Gravitation Potential Energy — Question 10

Question 10

Building a uniform sphere reveals its gravitational self-energy

Build a uniform sphere of final mass $M$ and radius $R$ by bringing thin shells from infinity. At assembly radius $r$, use $M(r)=M(r/R)^3$ and $dM=3Mr^2dr/R^3$. A shell contributes $dU=-GM(r)dM/r$.

1. Integrate the total self-energy.

2. If a fixed mass contracts between uniform states of radii $R$ and $R/2$, find the released gravitational energy.

Question 10 - Solution

Building a uniform sphere reveals its gravitational self-energy

(a) Derivation and calculation

A new shell interacts with the mass already assembled: $$dU=-\frac{GM(r)}r\,dM =-\frac G r\frac{Mr^3}{R^3}\frac{3Mr^2}{R^3}\,dr =-\frac{3GM^2}{R^6}r^4\,dr.$$ Integrating, $$\boxed{U=-\frac{3GM^2}{R^6}\left[\frac{r^5}5\right]_0^R =-\frac{3GM^2}{5R}}.$$

(b) Result and interpretation

For fixed mass contracting to $R_f=R/2$, $$U_i=-\frac{3GM^2}{5R},\qquad U_f=-\frac{6GM^2}{5R}.$$ $$\Delta U=-\frac{3GM^2}{5R},\qquad \boxed{E_{\mathrm{released}}=-\Delta U=\frac{3GM^2}{5R}}.$$ Energy partition among heat, radiation, and motion needs additional physics.

Physical check

The negative self-energy becomes more negative during contraction; assembly by shells counts each interaction once.