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/sci/ - Science & Math

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Anonymous Thu, Jun 13, 2013 09:16:48 No.5831492 [Reply] [Original]

Hey /sci/ a guy at work and I were talking about telescopes. I own a decent one that can see the planets and most Messier object. We were discussing why someone hasn't invented a telescope big/good enough that can see onto other planets in distant galaxies to see if there is life on those planets. How much do you believe one of these would cost and how big would it have to be?

>>	Anonymous Thu, Jun 13, 2013 09:19:02 No.5831496 >>5831492 About as large as the earth itself. Good luck finding one, OP.

Anonymous Thu, Jun 13, 2013 09:31:32 No.5831512

Bigger than earth. it has to satisfy the rayleigh criterion(look it up if you have any questions about it), which makes it so sin(theta)=1.22*(wavelength of light)/(diameter of telescope)
=>diameter=1.22*wavelength/sin(theta)
so say you wanted resolve something 5 meters long on planet Gliese 667Cc,which is 22 lightyears away, with a wavelength of 400nm(violet light)

Diameter=1.22*(400*10^-9)/(5/(22*9.46*10^15))=20 billion meters

earth's diameter is only 12.7 million meters

>>	Anonymous Thu, Jun 13, 2013 09:37:17 No.5831516 >>5831512 Holy hell.... well then I don't see anyone building a telescope that can do this for a LONG time. Thanks for the answer.

>>	Anonymous Thu, Jun 13, 2013 09:39:01 No.5831518 I'm speechless. Leave.

>>	Anonymous Thu, Jun 13, 2013 10:00:01 No.5831543 >>5831512 Bravo. Straightforward answer, knows what he's talking about, shows his work, doesn't insult. This is why people come to /sci/

Anonymous Thu, Jun 13, 2013 10:41:17 No.5831597

>>5831512
Nice.

>>5831492
Just curious, why planets in other galaxies? There are plenty in our galaxy, none of which (to my knowledge) have been directly imaged beyond a shitty 9x9 pixel quality. There's a more attainable goal for you that will be just as interesting and won't require a 20 billion meter telescope. But it's still far beyond our abilities at the moment.

>>	Anonymous Thu, Jun 13, 2013 10:43:47 No.5831599 >>5831597 lel nevermind, I didn't notice that >>5831512 was talking about a nearby planet in our galaxy.

Anonymous Thu, Jun 13, 2013 16:44:08 No.5832257

>>5831597
would be nice if we had 9*9 pixel quality.
unfortunately they're still effectively unresolvable (just as their hosting stars).

they seem to have a size > 1px on the images, cuz of natural pixel spreading proportional to the light flux in your CCD-camera.
And i think there was also some natural object noise contributing to the point-spread-function.
Not sure though and too lazy to look up

>>	Anonymous Thu, Jun 13, 2013 23:54:57 No.5832944 >>5831492 Last I heard, we don't have a visual telescope that is powerful enough to spot the Lunar Lander on the Moon. (Hubble might be, not sure.)

>>	Anonymous Thu, Jun 13, 2013 23:58:26 No.5832948 >>5831512 I have heard that you can effectively link several telescopes together and then do a lot of complicated math to effectively turn them into a single telescope with a very large diameter. Could this be useful in this situation?

>>	Anonymous Fri, Jun 14, 2013 00:12:42 No.5832961 >>5831512 >(5/(229.4610^15)) I don't understand how you substituted that for (sin x) in the equation.

>>	Anonymous Fri, Jun 14, 2013 01:01:12 No.5833013 >>5832948 You're thinking of a radio telescope

Anonymous Fri, Jun 14, 2013 01:08:27 No.5833022

>>5832948
With optical telescopes (visible light) it's not math it's physically linking telescopes (interferometry) with incredible precision. For the given example you would need to control the separation of the telescopes to one part in 10^18, that's 10 manometers in 10 billion meters.
Aperture synthesis could be used, but it wouldn't be easy. You cannot just have two telescopes a billion meters apart, you need to sample a multitude of separations at different angles. So lots of telescopes. By linking them you don't get a huge telescope, you get a telescope with a very high resolving power, if it has insufficient area you will see nothing. For resolving features at very small scales on a planet you will need a lot of collecting area, so lots of big telescopes linked together by a hugely complicated system of mirrors. Because interferometrs are have terrible sensitivity you would need much more area to get as much signal to noise as a filled aperture telescope.
It's not happening any time soon.

>>	Anonymous Fri, Jun 14, 2013 01:09:27 No.5833023 >>5833013 You can do it at other wavelengths, it's just harder.

>>	Anonymous Fri, Jun 14, 2013 01:10:13 No.5833024 >>5832961 Small angle approximation.

>>	Anonymous Fri, Jun 14, 2013 01:10:27 No.5833026 >>5833022 What about, say, some retarded number of Hubble-esque telescopes at various orbits away from the sun? So a whole load at Mercury's orbit, then more at Venus's, etc, etc.

>>	Anonymous Fri, Jun 14, 2013 01:19:26 No.5833044 >>5833024 The small angle approximation doesn't explain how or what reasoning you used to replace (sin x) with (size of object / distance of object).

Anonymous Fri, Jun 14, 2013 01:24:28 No.5833051

>>5833026
It would be totally impractical. An interferometer that large would have obscene resolving power as a result it would need incredible area to see anything. You would need far too many. The light from each telescope needs to be combined with the light from every other telescope, eventually there are no more photons and you cannot see anything. That will be the case here.

Your interferometer needs to be sensible, you can't have too many telescopes or you simply cannot combine the light. You cannot have to large an interferometer or you have no photons.

Interferometers have sensible uses but resolving tiny scales on exoplanets isn't happening. They have intersting uses in not resolving planets but allowing earth like planets to be spectrally analyed to the point we could find life, even this is decades away.

>>	Anonymous Fri, Jun 14, 2013 01:25:41 No.5833054 File: 858 KB, 240x228, ergh nigga stop that.gif [View same] [iqdb] [saucenao] [google] >>5831543 >This is why people come to /sci/

Anonymous Fri, Jun 14, 2013 01:31:12 No.5833059

>>5833044
Additionally, applying that formula to several other measurements that we know the result of seems to give nonsensical answers.
For example can view saturn with a telescope with
->1.22*(555*10^-9)/(116464000/(1,500,000,000))
gives a 'diameter' (of the lens?) of 8.72x10^-7m, which makes no sense.

I'm not sure if I misunderstand or you don't know what you're doing. All of the wiki results on Rayleigh criterion work using angular resolution, not distance as you did.

>>	Anonymous Fri, Jun 14, 2013 04:44:35 No.5833284 >>5833044 size of object / distance of object=tan(angle)~=sin(angle) It does explain it. >>5833059 Your numbers are nonsense, that's why you're getting nonsense answers. You've got size>distance. You can replace angles with the substitution above.

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