A space station consists of two living modules attached to a central hub on opposite sides of the hub by long corridors of equal length. Each living module contains $N$ astronauts of equal mass. The mass of the space station is negligible compared to the mass of the astronauts, and the size of the central hub and living modules is negligible compared to the length of the corridors. At the beginning of the day, the space station is rotating so that the astronauts feel as if they are in a gravitational field of strength $g$. Two astronauts, one from each module, climb into the central hub, and the remaining astronauts now feel a gravitational field of strength $g'$ . What is the ratio $g'/g$ in terms of $N$?[asy]
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path a = roundedpath((0,-0.3)--(4,-0.3)--(4,-1)--(5,-1)--(5,0),0.1);
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draw(scale(-1,-1)*a);
draw(scale(-1,+1)*a);
filldraw(circle((0,0),1),white,black);
filldraw(box((-2,-0.27),(2,0.27)),white,white);
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draw(arc((0,0),1.5,-150,-35),dashed,Arrow);[/asy]

$ \textbf{(A)}\ 2N/(N-1) $

$ \textbf{(B)}\ N/(N-1) $

$ \textbf{(C)}\ \sqrt{(N-1)/N} $

$ \textbf{(D)}\ \sqrt{N/(N-1)} $

$ \textbf{(E)}\ \text{none of the above} $
1. **Understanding the problem**: The space station is rotating to simulate gravity through centripetal force. Initially, each module has \(N\) astronauts, and the centripetal acceleration felt by the astronauts is \(g\). When two astronauts move to the central hub, the remaining astronauts feel a new centripetal acceleration \(g'\). We need to find the ratio \(\frac{g'}{g}\) in terms of \(N\).

2. **Centripetal acceleration**: The centripetal acceleration \(a_c\) is given by:
   \[
   a_c = \frac{v^2}{L}
   \]
   where \(v\) is the tangential velocity and \(L\) is the length of the corridor.

3. **Initial and final centripetal accelerations**: Let \(v_1\) be the initial tangential velocity and \(v_2\) be the final tangential velocity. Then:
   \[
   g = \frac{v_1^2}{L} \quad \text{and} \quad g' = \frac{v_2^2}{L}
   \]
   Therefore, the ratio of the accelerations is:
   \[
   \frac{g'}{g} = \frac{\frac{v_2^2}{L}}{\frac{v_1^2}{L}} = \frac{v_2^2}{v_1^2}
   \]

4. **Relating tangential velocity to angular velocity**: The tangential velocity \(v\) is related to the angular velocity \(\omega\) by:
   \[
   v = L\omega
   \]
   Thus:
   \[
   \frac{g'}{g} = \frac{(L\omega_2)^2}{(L\omega_1)^2} = \left(\frac{\omega_2}{\omega_1}\right)^2
   \]

5. **Conservation of angular momentum**: Since no external torques act on the system, the angular momentum is conserved. Let \(I_1\) and \(I_2\) be the moments of inertia before and after the astronauts move. Then:
   \[
   I_1\omega_1 = I_2\omega_2
   \]
   Therefore:
   \[
   \frac{\omega_2}{\omega_1} = \frac{I_1}{I_2}
   \]

6. **Calculating moments of inertia**: Initially, each module has \(N\) astronauts, so the total moment of inertia is:
   \[
   I_1 = 2 \times (mN L^2) = 2mNL^2
   \]
   After two astronauts move to the hub, each module has \(N-1\) astronauts, so the new moment of inertia is:
   \[
   I_2 = 2 \times (m(N-1) L^2) = 2m(N-1)L^2
   \]

7. **Substituting moments of inertia**:
   \[
   \frac{\omega_2}{\omega_1} = \frac{I_1}{I_2} = \frac{2mNL^2}{2m(N-1)L^2} = \frac{N}{N-1}
   \]
   Therefore:
   \[
   \frac{g'}{g} = \left(\frac{\omega_2}{\omega_1}\right)^2 = \left(\frac{N}{N-1}\right)^2
   \]

8. **Conclusion**:
   \[
   \boxed{\left(\frac{N}{N-1}\right)^2}
   \]