15
$\begingroup$

What is the farthest object which we can get a direct Detailed visual image of using visible light which appears more than just a dot and falls into one of the following categories:

  • Planet
  • Satellite
  • Star
  • Asteroids

I think Pluto is the farthest we've imaged visually using New Horizons.

Can the Hubble telescope take detailed images of, say, a star ?

$\endgroup$
5
  • 2
    $\begingroup$ Neither Voyager probe got anywhere near Pluto. Pluto was recently imaged by New Horizons. Voyager 2 imaged Uranus and Neptune. Voyager 1 only went to Jupiter and Saturn before leaving the plane of the ecliptic. Depending on what you mean by "imaging", we have now visually detected thousands of exoplanets. $\endgroup$
    – CuriousOne
    Commented Jun 9, 2016 at 5:39
  • $\begingroup$ What do you mean by direct visual image? $\endgroup$
    – ProfRob
    Commented Jun 9, 2016 at 9:52
  • 10
    $\begingroup$ Star? I see plenty of stars at night, and they are far... Shouldn't you remove "star" from your list? Or did I miss anything? $\endgroup$ Commented Jun 9, 2016 at 9:56
  • $\begingroup$ @OlivierGrégoire my interpretation (and John's as well) was that the image should be resolved, rather than point like. Stars are bright, but they are also small (well, compared to how far away they are). $\endgroup$
    – Kyle Oman
    Commented Jun 9, 2016 at 15:12
  • 1
    $\begingroup$ Just to give some perspective, Pluto has a larger angular size than any distant star. Here is the best image of Pluto Hubble could get -- basically a few pixels across, and it's been heavily modified to make it look like something other than random noise. Before New Horizons got pictures like this, there did not exist any better pictures of Pluto. $\endgroup$
    – user10851
    Commented Jun 9, 2016 at 23:49

2 Answers 2

23
$\begingroup$

To address your last point, there are several stars of which we have been able to resolve images i.e. see the star as more than just a featureless point. There is a list of these stars on Wikipedia (I love that they put the Sun at the top of the list - true but pedantic :-).

The farthest away of the stars in the list is Epsilon Aurigae at about 2000 light years, so this probably answers the main point in your question.

However there is some ambiguity in your phrase direct visual image. We can detect supernovae in distant galaxies, though they cannot be resolved and appear as a featureless point. I'm guessing you mean to exclude objects like this, in which case Epsilon Aurigae holds the crown.

$\endgroup$
7
  • 1
    $\begingroup$ I was a bit surprised to find that $\epsilon$-Aurigae is basically at the theoretical limit for modern equipment! $\endgroup$
    – Kyle Oman
    Commented Jun 9, 2016 at 8:29
  • $\begingroup$ Cool video showing actual pictures of the star. $\endgroup$ Commented Jun 9, 2016 at 8:32
  • $\begingroup$ Hm, I wonder if this was an optical interferometry image (LBT, Keck or VLT maybe?). That looks like impressive detail given the distance, size and available single-mirror equipment. $\endgroup$
    – Kyle Oman
    Commented Jun 9, 2016 at 8:37
  • 3
    $\begingroup$ But this is an interferometric reconstruction in infrared. Surely not a "direct visual image". $\endgroup$
    – ProfRob
    Commented Jun 9, 2016 at 10:02
  • 1
    $\begingroup$ @JanDvorak: see the Wikipedia article. The bottom line is that no-one knows. $\endgroup$ Commented Jun 10, 2016 at 4:30
19
$\begingroup$

I'll add a theoretical limit to the actual record put forward by John Rennie.

To image an object as more than a featureless "point source", it must be resolved by the telescope. The angular resolution $\theta$ of a telescope is:

$$\theta\sim1.22\frac{\lambda}{D_{\rm aperture}}$$

$\lambda$ is the wavelength of light, $D_{\rm aperture}$ is the diameter of the telescope. Smaller angular resolution is better. The angular size of a distant object is $\theta=\frac{L}{D}$, where $L$ is the size of the object (e.g. diameter) and $D$ is the distance. Putting this together and solving for distance:

$$D = \frac{LD_{\rm aperture}}{1.22\lambda}$$

The current largest optical telescopes are $\sim 10\,{\rm m}$ in diameter. For best results, we want to be observing at the blue end of the visible spectrum at about $400\,{\rm nm}$. This just leaves the size of the source. From your list, the largest object would be a star (conveniently also the brightest, so it's easier to see at large distances). The largest known star is UY Scuti at a whopping $1700\,{\rm R}_\odot = 1.2\times10^9\,{\rm km}$. This would give a maximum distance for current telescopes of $D=2600\,{\rm ly}$. This lines up nicely with John Rennie's figure. The maximum I give is to just barely resolve the star (e.g. 2 pixels), so a bit closer would give a properly resolved image. To do better requires a bigger telescope, the next generation of optical telescopes will be $\sim 30\,{\rm m}$ in diameter, so they could get to triple the distance. You could also try to reduce the wavelength, but UV and X-ray observations must be done from space, so the telescopes are necessarily smaller and in the end you're better off from the ground in the optical.

If you stretch the definition of star a bit to include planetary nebulae (note: nothing to do with planets!), these have been observed out to $20\,{\rm Mpc}$, about $20\,000$ times more distant than Epsilon Aurigae. I wasn't able to confirm whether these were resolved observations, but their diameters are typically $\sim 1\,{\rm ly}$, which is about $10\,000$ times larger than UY Scuti, so resolving a big one at those distances is plausible.

$\endgroup$
4
  • 2
    $\begingroup$ On a side note, where does the 1.22 constant come from? $\endgroup$
    – Carlos
    Commented Jun 9, 2016 at 9:24
  • $\begingroup$ The largest optical telescopes are 10 minutes diameter? $\endgroup$ Commented Jun 9, 2016 at 9:51
  • 1
    $\begingroup$ @Carlos there's a concise explanation here: en.wikipedia.org/wiki/Angular_resolution $\endgroup$
    – Kyle Oman
    Commented Jun 9, 2016 at 10:41
  • $\begingroup$ @immibis 10 metres. $\endgroup$
    – Kyle Oman
    Commented Jun 9, 2016 at 10:41

Not the answer you're looking for? Browse other questions tagged or ask your own question.