
Ask HN: What's the limits of planetary imaging? - forgottenacc56
Is there a limit for some reason of the optical pictures that we could get of planets in other systems?<p>What&#x27;s the best theoretical direct image picture possible of a planet in another system?
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dalke
FWIW,
[https://en.wikipedia.org/wiki/List_of_directly_imaged_exopla...](https://en.wikipedia.org/wiki/List_of_directly_imaged_exoplanets)
is a list of directly imaged exoplanets. These all appear as point sources. I
assume you want to be able to see something the size of a continent on a
planet?

There are different limits depending on your choice of technology. One is the
angular resolution.
[https://en.wikipedia.org/wiki/Angular_resolution](https://en.wikipedia.org/wiki/Angular_resolution)
. The wider the telescope diameter, the better the resolution.

Here's the best that Hubble could see of Pluto -
[http://www.nasa.gov/mission_pages/hubble/science/pluto-20100...](http://www.nasa.gov/mission_pages/hubble/science/pluto-20100204.html)
. A planet the size of Earth could be resolved to the same detail if it were
5.5x the distance as Pluto, or about 200-250 AU, which is about 30-35 light
hours away.

The nearest star system is light years away, so Hubble won't work.

It's possible to make a larger telescope. The key realization is that "D" in
the equation doesn't mean the entire telescope must be that far apart, only
that the two most distant points are D. This is called
[https://en.wikipedia.org/wiki/Interferometry](https://en.wikipedia.org/wiki/Interferometry)
. The Cambridge Optical Aperture Synthesis Telescope, for example, "was the
first long-baseline interferometer to obtain high-resolution images of the
surfaces of stars other than our sun."
[https://en.wikipedia.org/wiki/Cambridge_Optical_Aperture_Syn...](https://en.wikipedia.org/wiki/Cambridge_Optical_Aperture_Synthesis_Telescope)

Yellow light has a λ=580 nm wavelength. If your target resolution is 1000km
(the Earth is 6,371 km across, so a 6x6 image) at Proxima Centauri some 4.24
light years away, then:

    
    
      θ = 1000/4E13 = 1/4E10
      D = 1.22 λ/θ = 1.22*580*40 m = 28 km across
    

COAST, above, is 100 meters across.

This is the limit due to diffraction. The atmosphere makes it worse, so you
probably want this in space. I believe it's also a difficult engineering
problem, as the wavelengths needs to be in phase - it's not simply a matter of
having telescopes on both sides of the planet and combining the results.

There are other methods, like gravitational lensing. I don't know enough to
say anything about it.

------
PaulHoule
I don't know what the maximum resolution is, but one of the game changing
discoveries in SETI is that a probe can be send out 500 Au or so to use the
sun's gravitational lens. The efficiency of this is so high that it really is
plausible that somebody could pick up our radio signals.

