Animated Solution for Physics - Gravitation: Four identical particles of mass $M$ are located at the corners of a square of side $a$. What should be their speed, if each of them revolves under the influence of other's gravitational field in a circular orbit circumscribing the square ?

Imagine a beautifully symmetric cosmic dance: four identical stars, each of mass $M$, positioned perfectly at the corners of a vast square of side $a$. They are not stationary; they are sweeping through space in a perfect circular orbit that circumscribes their square formation. What keeps them in this delicate balance? It is the invisible, relentless pull of gravity they exert on one another.

In this problem, we are tasked with finding the exact orbital speed $v$ required to maintain this celestial choreography. Let's break down the physics step-by-step.

Before we calculate any forces, we must understand the geometry of the orbit. The four masses are moving in a circle that passes through all four corners of the square.

The center of this circular orbit is exactly the geometric center of the square. The radius $r$ of this orbit is the distance from the center to any corner. Since the diagonal of a square of side $a$ is $\sqrt{2}a$, the radius is simply half of that diagonal:

To find the speed of the particles, we need to focus on just one of them—let's pick the particle at corner B. Because of the perfect symmetry of the system, whatever happens to this particle is happening to all the others.

Particle B experiences three distinct gravitational pulls from its companions: 1. Pull from A: The mass at corner A is at a distance $a$. The force is $F_{BA}=\frac{G{M}^2}{a^2}$. 2. Pull from C: The mass at corner C is also at a distance $a$. The force is $F_{BC}=\frac{G{M}^2}{a^2}$. 3. Pull from D: The mass at the opposite corner D is at a distance equal to the diagonal, $\sqrt{2}a$. The force is $F_{BD}=\frac{G{M}^2}{(\sqrt{2}a{)}^2}=\frac{G{M}^2}{2a^2}$.

For the particle to move in a circle, it requires a net force directed exactly towards the center of the circle. We must resolve our three force vectors along the diagonal pointing towards the center.

The force $F_{BD}$ is already pointing straight along the diagonal towards the center. However, the forces $F_{BA}$ and $F_{BC}$ point along the edges of the square. Since the diagonal bisects the $90^{\circ }$ corner angle, the angle between the edge and the diagonal is $45^{\circ }$.

We take the components of $F_{BA}$ and $F_{BC}$ along the diagonal by multiplying them by $\cos 45^{\circ }$ (which is $\frac{1}{\sqrt{2}}$). The net force $F_{net}$ towards the center is:

Substituting our force expressions:

This net gravitational force is the sole provider of the centripetal force required to keep the mass $M$ moving in its circular orbit of radius $r$ at speed $v$.

We know the formula for centripetal force is $F_c=\frac{M{v}^2}{r}$. Equating the two forces gives us our master equation:

Substituting $r=\frac{a}{\sqrt{2}}$ and our expression for $F_{net}$:

Now, it is just a matter of careful algebra to isolate $v$. We cancel one $M$ and one $a$ from both sides, and divide by $\sqrt{2}$:

Let's evaluate the numerical constant inside the bracket. We know $\sqrt{2}\approx 1.414$, so $\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}\approx 0.353$.

Taking the square root of both sides to find the speed:

And there we have it! The delicate balance of gravity and inertia requires each particle to travel at exactly $1.16\sqrt{\frac{GM}{a}}$. This perfectly matches option (b).