From a solid sphere of mass M and radius R, a spherical portion of radius (R/2) is removed as shown

Q: Why do we subtract the potential of the removed part? Can't we just integrate over the remaining shape?

While you technically could integrate over the remaining shape, the math would be incredibly complex due to the asymmetric boundaries. The superposition principle is a powerful shortcut. Because gravitational potential is a scalar quantity, the potential of a whole object is exactly equal to the sum of the potentials of its parts. Therefore, Remaining = Complete - Removed.

Q: How did we get the mass of the removed sphere as $M/8$?

The problem states it's a solid sphere, implying uniform density. Density $\rho = \frac{\text{Mass}}{\text{Volume}}$. Since density is constant, Mass is proportional to Volume. Volume of a sphere is $\frac{4}{3}\pi R^3$, so Mass is proportional to $R^3$. The removed sphere has half the radius ($R/2$), so its mass is $(1/2)^3 = 1/8$ of the total mass $M$.

Q: Why is $r=0$ when calculating the potential for the removed sphere?

We are calculating the potential at point $P$. For the complete sphere, point $P$ is at a distance of $R/2$ from its center $O$. However, for the removed sphere, point $P$ is its *own* center. The distance from the center of the removed sphere to point $P$ is therefore zero.

Q: Can we use the formula $V = -\frac{GM}{r}$ here?

No! That formula is strictly for the potential *outside* a spherical body (or for a point mass). Point $P$ is *inside* the complete sphere, so we must use the internal potential formula: $V = -\frac{GM}{2R^3}(3R^2 - r^2)$.

Q: What if the reference point for zero potential was not at infinity?

The formula $V = -\frac{GM}{2R^3}(3R^2 - r^2)$ is derived specifically by integrating from infinity (where $V=0$) to the point $r$. If the reference point changes, a constant value would be added to all potential values. However, the superposition principle ($V_{\text{remaining}} = V_{\text{complete}} - V_{\text{removed}}$) would still hold perfectly true.

Question

From a solid sphere of mass $M$ and radius $R$, a spherical portion of radius $(R/2)$ is removed as shown in the figure. Taking gravitational potential $V = 0$ at $r = \infty$, the potential at the centre of the cavity thus formed is ($G =$ gravitational constant)

Options:

(A) $\frac{-GM}{2R}$

(B) $\frac{-GM}{R}$

(C) $\frac{-2GM}{3R}$

(D) $\frac{-2GM}{R}$

Sigma Percentile · Accepted Answer

### The Power of Superposition

Imagine you are tasked with finding the gravitational potential inside a piece of Swiss cheese. Calculating the potential of an irregular shape directly using integration is a mathematical nightmare. But physics offers us a beautiful, elegant shortcut: **The Principle of Superposition**.

Whenever you see a cavity in a continuous mass distribution, you should immediately think of it as a combination of two complete objects. The remaining mass is mathematically identical to a **complete solid sphere** plus a **phantom sphere of negative mass** occupying the cavity's space.

Therefore, the potential at any point $P$ is simply the algebraic sum of the potentials created by these two distinct spheres:

$$V_{	ext{remaining}} = V_{	ext{complete}} - V_{	ext{removed}}$$

### The Master Equation

Before we dive into the two phases of our calculation, we need our primary tool. The gravitational potential $V$ at an internal point (at a distance $r$ from the center) of a uniform solid sphere of mass $M$ and radius $R$ is given by:

$$V = -\frac{GM}{2R^3}(3R^2 - r^2)$$

This formula assumes that the potential at infinity is zero, which aligns perfectly with our problem's constraints.

### Phase 1: The Complete Sphere

Let's first pretend the cavity doesn't exist. We have a complete solid sphere of mass $M$ and radius $R$. We need to find the potential it creates at point $P$, which is the center of the cavity.

Looking at the geometry, point $P$ is located at a distance $r = \frac{R}{2}$ from the center $O$ of the main sphere. Let's substitute this into our master equation to find the potential $V_s$:

$$V_s = -\frac{GM}{2R^3}\left[3R^2 - \left(\frac{R}{2}ight)^2ight]$$

Squaring the distance gives us $\frac{R^2}{4}$. Subtracting this from $3R^2$ yields $\frac{11R^2}{4}$.

$$V_s = -\frac{GM}{2R^3}\left[\frac{11R^2}{4}ight] = -\frac{11GM}{8R}$$

This is the potential if the sphere were completely solid.

### Phase 2: The Phantom Sphere

Now, we must account for the mass that was removed to create the cavity. This removed portion is a smaller solid sphere of radius $\frac{R}{2}$.

First, what is its mass $M'$? Since the original sphere has a uniform density, mass is directly proportional to volume, and volume is proportional to the cube of the radius ($R^3$).

$$M' = M \left(\frac{R/2}{R}ight)^3 = M \left(\frac{1}{2}ight)^3 = \frac{M}{8}$$

Next, we need the potential $V_c$ created by this removed sphere at point $P$. Here is the crucial insight: **Point $P$ is the exact center of the removed sphere!** Therefore, for this specific calculation, the distance from the center is $r = 0$.

Substituting $r = 0$, mass $M' = \frac{M}{8}$, and radius $R' = \frac{R}{2}$ into our master equation:

$$V_c = -\frac{3G M'}{2 R'} = -\frac{3G (M/8)}{2 (R/2)} = -\frac{3GM}{8R}$$

### The Grand Finale

We have the potential of the complete sphere and the potential of the removed sphere. It's time to bring them together using the superposition principle. We subtract the potential of the removed part from the complete part:

$$V_P = V_s - V_c$$

$$V_P = -\frac{11GM}{8R} - \left(-\frac{3GM}{8R}ight)$$

Be very careful with the double negative here! It becomes a positive:

$$V_P = -\frac{11GM}{8R} + \frac{3GM}{8R} = -\frac{8GM}{8R}$$

And in a moment of mathematical elegance, the $8$ cancels out perfectly, leaving us with our final answer:

$$V_P = -\frac{GM}{R}$$

This matches option (b). The beauty of superposition turns a complex geometric void into a simple subtraction problem!

Select Answer:

Visualized Solution

\text{Principle of Superposition}

\text{Potential of a Solid Sphere}

\text{Potential due to Complete Sphere } (V_s)

\text{Calculating } V_s

\text{Mass of Removed Sphere } (M')

\text{Calculating } M'

\text{Potential due to Removed Sphere } (V_c)

\text{Calculating } V_c

\text{Superposition Principle}

\text{Final Answer}

\text{The Way Forward}

The Sigma Insight: Gravitational Potential and Potential Energy

The Power of Superposition

The Master Equation

Phase 1

Phase 2

The Grand Finale

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