What kind of telescope to view Kupier objects in detail?

[ROXunreal]

Has anyone ever done the math? What diameter space telescope would be needed to visually observe celestial bodies at extreme distances, such as Eris or Sedna, and get images that resolve surface detail, or god forbid look like those of moons imaged by Voyager probes? I know it's currently completely implausible to make or bring anything of the sort to space, I'm asking hypothetically. This is the best Hubble's 2.4m mirror can do on Pluto, and this is likely with tons of post-processing:

Edited January 20, 2015 by ROXunreal

[K^2]

Depends on resolution you want to get. The ratio of the "pixel" real size on the image you get to the distance to the object is the same as ratio of light wavelength (about 0.5 microns) to the diameter of the objective lens or mirror. So if you want to be able to see 1mm from 1km away, you need a telescope with a 0.5m mirror. 0.5m / 0.5 microns = 1km / 1mm. You can do the math for all of the objects you are interested in.

There are astronomical interferometers that can give you better resolution, because for them, it's distance between telescopes in array that gives you resolving power rather than mirror diameter on any one telescope. But without optical processing, these will not have the same versatility as radio arrays.

[Sirrobert]

Is that vague roundish blueish shape Pluto?

[Moar Boosters]

The whole thing is pluto. They had to use some technical wizardry to get pictures as good as they are.

It's explained here.

http://www.nasa.gov/mission_pages/hubble/science/pluto-20100204.html

[Crimson Sunrise]

Most 'extreme long' distance telescopes are not even optical ones. Consider all the trouble they had to replace Hubble's main lens. Imagine the complex image sensors they would need and all to register objects light years away or something like that.

[Bill Phil]

none....

Seriously, that hubble shot is from orbit. What is better than a telescope that powerful that's above the atmosphere (for the most part)?

A more powerful telescope even further out!

[Hannu]

Depends on resolution you want to get. The ratio of the "pixel" real size on the image you get to the distance to the object is the same as ratio of light wavelength (about 0.5 microns) to the diameter of the objective lens or mirror. So if you want to be able to see 1mm from 1km away, you need a telescope with a 0.5m mirror. 0.5m / 0.5 microns = 1km / 1mm. You can do the math for all of the objects you are interested in.

This is diffraction limit based of fundamental properties of electromagnetic radiation. Distance to Sedna is about 90 AU. If you want to see details of 1 km you need about 7 km wide mirror. It is clearly impossible with currently known technology (including interferometric methods). It is cheaper and faster to send a probe.

[ROXunreal]

That's kind of what I was interested in knowing. I knew it was going to be a ludicrous diameter, I just wanted to know how ludicrous

[GreeningGalaxy]

I'll be curious to see what the Webb telescope can manage for Pluto and friends. Sure, clearly it's not got the mirror size you would need, but it does have that nifty cryogenic cooling system. I have no idea what that counts for, or if Pluto will even look like anything much in the infrared, but hey.

[Camacha]

Is that vague roundish blueish shape Pluto?

Blue? One of us needs to recalibrate his monitor for sure.

[-Velocity-]

That's kind of what I was interested in knowing. I knew it was going to be a ludicrous diameter, I just wanted to know how ludicrous

It's basic trigonometry. If you wanted to resolve features, say, 1 km across on Pluto, that means your telescope has a resolving power of 1x10^3 meters / 6x10^12 meters = 1.67x10^-10 radians (this is the small angle approximation for tangent that you should hopefully remember from high school).

The only other equation you need is to apply the Rayleigh resolution criterion (http://en.wikipedia.org/wiki/Angular_resolution)-

Resolution = 1.22*wavelength/telescope aperture. For visible light, 500 nm is a good middle-of-the-spectrum number for wavelength. So, it's

1.67x10^-10 = 1.22*5x10-7/d

d = 3660 meters = 3.66 km.

That's a big mirror.

[Newt]

While it would be impractical to use on objects as close as the Kuiper Belt, we do have an incredibly powerful lens rather close by: the Sun. There have been proposals for getting to the Sun's focus, so as to be able to map exoplanets in nearby star systems.

Yes. Map exoplanets.

Of course, getting to the focus would likely take a considerable length of time in itself, but no where near as long as getting to another system.

Link to basic description of FOCAL mission

Link to more recent news article