A spherically symmetric gravitational system of particles has a mass density rho = rho_0 for r

Q: Why does the mass outside the radius $r$ not exert any gravitational force on the test mass when it is inside the sphere?

This is a direct consequence of Newton's Shell Theorem. For any point inside a uniform spherical shell, the gravitational forces exerted by different parts of the shell in opposite directions cancel each other out perfectly, resulting in a net force of zero.

Q: Is the velocity graph continuous at the boundary $r = R$?

Yes, the velocity is continuous at $r = R$. If you substitute $r = R$ into the inner formula $v = \sqrt{\frac{4}{3}\pi G \rho_0} r$ and use $M = \frac{4}{3}\pi R^3 \rho_0$, you get $v = \sqrt{\frac{GM}{R}}$, which is exactly the same as the outer formula evaluated at $r = R$.

Q: Why is the curve for $r > R$ not a straight line?

Outside the sphere, the speed $v$ is proportional to $1/\sqrt{r}$. Since the exponent of $r$ is negative and fractional, the relationship is non-linear, resulting in a curve that decreases and asymptotically approaches the $r$-axis.

Q: Would the result be different if the test mass was extremely heavy?

Yes. If the test mass $m$ were comparable to the mass of the sphere, we would have to consider the motion of both bodies around their common center of mass (two-body problem). However, the problem specifies a 'test mass', which implies $m \ll M$, so the sphere remains stationary at the center.

Q: What would happen to the graph if the sphere was hollow?

If the sphere was a hollow shell of radius $R$, the gravitational force inside ($r R$), the force and velocity profile would remain exactly the same as a solid sphere of the same mass.

Question

A spherically symmetric gravitational system of particles has a mass density $ho = \begin{cases} ho_0 & 	ext{for } r \le R \ 0 & 	ext{for } r > R \end{cases}$, where $ho_0$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed $v$ as a function of distance $r$ from the centre of the system is represented by

Sigma Percentile · Accepted Answer

Imagine you are a cosmic explorer, piloting a tiny probe through a mysterious, uniform dust cloud of radius $R$ floating in deep space.

As you orbit around the center of this cloud, you notice something fascinating: your orbital speed changes depending on how deep you are.

This is not just a sci-fi scenario; it is a classic JEE problem that tests your fundamental understanding of gravity, shell theorem, and circular dynamics.

Let us embark on this journey to understand how the orbital speed $v$ behaves as a function of distance $r$ from the center.

### The Cosmic Dance of Gravity

To understand any orbital motion, we must first look at the forces at play.

When an object of mass $m$ moves in a circular orbit of radius $r$ with speed $v$, it experiences a centripetal acceleration directed toward the center.

This acceleration requires a force, which in this case is provided entirely by the gravitational attraction of the mass distribution.

Therefore, our starting point is the master equation of circular orbits:

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

Here, $F_g$ is the gravitational force acting on the test mass $m$ at a distance $r$ from the center.

### Journey to the Center of the Earth: The Inside Story ($r \le R$)

Let us first dive deep inside the sphere, where $r \le R$.

You might think that all the mass of the sphere pulls on our probe, but Isaac Newton's famous **Shell Theorem** tells us otherwise.

According to the Shell Theorem, if you are inside a spherically symmetric shell of mass, the net gravitational force exerted by that shell on you is exactly zero!

This means that only the mass enclosed within a sphere of radius $r$ exerts a net gravitational pull on our test mass.

The mass outside this radius $r$ has no effect on your motion because the pulls from different parts of the outer shell perfectly cancel each other out.

So, let us calculate this enclosed mass, which we will call $M_{	ext{enclosed}}$.

Since the sphere has a uniform mass density $ho_0$, the enclosed mass is simply the volume of the sphere of radius $r$ multiplied by the density:

$$M_{	ext{enclosed}} = \frac{4}{3}\pi r^3 ho_0$$

Now, we can write the gravitational force using Newton's law of gravitation:

$$F_g = \frac{G m M_{	ext{enclosed}}}{r^2}$$

Substituting our expression for $M_{	ext{enclosed}}$:

$$F_g = \frac{G m \left(\frac{4}{3}\pi r^3 ho_0ight)}{r^2} = \frac{4}{3}\pi G m ho_0 r$$

Notice how the force is directly proportional to the distance $r$.

