Consider a spherical gaseous cloud of mass density rho(r) in free space where r is the radial distan

Q: Why do we only consider the enclosed mass $M$ and not the entire cloud?

According to Newton's Shell Theorem, a spherically symmetric mass distribution exerts a gravitational force on an object outside it as if all its mass were concentrated at the center. The mass outside the radius $r$ exerts zero net gravitational force on the particle.

Q: Why is the kinetic energy $K$ constant for all particles?

The problem explicitly states that all particles are moving in circular orbits with the *same* kinetic energy $K$. This is a specific given condition of this particular gaseous cloud model, not a general rule for all galaxies or clouds.

Q: How did we get $dM = \rho(r) 4\pi r^2 dr$?

Imagine a very thin spherical shell of radius $r$ and thickness $dr$. Its surface area is $4\pi r^2$. Multiplying this area by the tiny thickness $dr$ gives its volume. Multiplying this volume by the local mass density $\rho(r)$ gives the mass of that thin shell, $dM$.

Q: Why did we differentiate $M$ with respect to $r$?

We found an expression for the *total* enclosed mass $M$ up to radius $r$. To find the *local* density $\rho(r)$ at exactly radius $r$, we need to see how much mass is added when we increase the radius by a tiny amount $dr$. This is exactly what the derivative $\frac{dM}{dr}$ tells us.

Q: What is the difference between mass density $\rho(r)$ and number density $n(r)$?

Mass density $\rho(r)$ tells you how much *mass* (in kg) is present in a unit volume. Number density $n(r)$ tells you how many *particles* are present in a unit volume. Since each particle has mass $m$, dividing the total mass in a volume by $m$ gives the number of particles.

Question

Consider a spherical gaseous cloud of mass density $\rho(r)$ in free space where $r$ is the radial distance from its center. The gaseous cloud is made of particles of equal mass $m$ moving in circular orbits about the common center with the same kinetic energy $K$. The force acting on the particles is their mutual gravitational force. If $\rho(r)$ is constant in time, the particle number density $n(r) = \rho(r)/m$ is :
[$G$ is universal gravitational constant]

Options:

(A) $\frac{K}{\pi r^2 m^2 G}$

(B) $\frac{K}{6\pi r^2 m^2 G}$

(C) $\frac{3K}{\pi r^2 m^2 G}$

(D) $\frac{K}{2\pi r^2 m^2 G}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a vast, spherical cloud of gas floating in the emptiness of space. This isn't just any cloud; it's a dynamic system where every single particle of mass $m$ is orbiting the common center in perfect circles.

The problem gives us a fascinating constraint: every particle, regardless of its orbital radius $r$, possesses the exact same kinetic energy $K$. Our goal is to uncover the hidden structure of this cloud—specifically, how the number of particles per unit volume, $n(r)$, changes as we move outward from the center.

To begin, let's focus on a single particle of mass $m$ orbiting at a distance $r$. According to Newton's Shell Theorem, the gravitational force acting on this particle is determined solely by the mass enclosed within its orbit. Let's call this enclosed mass $M$.

### The Master Equation

For our particle to maintain its circular orbit, it requires a centripetal force. In this isolated system, the only force available to play this role is the mutual gravitational attraction from the enclosed mass $M$.

We can set up our fundamental dynamic equation by equating the gravitational force to the required centripetal force:

$$ \frac{GMm}{r^2} = \frac{mv^2}{r} $$

This equation is the bridge between the geometry of the orbit and the mass distribution of the cloud.

### The Kinetic Energy Connection

The problem states that the kinetic energy $K$ is constant for all particles. We know the formula for kinetic energy is $K = \frac{1}{2}mv^2$.

Let's manipulate our force equation to reveal this kinetic energy term. By multiplying both sides by $r$, we get:

$$ \frac{GMm}{r} = mv^2 $$

Now, we can substitute $mv^2$ with $2K$:

$$ \frac{GMm}{r} = 2K $$

This is a beautiful result! It directly links the enclosed mass $M$ to the constant kinetic energy $K$. Let's isolate $M$ to see how the enclosed mass grows with the radius:

$$ M = \frac{2Kr}{Gm} $$

Notice that since $K$, $G$, and $m$ are all constants, the enclosed mass $M$ increases linearly with the radius $r$.

### Unveiling the Mass Distribution

We have the total enclosed mass $M(r)$, but we need the local mass density $\rho(r)$. To find this, we must look at how the mass changes as we increase the radius by a tiny amount $dr$.

Let's differentiate our mass equation with respect to $r$:

$$ dM = \frac{2K}{Gm} dr $$

This $dM$ represents the mass of a very thin spherical shell of thickness $dr$ located at radius $r$.

Geometrically, the mass of this thin shell can also be expressed as its volume multiplied by the local density $\rho(r)$. The volume of a thin spherical shell is its surface area ($4\pi r^2$) times its thickness ($dr$):

$$ dM = \rho(r) \cdot 4\pi r^2 dr $$

Now, we have two distinct expressions for the same tiny mass $dM$. Let's equate them to solve for the density:

$$ \rho(r) \cdot 4\pi r^2 dr = \frac{2K}{Gm} dr $$

The $dr$ terms cancel out perfectly, leaving us with:

$$ \rho(r) = \frac{2K}{4\pi r^2 Gm} $$

Simplifying the fraction, we arrive at the mass density profile of the cloud:

$$ \rho(r) = \frac{K}{2\pi r^2 Gm} $$

### Final Calculation

We are almost there! The question asks for the **particle number density**, $n(r)$, which is the number of particles per unit volume.

Since we know the total mass per unit volume ($\rho(r)$) and the mass of a single particle ($m$), we can find the number density by simply dividing the mass density by the mass of one particle:

$$ n(r) = \frac{\rho(r)}{m} $$

Substituting our expression for $\rho(r)$:

$$ n(r) = \frac{K}{2\pi r^2 m^2 G} $$

This matches option (D). The inverse-square dependence on $r$ tells us that the cloud is much denser at the center and thins out rapidly as we move towards the edges.

Select Answer:

Visualized Solution

Visualizing the System

Gravitational Force as Centripetal Force

Introducing Kinetic Energy

Substituting Kinetic Energy

Expression for Enclosed Mass

Differentiating Mass

Mass of a Spherical Shell

Equating the Mass Expressions

Solving for Mass Density

Calculating Particle Number Density

The Sigma Insight: Orbital Motion of a Satellite

Analyzing the Setup

The Master Equation

The Kinetic Energy Connection

Unveiling the Mass Distribution

Final Calculation

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