Q.The gravitational force between a hollow spherical shell (of radius $R$ and uniform density) and a point mass is $F$. Show the nature of $F$ vs $r$ graph where $r$ is the distance of the point from the centre of the hollow spherical shell of uniform density.

Concept: Gravitational field of a spherical shell

A uniform hollow spherical shell produces a gravitational field that depends on whether the point mass lies inside or outside the shell. Two results govern the behavior:

1. Outside the shell ($r > R$): The shell behaves as if all its mass $M$ were concentrated at the center. The force follows the inverse-square law:

As $r$ increases beyond $R$, $F$ decreases as $1/r^2$.

2. Inside the shell ($r < R$): By the shell theorem, the net gravitational force on any point inside a uniform spherical shell is exactly zero. Every mass element of the shell is balanced by another on the opposite side, so: $F = 0$ …

Why the Shell Behaves This Way

The gravitational field of a hollow spherical shell has a beautiful symmetry that leads to two remarkable results. Every bit of mass on the shell pulls on a point mass, but when you're inside the shell, the nearby portions pull harder while the far portions have more mass—these effects cancel perfectly by symmetry. When you're outside, the entire shell's mass acts as if concentrated at the center.

These aren't approximations; they're exact consequences of the inverse-square law and spherical symmetry, first proved by Newton himself.

Building the $F$ vs $r$ Graph

1. Inside the shell ($r < R$):

Place a point mass anywhere inside the hollow shell. Draw a narrow cone from that point; it intersects the shell at two patches, one near and one far. The near patch subtends a solid angle $d\Omega$, and its area scales as $r_1^2 d\Omega$ (where $r_1$ is its distance from the point). The far patch has area $r_2^2 d\Omega$. Since the shell has uniform surface density $\sigma$, the masses are proportional to these areas.

The gravitational force from the near patch goes as $\frac{m_1}{r_1^2} \propto \frac{r_1^2}{r_1^2} = 1$, and from the far patch as $\frac{m_2}{r_2^2} \propto \frac{r_2^2}{r_2^2} = 1$. But they pull in opposite directions. When you integrate over all such cone pairs covering the entire shell, the symmetry ensures complete cancellation.

2. On the surface ($r = R$):

Right at the surface, the point mass sits on the boundary. Rigorously, the interior result extends up to $r = R$ from below, so $F(R^-) = 0$. From outside, we'll see that $F(R^+) = \frac{GM m}{R^2}$ where $M$ is the shell's total mass. There's a discontinuity at $r = R$.

3. Outside the shell ($r > R$):

For any point outside, every mass element $dm$ on the shell is at some distance from the point. The shell theorem (derived by integrating over spherical coordinates) shows that the net gravitational force is identical to what you'd get if all the mass $M$ were concentrated at the center:

This is an inverse-square law, exactly like a point mass.

4. Behavior as $r \to \infty$:

Far from the shell, $F \to 0$ as $1/r^2$, asymptotically approaching the horizontal axis.

5. Behavior as $r \to R^+$:

Just outside the surface, $F$ reaches its maximum value $\frac{GMm}{R^2}$, then decreases. …

Step 1: split via shell theorem into inside (r<R) and outside (r>R). Step 2: inside, opposite patches' contributions cancel exactly -- F=0 for all r<R. Step 3: outside, the shell acts as if its mass were concentrated at the centre -- F=GMm/r^2. Step 4: at r=R there's a genuine discontinui …