Animated Solution for Physics - Gravitation: A test particle is moving in a circular orbit in the gravitational field produced by mass density $\rho (r)=\frac{K}{r^2}$. Identify the correct relation between the radius $R$ of the particle's orbit and its period $T$

Imagine a vast, sprawling cloud of cosmic dust where the density isn't uniform. Instead of being packed tightly at the center like a star, the density of this cloud decreases as you move outward, specifically following the rule $\rho (r)=\frac{K}{r^2}$.

Now, picture a small test particle of mass $m$ gracefully orbiting within this cloud in a perfect circle of radius $R$. Our mission is to uncover the relationship between the time it takes to complete one orbit (the period $T$) and the radius of that orbit ($R$).

To find the gravitational pull on our particle, we don't need to worry about the entire cloud. Thanks to Newton's Shell Theorem, any mass located outside the particle's orbit exerts zero net gravitational force on it. We only care about the mass enclosed within the sphere of radius $R$.

To calculate this enclosed mass $M$, we imagine the cloud as a series of thin, concentric spherical shells. Let's take one such shell at a distance $r$ from the center, with a tiny thickness $dr$.

The volume of this thin shell is its surface area multiplied by its thickness: $dV=4\pi r^{2}dr$.

The mass of this specific shell is simply its volume times the density at that distance:

Notice the mathematical magic happening here? The $r^2$ in the denominator of the density perfectly cancels out the $r^2$ in the surface area formula!

To find the total enclosed mass $M$, we integrate this from the center ($0$) to the orbit's radius ($R$):

For our test particle to maintain its perfect circular orbit, the inward gravitational pull must exactly provide the required centripetal force.

Setting gravitational force equal to centripetal force ($F_g=F_c$):

Now, let's substitute the enclosed mass $M=4\pi KR$ that we just calculated:

Simplifying the left side gives:

Look closely at the equation above. The mass of the particle $m$ cancels out, which is standard for orbits. But incredibly, the radius $R$ also completely cancels out from both sides!

This is a profound result. It tells us that the orbital velocity $v$ is a constant, entirely independent of how large or small the orbit is. Whether the particle is orbiting close to the center or far out at the edges of the cloud, it travels at the exact same speed!

Finally, we know that the time period $T$ is the total distance of one orbit (the circumference) divided by the orbital speed:

Substituting our constant velocity:

By rearranging this, we isolate the variables from the constants:

Since $\pi$, $G$, and $K$ are all constants, the ratio $\frac{T}{R}$ is a constant. This is a fascinating deviation from Kepler's Third Law ($T^2\propto R^3$), which only applies when the central mass is a point source. Here, the extended mass distribution fundamentally alters the orbital dynamics!