Irodov 1.211: Every Shell Attracts as Concentrated – Amit & Deepali Sinha Foundation

Problem Irodov 1.211

A point particle of mass $m$ sits a distance $r$ from the center of a uniform solid sphere of mass $M$ , outside the sphere entirely (so $r$ exceeds the sphere’s own radius). Find (a) the gravitational potential energy of the particle-sphere system, and (b) the magnitude and direction of the gravitational force the sphere exerts on the particle.

The particle m outside the sphere M, at distance r from its center, with the attractive force F it feels. — The particle $m$ outside the sphere $M$ , at distance $r$ from its center, with the attractive force $\vect F$ it feels.

Given

$m$
$M$
$r$ (outside the sphere)

Find

(a)

U

, (b)

\vect F

Solution

Step 1Every point of one shell’s belt is equidistant from the particle

One thin shell of radius ρ, in cross section through its center O and the particle at A. The ring at colatitude θ, drawn as two thick arcs through P, lies a distance l from A. — One thin shell of radius $\rho$ , in cross section through its center $O$ and the particle at $A$ . The ring at colatitude $\theta$ , drawn as two thick arcs through $P$ , lies a distance $l$ from $A$ .

Build the sphere out of thin concentric shells, radius $\rho$ running from $0$ to the sphere’s own radius, each of mass $\delta M$ . Within one shell, split off the ring (“belt”) at colatitude $\theta$ from the axis $OA$ : by the ring’s rotational symmetry about that axis, every point of it, such as $P$ in the figure, is the same distance $l$ from $A$ . The triangle $OAP$ fixes that distance.

l^2 = \rho^2 + r^2 - 2\rho r\cos\theta \tag{1}

Step 2The belt’s mass and its potential energy with the particle

The shell carries its mass $\delta M$ spread evenly over its whole area $4\pi\rho^2$ , and the belt at colatitude $\theta$ is a strip of circumference $2\pi\rho\sin\theta$ and width $\rho\,d\theta$ . Dividing the strip’s area by the shell’s area gives the fraction of $\delta M$ that the belt holds.

\begin{align*}dS_{\text{belt}} &= (2\pi\rho\sin\theta)(\rho\,d\theta) \\ &= 2\pi\rho^2\sin\theta\,d\theta \\ \frac{dS_{\text{belt}}}{4\pi\rho^2} &= \frac{2\pi\rho^2\sin\theta\,d\theta}{4\pi\rho^2} \\ &= \frac{\sin\theta}{2}\,d\theta\end{align*}

That fraction of the shell’s mass is the belt’s own mass.

\delta M_{\text{belt}} = \frac{\delta M}{2}\sin\theta\,d\theta \tag{2}

Every part of the belt sits at the same distance $l$ from the particle, the positive square root of (1), so the belt interacts with $m$ exactly as a point mass $\delta M_{\text{belt}}$ placed at that one distance.

\begin{align*}l &= \sqrt{\rho^2 + r^2 - 2\rho r\cos\theta} \\ dU &= -\frac{\gamma m\,\delta M_{\text{belt}}}{l}\end{align*}

Putting (2) in place of $\delta M_{\text{belt}}$ leaves the belt’s contribution carrying the angle explicitly.

dU = -\frac{\gamma m\,\delta M}{2l}\sin\theta\,d\theta

(3)

Step 3Integrate the belt over one whole shell

The contribution (3) still carries both $l$ and $\theta$ , and the two are tied together by the triangle: differentiating (1) at fixed $\rho$ and $r$ trades one for the other.

\begin{align*} 2l\,dl &= 2\rho r\sin\theta\,d\theta \\ \sin\theta\,d\theta &= \frac{l\,dl}{\rho r} \tag{4}\end{align*}

Substituting (4) into (3) cancels the $l$ in the denominator and leaves a constant integrand. As $\theta$ sweeps from $0$ to $\pi$ the distance $l$ runs from $r-\rho$ to $r+\rho$ , since the particle lies outside this shell.

\begin{align*}dU &= -\frac{\gamma m\,\delta M}{2l}\cdot\frac{l\,dl}{\rho r} \\ &= -\frac{\gamma m\,\delta M}{2\rho r}\,dl \\ \delta U &= -\frac{\gamma m\,\delta M}{2\rho r}\int_{r-\rho}^{r+\rho} dl \\ &= -\frac{\gamma m\,\delta M}{2\rho r}\Bigl[\,l\,\Bigr]_{r-\rho}^{r+\rho} \\ &= -\frac{\gamma m\,\delta M}{2\rho r}\bigl[(r+\rho)-(r-\rho)\bigr] \\ &= -\frac{\gamma m\,\delta M}{2\rho r}\cdot 2\rho\end{align*}

\delta U = -\frac{\gamma m\,\delta M}{r}

(5)

Step 4Sum over every shell, then differentiate for the force

(5) does not depend on the shell’s own radius $\rho$ at all, so every shell making up the sphere contributes at the same rate per unit of its own mass. Summing the shells therefore only adds their masses, and those add to the sphere’s total mass $M$ .

\begin{align*}\sum_i \delta M_i &= M \\ U &= \sum_i \delta U_i \\ &= -\frac{\gamma m}{r}\sum_i \delta M_i\end{align*}

U = -\frac{\gamma m M}{r}

(6)

The radial force is minus the derivative of this energy with respect to $r$ , the particle’s only coordinate that matters by spherical symmetry, so (6) is differentiated once.

\frac{\partial U}{\partial r} = \frac{\partial}{\partial r}\left(-\frac{\gamma m M}{r}\right) = \frac{\gamma m M}{r^2}

F_r = -\frac{\partial U}{\partial r} = -\frac{\gamma m M}{r^2}

(7)

the minus sign meaning $\vect F$ points from the particle back toward the sphere’s center, i.e. it is attractive.

Check

(5) carries no dependence on the shell’s own radius $\rho$ at all, so it must survive two limits actually worth evaluating. As $\rho\to0$ the shell shrinks onto a point mass $\delta M$ sitting at the center, and $-\gamma m\,\delta M/r$ is exactly that point mass’s potential energy with $m$ , as it must be. As $\rho\to r$ the particle closes in on the shell’s own surface, and the formula still reads $-\gamma m\,\delta M/r$ : the shell attracts as its mass concentrated at the center right up to its own surface, not just from far away. Neither limit is automatic in general; both hold here only because the belt integration in Step 3 made every $\rho$ -dependence cancel exactly.

Remark

Step 4 never used that the sphere is uniform in any way beyond “each shell has the same density all over its own surface”: the radial density could vary from shell to shell in any way at all and the argument would be unchanged, since $\delta U=-\gamma m\,\delta M/r$ held for one shell regardless of $\rho$ . That is Newton’s shell theorem in full: any spherically symmetric body attracts an external particle exactly as if the whole of its mass were concentrated at its center.