Animated Solution for Physics - Gravitation: The mass density of a spherical galaxy varies as $\frac{K}{r}$ over a large distance $r$ from its centre. In that region, a small star is in a circular orbit of radius $R$. Then, the period of revolution $T$ depends on $R$ as

Imagine you are an astronomer observing a distant, massive spherical galaxy. Unlike a solid planet where the density might be relatively uniform, this galaxy is a sprawling cloud of stars and gas. Its mass density isn't constant; it thins out as you move further from the center. Specifically, the problem tells us that the density $\rho (r)$ varies inversely with the distance $r$ from the center, given by the relation $\rho (r)=\frac{K}{r}$, where $K$ is a constant.

Now, picture a small star orbiting this galaxy in a perfect circle of radius $R$. Our mission is to figure out how the time period of this star's revolution, $T$, depends on its orbital radius $R$.

To understand the gravitational pull on the star, we must first determine the total mass of the galaxy enclosed within the star's orbit. According to Newton's Shell Theorem, only the mass inside the radius $R$ contributes to the net gravitational force on the star. The mass outside this radius exerts forces that perfectly cancel each other out.

Since the density is variable, we cannot simply multiply density by the total volume. We must build the mass up layer by layer. Imagine a thin spherical shell of radius $r$ and an infinitesimally small thickness $dr$. The volume of this shell is its surface area multiplied by its thickness: $dV=4\pi r^{2}dr$.

The mass of this tiny shell, $dm$, is its density multiplied by its volume:

Notice how beautifully the math simplifies. One $r$ cancels out, leaving us with:

To find the total enclosed mass $M$, we integrate this elemental mass from the center ($r=0$) all the way out to the star's orbit ($r=R$):

Evaluating this gives us the total effective mass pulling on our star:

For the star to maintain its circular orbit, the gravitational force pulling it inward must perfectly provide the required centripetal force. Let $m$ be the mass of the star and $v$ be its orbital velocity. We equate the two forces:

The mass of the star $m$ cancels out, which is why the mass of an orbiting body doesn't affect its orbit. Solving for $v^2$, we get:

Now, we substitute the enclosed mass $M=2\pi K{R}^2$ that we just calculated:

Taking the square root, we find the orbital velocity:

Finally, we need to find the time period $T$. The time period is simply the total distance traveled in one orbit (the circumference) divided by the orbital velocity:

Substituting our expression for $v$:

We can separate the constants from the variables to see the proportionality clearly. The constants $2\pi$, $G$, and $K$ don't affect the relationship between $T$ and $R$.

Since $\frac{R}{\sqrt{R}}=\sqrt{R}$, we have:

To match the options provided in the question, we square both sides of this proportionality:

This elegant result shows that for a galaxy with a density profile of $\frac{K}{r}$, the square of the orbital period is directly proportional to the radius, unlike Kepler's Third Law ($T^2\propto R^3$) which applies to solar systems where almost all the mass is concentrated at a single central point!