\begin{table}%t1
%%%\begin{center}

\par
\caption{\label{tble}Characteristics of a 7 $M_\odot$ star rotating at critical speed.}
\begin{tabular}{cccc} 
%\begin{tabular*}{.5\textwidth}{@{\extracolsep{\fill}}|*{4}{c|}} 
   \hline    \hline 
   Parameter & ZAMS & $X_{\rm c}=0.30$ & $X_{\rm c}=0.02$  \\ 
   \hline
  {\it No rotation} \hskip 4.5pc\strut & &  &  \\  
  radius ($R_\odot$) & 4.348 &4.953 & 5.857 \\
  quadrupolar moment $J_2$ & 3.370 $\times$ 10$^{-3}\!\!\!$& 2.766 $\times$  10$^{-3}\!\!\!\!$ & 2.013 $\times$  10$^{-3}\!\!\!\!$ \\    
       \hline  
   {\it     Uniform rotation} \hskip 3pc\strut & &  &  \\  
    quadrupolar moment $J_2$ & 8.411  $\times$ 10$^{-3}\!\!\!$ &7.136 $\times$  10$^{-3}\!\!\!\!$ & 5.171  $\times$ 10$^{-3}\!\!\!\!$ \\
  mean radius $R_0$ & 5.125 &5.803 & 6.852 \\   
    equatorial radius $R_{\rm E}$ & 5.821 &6.591 & 7.783 \\   
   polar radius $R_{\rm P}$ & 3.814 &4.329 & 5.133 \\  
   flattening $R_{\rm E}/R_{\rm P}$ & 1.526 &1.522 & 1.516 \\    
   \hline   
     {\it  Differ.  rotation} \quad $\Omega_{\rm c}/\Omega_s=4$ & &  &  \\  
    quadrupolar moment $J_2$ & 1.900 $\times$  10$^{-2}\!\!\!$ &1.550  $\times$ 10$^{-2}\!\!\!\!$ & 1.123  $\times$ 10$^{-2}\!\!\!\!$ \\
  mean radius $R_0$ & 5.773 &6.671 & 7.902 \\   
    equatorial radius $R_{\rm E}$ & 6.557 &7.577 & 8.846 \\   
   polar radius $R_{\rm P}$ & 4.204 &4.889 &5.846 \\  
   flattening $R_{\rm E}/R_{\rm P}$ & 1.560 & 1.550 & 1.535 \\    
   \hline 
\end{tabular}
  \tablefoot{$J_2$ is the quadrupolar moment defined in Eq.~(\ref{expansion}). $R_{\rm E}$, $R_{\rm P}$ and  $R_0$ are respectively the equatorial, polar and mean radius. Taking the quadrupolar moment into account increases the  flattening $R_{\rm E}/R_{\rm P}$ of the stellar surface beyond the value 1.50 of the Roche model, and even more so when the rotation is non-uniform (with the profile of Eq. (\ref{rot-profile})); this is illustrated here with a center-to-surface contrast of~4. In the ``no rotation'' case, the non rotating model was used as reference, while uniform rotation ($h(x)=1$) was assumed when solving the Poisson   Eq.~(\ref{diff-eq}).}
\end{table}