Solved thermodynamics exercise

Assembling a uniform gravitating sphere

Exercise 15 · Lesson 4 — The First Law of Thermodynamics

gravitation
interaction energy
uniform sphere
extensivity
long range

Statement

A sphere of radius $R$ contains a mass $M$ distributed uniformly, with density $ρ_{m}$ . We wish to calculate its gravitational potential energy $U_{grav}$ , taking the zero of energy when its mass elements are infinitely far apart.

Imagine assembling the sphere layer by concentric layer, keeping the same density. At the stage where the radius has reached $r$ , the mass already assembled is $M (r)$ , and a shell of mass $d M$ is brought in from infinity. We take as given that the potential produced at its surface by the inner mass is $- GM (r) / r$ . The calculation concerns only the gravitational energy; no mechanical equilibrium of the sphere is assumed.

Draw the sphere already assembled and the shell added between $r$ and $r + d r$ . Express $ρ_{m}$ , $M (r)$ and $d M$ in terms of $M, R, r$ and $d r$ .
Express the interaction energy of the shell with the inner sphere, then integrate to obtain $U_{grav}$ . Should the result be divided by two?
The sphere contains $N$ constituents of mass $m$ , with $M = N m$ . At fixed density $ρ_{m}$ , express $R$ in terms of $N$ , then determine how $U_{grav}$ and $U_{grav} / N$ depend on $N$ .
Compare the energy of a sphere of $2 N$ constituents with the sum of the energies of two spheres of $N$ constituents each, infinitely far apart, all at the same density. Comment on the sign of the difference and on extensivity.

Hint

The shell has volume

4 π r^{2} d r

. Its interaction energy with the mass already in place is

d U_{grav} = - GM (r) d M / r

. In this counting, each pair is encountered when its outer member is added.

Detailed solution

Solution

Question 1. The diagram shows a ball of radius

r

and mass

M (r)

, surrounded by a layer of thickness

d r

; the final radius

R

is marked. The uniformity of the density gives

ρ_{m} = \frac{M}{4 π R ^{3} /3} = \frac{3 M}{4 π R ^{3}}, M (r) = ρ_{m} \frac{4 π r ^{3}}{3} = M \frac{r ^{3}}{R ^{3}} .

The added mass is therefore

d M = 4 π ρ_{m} r^{2} d r = \frac{3 M}{R ^{3}} r^{2} d r .

Question 2. The gravitational potential is an energy per unit mass. Multiplying it by the mass brought in gives

d U_{grav} = - \frac{GM ( r )}{r} d M = - \frac{G}{r} \frac{M r ^{3}}{R ^{3}} \frac{3 M r ^{2}}{R ^{3}} d r = - \frac{3 G M ^{2}}{R ^{6}} r^{4} d r .

The integral runs over all assembly radii:

U_{grav} = - \frac{3 G M ^{2}}{R ^{6}} \int_{0}^{R} r^{4} d r = - \frac{3 G M ^{2}}{R ^{6}} [\frac{r ^{5}}{5}]_{0}^{R} = - \frac{3 G M ^{2}}{R ^{6}} \frac{R ^{5}}{5} = - \frac{3 G M ^{2}}{5 R} .

Each interaction between a new layer and an inner layer is counted only once, so no factor of $1/2$ should be added. The self-energy of an infinitesimal shell is of order $(d M)^{2}$ and does not contribute to the integral in the continuum limit. The negative sign reflects the attraction: separating the masses requires an input of energy. Finally, $G M^{2} / R$ does indeed have the unit of a joule.

Question 3. At fixed density,

\frac{4 π R ^{3}}{3} = \frac{N m}{ρ _{m}}, R = (\frac{3 m}{4 π ρ _{m}})^{1/3} N^{1/3} .

Substituting $M$ and $R$ into the energy,

U_{grav} = - \frac{3}{5} G m^{2} (\frac{4 π ρ _{m}}{3 m})^{1/3} \frac{N ^{2}}{N ^{1/3}} = - a_{g} N^{5/3},

where $a_{g} = \frac{3}{5} G m^{2} (4 π ρ_{m} / (3 m))^{1/3} > 0$ is independent of $N$ . The energy per constituent is therefore $U_{grav} / N = - a_{g} N^{2/3}$ . Its absolute value grows with the size of the system: it does not remain intensive.

Question 4. The large sphere has energy

U_{grav} (2 N) = - a_{g} 2^{5/3} N^{5/3}

. For the two infinitely distant spheres, the mutual interaction is zero and the sum is

2 U_{grav} (N) = - 2 a_{g} N^{5/3}

. The difference is

U_{grav} (2 N) - 2 U_{grav} (N) = - a_{g} (2^{5/3} - 2) N^{5/3} < 0.

The large sphere is more tightly gravitationally bound. This comparison of states does not by itself describe the dynamics of merging them. It shows that $U_{grav} (2 N) \neq = 2 U_{grav} (N)$ : gravitational energy is not extensive. Energy conservation is not called into question; extensivity is an additional assumption that fails for this long-range interaction.