How big of a telescope to view Gliese 581g in great detail? I've always been interested in mega sized telescopes and how big they would have to be to see exoplanets in great detail. So, Gliese 581g is 22 light years away. Could we build a telescope in space that could see Gliese 581 g with the same detail as we can see Earth from the moon? Or would there be too much interference?
And how would it scale? How much bigger would the telescope have to be if we doubled the distance to 40 light years? 60? 100?
 A: First let's quantify what kind of resolution we have of Earth from the moon?  This can be calculated.  The distance (range) from the Earth to the moon is is $R_\text{EM} \approx384,400,000$ meters.  The angular resolution of the human eye is $\theta_\text{eye}\approx.07^o \approx .0012 \text{ radians}$.  The spatial resolution of the earth viewed by the naked eye on the moon is:
$$ \text{Ground Sample Distance (GSD)} = R_\text{EM} \theta_\text{eye} = 460 \text{km}$$
We can then back out the necessary telescope size to achieve the same resolution for observing an exo-planet.  The planet you mentioned is 22 light years away:
$$ R_\text{G581} \approx 22 \text{ light years} \times \frac{9.5 \times 10^{15}\text{ meters}}{1\text{ light year}} \approx 2\times 10^{17} \text{ meters}$$
Suppose we want the same resolution: 460 km.   This requires an angular resolution of:
$$\theta \approx \frac{\text{GSD}}{R_\text{EM}} = \frac{460000}{2\times 10^{17}} \approx 2.3\times 10^{-12} \text{ Radians}$$
If you have a diffraction limited telescope, operating at wavelength $\lambda = 500 \text{ nm}$ (visible), and the entrance aperture diameter $D$, the angular resolution is given by:
$$ \sin\theta\approx\theta \approx \frac{1.22 \lambda}{D}= \frac{\text{GSD}}{R_\text{G581}}$$
Solving for entrance aperture, you have
$$ D = \frac{1.22 \lambda}{\theta} = \frac{1.22\lambda R_\text{G581}}{\text{GSD}} \approx \frac{1.22\times 500\times 10^{-9}\text{ m}\times 2\times 10^{17}\text{ m}}{460\times 10^3\text{ m}} \approx 2.6\times 10^5\text{ meters} $$
So your objective lens would be nominally 260 kilometers in diameter.
The objective size in this case is directly proportional to the distance to your exoplanet.  Doubling the distance would result in doubling the necessary entrance aperture diameter.
This all assumes that you have a perfect optical system (diffraction limited) and you are only limited by the laws of physics.  
