Irodov 1.212: Opposite Shell Patches Cancel Inside – Amit & Deepali Sinha Foundation

Problem Irodov 1.212

A particle sits at an arbitrary point inside the hollow region enclosed by a uniform spherical layer of matter, that is, in the cavity bounded by the layer’s inner surface. The layer is the region between two concentric spherical surfaces, filled with matter of the same density throughout (an infinitesimally thin shell, its whole mass on one surface, is the limiting case). Demonstrate that the net gravitational force this layer exerts on the particle is zero, wherever in the cavity the particle happens to be.

The uniform spherical layer in cross section, the matter lying between its inner and outer surfaces (both centered on O), with the particle A at an arbitrary point of the cavity inside. — The uniform spherical layer in cross section, the matter lying between its inner and outer surfaces (both centered on $O$ ), with the particle $A$ at an arbitrary point of the cavity inside.

Given

a uniform spherical layer (a thin shell first)
a particle $A$ anywhere in the cavity

Find

demonstrate that the layer’s net force on

A

vanishes

Solution

Step 1Cut the shell with a thin double cone from the particle

The double cone from the particle A cuts the shell of radius R centered on O at the near patch P_1 of area dS_1 and the far patch P_2 of area dS_2. Its edge rays are dashed and the base angles α equal. — The double cone from the particle $A$ cuts the shell of radius $R$ centered on $O$ at the near patch $P_1$ of area $dS_1$ and the far patch $P_2$ of area $dS_2$ . Its edge rays are dashed and the base angles $\alpha$ equal.

Draw a narrow cone of solid angle $d\Omega$ with its vertex at $A$ , opening in both directions along one line (a “double cone”), as the figure shows. Because $A$ is inside the shell, this line crosses the spherical surface twice, once nearby at distance $r_1$ (cutting a small patch $dS_1$ ) and once farther away at distance $r_2$ (cutting $dS_2$ ).

Step 2The two patches have areas in the ratio of the squares of their distances

Let $O$ be the shell’s center and $\alpha_1,\alpha_2$ the angles the outward radii $OP_1,OP_2$ make with the cutting line at the two patches. The triangle $OP_1P_2$ is isosceles ( $OP_1=OP_2=R$ , the shell’s radius), so its base angles are equal.

\alpha_1 = \alpha_2 \equiv \alpha

A patch seen from $A$ under solid angle $d\Omega$ and tilted by $\alpha$ from face on has area $dS = r^2\,d\Omega/\cos\alpha$ , so with the same $\alpha$ on both sides,

\begin{align*}dS_1 &= \frac{r_1^2\,d\Omega}{\cos\alpha} \\ dS_2 &= \frac{r_2^2\,d\Omega}{\cos\alpha}\end{align*}

\frac{dS_1}{r_1^2} = \frac{dS_2}{r_2^2} = \frac{d\Omega}{\cos\alpha}

(1)

Step 3Equal masses seen from

A

give equal forces

With uniform surface density $\sigma$ , the two patches carry mass $\sigma\,dS_1$ and $\sigma\,dS_2$ , and (1) makes their ratio the same as the ratio of the squares of their distances.

\frac{\sigma\,dS_1}{r_1^2} = \frac{\sigma\,dS_2}{r_2^2}

The two patches' pulls on the particle, d F_1 toward the near patch and d F_2 toward the far one, both along the single line through A and drawn equal in length. — The two patches’ pulls on the particle, $d\vect F_1$ toward the near patch and $d\vect F_2$ toward the far one, both along the single line through $A$ and drawn equal in length.

Each patch attracts the particle (mass $m$ ) with force $\gamma m$ times its own mass over the square of its own distance, so the two magnitudes are built from exactly this ratio.

\begin{align*}dF_1 &= \frac{\gamma m\,\sigma\,dS_1}{r_1^2} \\ dF_2 &= \frac{\gamma m\,\sigma\,dS_2}{r_2^2}\end{align*}

dF_1 = dF_2

(2)

Step 4The two forces cancel, patch by patch, over the whole shell

$P_1$ , $A$ and $P_2$ lie on one line with $A$ between the two patches, so the pull toward $P_1$ and the pull toward $P_2$ point in exactly opposite directions; (2) makes them equal in size as well, so this pair contributes nothing to the net force. Sweeping the double cone’s direction over every line through $A$ pairs up the entire shell this way, with nothing left over and nothing counted twice.

\vect F = \vect 0

(3)

Check

If $A$ sits exactly at the shell’s center, every $r_1=r_2=R$ and every $\alpha\to0$ (the radii meet the shell face on), so (1) trivially gives $dS_1=dS_2$ and every direction is paired with its own opposite by symmetry alone; the general argument above, tilt factor $\cos\alpha$ and all, reduces to that trivial case as $A\to O$ , as it must.

Remark

Together with the previous problem this completes the shell theorem: a uniform shell exerts no force at all on anything inside it, and behaves as a point mass at its center for anything outside. For a solid sphere built of many such shells, a particle at radius $r$ inside it therefore feels only the shells with radius less than $r$ ; the shells farther out contribute exactly zero, whatever their thickness or how the density varies from shell to shell.