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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?

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BTW, it's not clear that Gliese 581g even exists. When it comes to exoplanets, the more "habitable" they are, the greater the chance their very existence is a false positive, given that astronomers too can be over-enthusiastic at times. –  Chris White Jun 16 '13 at 15:49
    
I think that this would be more appropriate as a comment, but unfortunately I can't comment on questions yet. Regardless, here is the link to a paper refuting the existence of planets 581f and 581g that might be interesting. arxiv.org/abs/1109.2505 –  Sigma Jun 16 '13 at 20:35
    
Also, here is a link to the original paper claiming the discovery of planets 581f and 581g aanda.org/articles/aa/abs/2007/27/aa7612-07/aa7612-07.html –  Sigma Jun 16 '13 at 20:39

1 Answer 1

up vote 10 down vote accepted

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.

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To put this in a different perspective, that's about 160 miles in diameter, which if placed on Massachusetts would be a dish extending from Boston to the New York border. Get your grant application in now! –  Olin Lathrop Jun 16 '13 at 22:17
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with space-based interferometry, such narrow optical resolution is possible, by deploying many smaller telescopes and have then interfere in a single imager. Still very constly, but this was the plan of the Terrestrial Planet Finder (en.wikipedia.org/wiki/Terrestrial_Planet_Finder) before it was ditched so the US could 'bail out' the bank barons and patricians that elevated private gains and public losses to an state-sanctioned form of swindle –  lurscher Jun 16 '13 at 22:36

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