Now, let us plug this force back into our master orbital equation:

$$\frac{mv^2}{r} = \frac{4}{3}\pi G m ho_0 r$$

The mass of the probe $m$ cancels out beautifully from both sides.

Multiplying both sides by $r$, we get:

$$v^2 = \frac{4}{3}\pi G ho_0 r^2$$

Taking the square root of both sides:

$$v = \sqrt{\frac{4}{3}\pi G ho_0} \cdot r$$

Since $G$ and $ho_0$ are constants, we find that:

$$v \propto r$$

This is a stunning result! Inside the uniform sphere, your orbital speed increases **linearly** with distance.

At the very center ($r = 0$), the speed is zero, and it reaches its maximum value at the surface ($r = R$).

### Breaking Free: The Outside Story ($r > R$)

Now, let us fly out of the sphere into the empty space beyond, where $r > R$.

Once you are outside, the entire mass of the sphere lies within your orbit.

According to the second part of the Shell Theorem, a spherically symmetric body attracts external objects as if all its mass were concentrated at its center.

So, the total mass $M$ of the sphere is constant and is given by:

$$M = \frac{4}{3}\pi R^3 ho_0$$

The gravitational force on our probe at a distance $r$ is:

$$F_g = \frac{G M m}{r^2}$$

Let us equate this to the centripetal force:

$$\frac{mv^2}{r} = \frac{G M m}{r^2}$$

Again, the mass $m$ cancels out, and one factor of $r$ in the denominator cancels:

$$v^2 = \frac{G M}{r}$$

Taking the square root:

$$v = \sqrt{\frac{G M}{r}}$$

Since $G$ and $M$ are constants, we get:

$$v \propto \frac{1}{\sqrt{r}}$$

This is the classic Keplerian relation for orbits around a point mass!

As you move further away from the sphere, your orbital speed decreases non-linearly, asymptotically approaching zero as $r$ goes to infinity.

### Stitching the Pieces Together: The Complete Graph

Let us summarize what we have discovered:

1. For $r \le R$, the speed $v$ is directly proportional to $r$ ($v \propto r$). This is represented by a straight line passing through the origin.

2. For $r > R$, the speed $v$ is inversely proportional to the square root of $r$ ($v \propto 1/\sqrt{r}$). This is represented by a curve that slopes downwards.

At the boundary $r = R$, both formulas yield the same value:

$$v(R) = \sqrt{\frac{G M}{R}}$$

This means the transition is continuous, with a peak at $r = R$.

Looking at the options, **Option (c)** is the only graph that correctly shows a linear increase up to $r = R$ followed by a non-linear decrease.

### What If? The Way Forward

As an elite physicist, you should always ask: 'What if the conditions change?'

What if the density of the sphere was not uniform, but instead decreased with distance, say $ho(r) = \frac{K}{r}$?

In that case, the enclosed mass would be:

$$M_{	ext{enclosed}} = \int_0^r 4\pi x^2 ho(x) dx = \int_0^r 4\pi x^2 \frac{K}{x} dx = 2\pi K r^2$$

Equating centripetal force to gravitational force:

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

$$v^2 = 2\pi G K r \implies v \propto \sqrt{r}$$

In this case, the speed inside would increase as the square root of $r$ rather than linearly!

This kind of critical thinking is what separates top rankers from the rest. Keep exploring, keep questioning, and let the beauty of physics guide you!

Select Answer:

Visualized Solution

Visual Anchor: Spherically Symmetric System

Logic Bridge: Condition for Circular Orbit

Case 1: Inside the Sphere ( $r \leq R$ )

Enclosed Mass Calculation

Gravitational Force Inside

Orbital Speed Inside ( $r \leq R$ )

Case 2: Outside the Sphere ( $r > R$ )

Total Mass of the Sphere

Gravitational Force Outside

Orbital Speed Outside ( $r > R$ )

Conclusion: The Complete Graph

The Way Forward: Non-Uniform Density

The Sigma Insight: Orbital Motion of a Satellite

The Cosmic Dance of Gravity

Journey to the Center of the Earth

Breaking Free

Stitching the Pieces Together

What If? The Way Forward

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