Lower limit of the size of the Universe? (WMAP) The measurement of the WMAP satellite resulted a planar geometry of the universe with a 0.4% uncertainity (http://en.wikipedia.org/wiki/Shape_of_the_universe).
If there is a little deviation from the measured zero curvature, I think it could give a lower limit to the size of the universe (in case of positive curveture and spherical geometry). How big is it? How could it be calculated?
 A: From the Friedmann equations, you can derive that
$$
\dot{R}^2 - \frac{8\pi}{3}G\rho R^2 = -k c^2,
$$
where $\rho$ is the total density of the universe and $k$ is a constant that determines the shape of the universe: $k=-1,0,1$ for an open, flat and closed universe, respectively. If the universe is a hypersphere ($k=1$), then $R$ can be thought of as its 'radius'.
Because the right-hand side is a constant, it is also equal to the present-day values
$$
\dot{R}_0^2 - \frac{8\pi}{3}G\rho_0 R_0^2 = -k c^2,
$$
or
$$
\frac{\dot{R}_0^2}{R_0^2} - \frac{8\pi}{3}G\rho_0 = -\frac{kc^2}{R_0^2},
$$
and, introducing the Hubble constant $H_0=\dot{R}_0/R_0$, we get
$$
H_0^2 - \frac{8\pi}{3}G\rho_0 = -\frac{kc^2}{R_0^2}.
$$
If $k=0$, we have a flat universe, and the corresponding density equals the so-called critical density 
$$
\rho_{c,0} = \frac{3H_0^2}{8\pi G}.
$$
The general case can thus be written in the form
$$
H_0^2\left(1 - \frac{\rho_0}{\rho_{c,0}}\right) = -\frac{kc^2}{R_0^2}.
$$
Finally, the factor between brackets is denoted as $\Omega_{K,0}$, so that
$$
H_0^2\,\Omega_{K,0} = -\frac{kc^2}{R_0^2}.
$$
In case of a universe with positive curvature, $k=1$ and $\Omega_{K,0}$ is negative, so that
$$
R_0 = \frac{c}{H_0\sqrt{-\Omega_{K,0}}}.
$$
The nine-year WMAP value for $\Omega_{K,0}$ is (see the last table on the wiki page)
$$
\begin{align}
\Omega_{K,0} &= -0.037^{+0.044}_{-0.042}\qquad&&\text{(WMAP only)},\\
&= - 0.0027^{+0.0039}_{-0.0038}\qquad&&\text{(WMAP + other obs.)},
\end{align} 
$$
and $H_0 = 70\;\text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}$. So we find
$$
\begin{align}
R_0 &\approx 22.3\;\text{Gpc}\approx 72.7\;\text{billion ly}\qquad&&(\text{for $\Omega_{K,0} = -0.037$}),\\
R_0&\approx 82.5\;\text{Gpc}\approx 269\;\text{billion ly}\qquad&&(\text{for $\Omega_{K,0} = -0.0027$}).
\end{align} 
$$
This can be interpreted as the radius of the universe if it is a hypersphere, although the topology of the universe could be more complicated. The latest Plank results put even tighter constraints on the curvature of the universe (see page 40 in this paper).
