Animated Solution for Physics - Gravitation: A solid sphere of uniform density and radius $4$ units is located with its centre at the origin $O$ of coordinates. Two spheres of equal radii $1$ unit, with their centres at $A(-2,0,0)$ and $B(2,0,0)$ respectively, are taken out of the solid leaving behind spherical cavities as shown in figure.

When dealing with gravitational systems that have cavities or missing mass, direct integration can quickly become a mathematical nightmare.

Fortunately, physics offers us a beautiful shortcut: The Principle of Superposition.

Instead of viewing a cavity as "empty space," we can mathematically treat it as a region containing negative mass superimposed on a fully complete, solid sphere.

This conceptual shift transforms a complex geometry problem into a simple exercise of vector addition and scalar summation.

Let's dive deep into how this elegant principle helps us analyze the gravitational field and potential of a sphere with symmetric cavities.

Let's first look at the center of our coordinate system, the origin $O(0,0,0)$.

By spherical symmetry, a complete, uniform solid sphere of radius $R=4$ produces absolutely zero gravitational field at its own center:

Now, we introduce the two cavities at $A(-2,0,0)$ and $B(2,0,0)$.

Using the negative mass concept, we place a negative mass $-m$ at $A$ and another negative mass $-m$ at $B$.

Since gravitational field vectors point towards the mass creating them, a negative mass will produce a field pointing away from its center.

Therefore, the negative mass at $A$ produces a field at the origin pointing in the $+x$ direction:

Symmetrically, the negative mass at $B$ produces a field at the origin pointing in the $-x$ direction:

When we sum these contributions, they cancel each other out perfectly:

Thus, the gravitational field at the origin is exactly zero, making option (a) correct.

Now, let's move to the center of the right cavity, point $B(2,0,0)$.

Does the field vanish here?

At $x=2$, we are inside the complete sphere, so it exerts a non-zero gravitational pull directed towards the origin:

The negative mass at $B$ produces zero field at its own center:

However, the negative mass at $A(-2,0,0)$ is at a distance of $4$ units from $B$, and it exerts a field pointing in the $+x$ direction:

Since ${\vec{E}}_{\text{complete}}(B)$ and ${\vec{E}}_A(B)$ do not have equal magnitudes, they cannot cancel each other out.

Therefore, the net gravitational field at point $B$ is non-zero, which disproves option (b).

Let's look at options (c) and (d), which describe circles in the $y-z$ plane:

Since these equations have no $x$ term, they lie entirely in the plane $x=0$.

This plane is a perpendicular bisector to the line segment $AB$ connecting the centers of the two cavities.

Let's choose any arbitrary point $P(0,y,z)$ on one of these circles.

Its distance to the origin $O(0,0,0)$ is:

Since $y^2+z^2$ is constant for all points on a given circle, $d_O$ is completely constant.

Now, let's find the distance from $P$ to the center of cavity $A(-2,0,0)$:

Similarly, the distance to the center of cavity $B(2,0,0)$ is:

Notice something incredible here!

For any point on the circle, the distances $d_O$, $d_A$, and $d_B$ are not only constant, but $d_A$ is also perfectly equal to $d_B$:

Since gravitational potential is a scalar quantity depending only on distance:

Because $d_O$, $d_A$, and $d_B$ are constant for all points on the circle, the net potential $V_{\text{net}}(P)$ must be absolutely constant along the entire circle.

This elegant symmetry holds true for both circles ($R=6$ and $R=2$).

Thus, options (c) and (d) are both correct.

By leveraging the power of superposition and geometric symmetry, we solved this advanced problem with minimal algebraic effort: - The symmetric placement of cavities ensures the field cancels at the origin (Option a). - The perpendicular bisector plane ($x=0$) guarantees equidistant paths to both cavities, keeping the potential uniform along any concentric circle in this plane (Options c and d).