In this isolated system, where does this force come from? It is provided entirely by the mutual gravitational attraction between the two masses.

The gravitational force $F_g$ between the two particles is given by Newton's Law of Universal Gravitation:
$$F_g = \frac{G m_1 m_2}{r^2}$$

Substituting our specific values ($m_1 = m_2 = m$ and $r = 2R$), we get:
$$F_g = \frac{G \cdot m \cdot m}{(2R)^2} = \frac{G m^2}{4R^2}$$

Now, this gravitational force acts as the centripetal force $F_c$ for each particle. The formula for centripetal force is:
$$F_c = \frac{m v^2}{R}$$

Notice that we use $R$ here, not $2R$, because each particle is moving in a circle of radius $R$ around the center of mass.

### Final Calculation

Equating the gravitational force to the centripetal force, we set up our master equation:
$$\frac{G m^2}{4R^2} = \frac{m v^2}{R}$$

Now, we carefully simplify the expression. We can cancel one mass $m$ and one radius $R$ from both sides:
$$\frac{G m}{4R} = v^2$$

Taking the square root of both sides, we find the orbital speed $v$ of each particle:
$$v = \sqrt{\frac{Gm}{4R}}$$

And there we have it! The speed of each particle with respect to their center of mass is $\sqrt{\frac{Gm}{4R}}$. This elegant result shows how mass, gravity, and orbital radius are intimately connected in the mechanics of the universe.

Animated Solution for Physics - Gravitation: Two particles of equal mass $m$ go around a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is

Select Answer:

Visualized Solution

\text{Visualizing the Orbit}

\text{Gravitational Force}

\text{Equating Forces}

\text{Solving for } v

\text{Final Speed}

\text{What if masses were different?}

The Sigma Insight: Gravitational Force

Analyzing the Setup

The Master Equation

Final Calculation

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