Irodov 1.215: A Cavity's Field from Two Spheres – Amit & Deepali Sinha Foundation

Problem Irodov 1.215

A large solid sphere of uniform density $\rho$ has a smaller spherical cavity hollowed out from somewhere inside it. Let $\vect l$ be the vector from the sphere’s own center to the cavity’s center, so that $l$ , the magnitude of $\vect l$ , is the distance between the two centers. Find the strength $G$ of the gravitational field at an arbitrary point inside the cavity, in terms of $\rho$ and $\vect l$ .

The uniform sphere of density ρ (center O) with the spherical cavity (center O') hollowed out of it; the two centers are a distance l apart. — The uniform sphere of density $\rho$ (center $O$ ) with the spherical cavity (center $O'$ ) hollowed out of it; the two centers are a distance $l$ apart.

Given

$\rho$ , $\vect l = \overrightarrow{OO'}$

Find

The field

\vect G

at any point inside the cavity

Solution

Step 1Write the cavity as a superposition of two complete spheres

A sphere with a cavity is the same as a complete uniform sphere of density $\rho$ (center $O$ ) together with a second sphere of the cavity’s size but density $-\rho$ (center $O'$ ), canceling the missing material there; by superposition the true field anywhere is the sum of the two spheres’ fields taken separately,

\vect G(\vect r) = \vect g_\rho(\vect r) + \vect g_{-\rho}(\vect r), \tag{1}

where $\vect g_\rho$ is the field of the complete sphere of density $\rho$ at $O$ , and $\vect g_{-\rho}$ is the field of the complete sphere of density $-\rho$ and the cavity’s radius, at $O'$ .

Step 2Recall the field inside a uniform sphere

By the shell theorem established in Problem 1.214, the material outside radius $r$ pulls a point at $r$ with zero net field, so the field there is set by the enclosed mass $m(r)$ alone, which acts as a point mass at the center. Taking a uniform sphere of density $\rho$ , the enclosed mass, the magnitude of the field it produces at radius $r$ , and the same field written as a vector from its own center are

\begin{align*} m(r) &= \frac{4}{3}\pi r^3\rho, \\ g(r) &= \frac{\gamma\,m(r)}{r^2} = \frac{\gamma}{r^2}\,\frac{4}{3}\pi r^3\rho = \frac{4}{3}\pi\gamma\rho\,r, \\ \vect g(\vect r) &= -\frac{4}{3}\pi\gamma\rho\,\vect r. \tag{2}\end{align*}

The field grows linearly with $r$ , and the minus sign carries its direction, toward that sphere’s own center. The density enters only as a linear factor, so the same relation serves the second sphere of Step 1, whose density is negative.

Step 3Add the two fields at a point inside the cavity

Let $\vect r$ be measured from $O$ and $\vect r' = \vect r - \vect l$ from $O'$ , so $\vect r'$ is the position relative to the cavity’s center; applying (2) to each sphere and substituting into the superposition (1) gives the two contributions at any point inside the cavity,

\vect G = -\frac{4}{3}\pi\gamma\rho\,\vect r \;-\;\Big(-\frac{4}{3}\pi\gamma\rho\,\vect r'\Big) = -\frac{4}{3}\pi\gamma\rho\,(\vect r - \vect r'). \tag{3}

Step 4Simplify using the geometry of the two centers

By construction $\vect r - \vect r' = \vect l$ for every point, inside the cavity or not: it is simply the fixed offset between the two centers. Substituting into (3) removes the position entirely,

\vect G = -\dfrac{4}{3}\pi\gamma\rho\,\vect l

(4)

Inside the cavity the field is the same vector G at every point, of magnitude 43πγρ l and directed from O' toward O, opposite to l. — Inside the cavity the field is the same vector $\vect G$ at every point, of magnitude $\frac43\pi\gamma\rho\,l$ and directed from $O'$ toward $O$ , opposite to $\vect l$ .

Check

The position-dependent parts of the two fields in (3), $-\frac43\pi\gamma\rho\,\vect r$ and $+\frac43\pi\gamma\rho\,\vect r'$ , cancel exactly because $\vect r-\vect r' = \vect l$ is the same fixed vector at every point of the cavity; that cancellation is why the field inside the cavity comes out uniform, not the other way round. As a limiting case, $l=0$ (a cavity concentric with the sphere) makes $\vect l$ vanish and the field with it, matching the familiar result that the field is zero everywhere inside a hollow concentric shell.

Remark

The cavity’s own radius never enters the boxed result: only $\rho$ and the offset $\vect l$ do, so any spherical cavity carved a distance $l$ from the center, whatever its size, produces the same uniform field inside it. That radius independence holds only for the field inside the cavity; the exterior gravity anomaly a surface survey measures is not independent of the cavity’s radius, scaling as $a^3/d^2$ for a cavity of radius $a$ at depth $d$ , which is exactly how microgravity surveys size a buried void.