Consider a universe filled with dust of uniform density $\rho(t)$ and consider a spherical shell of mass $m$ and radius $r(t)$. This shell expands along with the universe, always containing the same dust particles. So, $\rho(t)r(t)^3$ is constant.

The total mechanical energy of the shell is $K + U = E$. As the shell expands, the gravitational pull from the mass inside causes $K$ to decrease while the gravitational P.E. $U$ increases. But, by energy conservation $E$ is constant. We write

(1)The constant $k$ has dimensions of $(length)^{-2}$ and $\bar{\omega} = r(t_{0})$ is the present radius of the shell. Therefore,

(2)where $M_{r} = \frac{4}{3}\pi r^{3}\rho$ is the mass *interior* to the shell and $v$ is the recessional speed of the shell. Thus we have

The constant $k$ determines the overall geometry of the universe

- $k > 0$: total energy < 0, universe is
*bounded*or**closed**. The expansion will someday halt and reverse itself. - $k < 0$: total energy > 0, universe is
*unbounded*or**open**. The expansion will continue forever. - $k = 0$: total energy = 0, universe is
**flat**(neither open nor closed). The expansion will continue to*slow down*coming to a halt only as $t \rightarrow \infty$ and the universe is infinitely dispersed.