Irodov 1.217: Assembling a Sphere Against Gravity – Amit & Deepali Sinha Foundation

Problem Irodov 1.217

Find the proper (self) gravitational potential energy, the energy needed to assemble the body from matter dispersed at infinity, of each of the following mass distributions, both of total mass $m$ and radius $R$ .

A thin uniform hollow spherical shell.
A uniform solid sphere.

(a) The thin uniform spherical shell of mass m and radius R, with one increment dm' of the arriving matter marked. Part (b) is the same radius R filled solid with matter of the same total mass m. — (a) The thin uniform spherical shell of mass $m$ and radius $R$ , with one increment $dm'$ of the arriving matter marked. Part (b) is the same radius $R$ filled solid with matter of the same total mass $m$ .

Given

(a) $m$ , $R$ (thin shell)
(b) $m$ , $R$ (solid sphere)

Find

The proper (self) gravitational potential energy in each case

Solution

Step 1(a) The cost of laying one increment onto a partly built shell

Build the shell by bringing its matter in from infinity a little at a time; the increment $dm'$ marked on the shell is one such arrival, and the mass $m'$ already present is taken as a uniform shell of radius $R$ . By the shell theorem such a shell pulls on anything at or outside its own radius exactly as a point mass at the center would, so wherever on the surface the increment lands it falls through the same potential of the mass $m'$ already in place. The surface potential, the work the field does on the increment, and the energy that work takes out of the configuration are

\begin{align*} \varphi(R) &= -\frac{\gamma m'}{R}, \\ dA &= -\varphi(R)\,dm' = \frac{\gamma m'\,dm'}{R}, \\ dU &= -dA = -\frac{\gamma m'\,dm'}{R}. \tag{1}\end{align*}

Step 2Assemble the shell from zero mass up to

m

Nothing on the right of (1) changes during the assembly except $m'$ itself, since the radius $R$ is fixed from the first increment to the last. The total energy is therefore the integral of that increment as $m'$ runs from an empty shell to the full mass $m$ :

U = -\frac{\gamma}{R}\int_0^m m'\,dm' = -\frac{\gamma}{R}\left[\frac{{m'}^2}{2}\right]_0^m .

U = -\dfrac{\gamma m^2}{2R}

(2)

Step 3(b) Build the solid sphere shell by shell

The solid sphere subdivided into concentric shells for the assembly of this step: the shell at radius r, thickness dr, is brought in onto the material already inside it. — The solid sphere subdivided into concentric shells for the assembly of this step: the shell at radius $r$ , thickness $dr$ , is brought in onto the material already inside it.

Assemble the solid sphere the same way, but now in concentric layers: at each stage a uniform ball of radius $r$ is already in place and the next layer of thickness $dr$ comes in from infinity onto it. Uniformity fixes the mass already assembled as the corresponding fraction of the total, and the mass of the next layer is the differential of that:

\begin{align*} m(r) &= m\,\frac{r^3}{R^3}, \tag{3} \\ dm &= \frac{d m(r)}{dr}\,dr = \frac{3m r^2}{R^3}\,dr. \tag{4}\end{align*}

Bringing that layer in repeats (1) with the assembled mass $m(r)$ in the role of $m'$ and the current radius $r$ in the role of $R$ :

dU = -\frac{\gamma\,m(r)\,dm}{r}. \tag{5}

Step 4Integrate over the whole sphere

Substituting (3) and (4) into (5) and letting $r$ run from the center out to the surface collects every layer of the finished sphere.

\begin{align*}U &= -\gamma\int_0^R \frac{m(r)\,dm}{r} \\ &= -\gamma\int_0^R \frac{1}{r}\cdot m\,\frac{r^3}{R^3} \cdot\frac{3m r^2}{R^3}\,dr \\ &= -\frac{3\gamma m^2}{R^6}\int_0^R r^4\,dr \\ &= -\frac{3\gamma m^2}{R^6}\left[\frac{r^5}{5}\right]_0^R = -\frac{3\gamma m^2}{R^6}\cdot\frac{R^5}{5}.\end{align*}

U = -\dfrac{3}{5}\,\dfrac{\gamma m^2}{R}

(6)

Check

The solid sphere’s energy is $\frac{6}{5}$ times the shell’s, $\frac{3/5}{1/2}=\frac65$ , not merely proportionally larger: every inner layer of the solid sphere is assembled against a radius smaller than $R$ , hence against a deeper potential per unit mass, so pulling it apart again costs more than the shell, all of whose mass sits at $R$ .

Remark

Both results scale as $\gamma m^2/R$ , the only combination of the given quantities with the units of energy; the numerical prefactor, $\frac12$ for a shell and $\frac35$ for a solid sphere, is what actually distinguishes the two mass distributions.