Wyvyn S. answered • 11/07/24
Ad Astra Per Aspera
There's a particular equation used to measure the angular resolution of a telescope. In short, the angular resolution can be thought of the "pixelation" of the optical image produced by a telescope. So, a higher angular resolution means one of those "pixels" is smaller, and the distance between two points contains more pixels to discern.
The equation is like this; X = 1.22 * W / D, where X is the angular resolution in Radians, W is the wavelength of the light you're looking for, and D is the diameter of the telescope aperture. (I don't have an answer for where 1.22 pops into the equation, but it has to do with how much power per area can be collected, and it's more accurately a limit that ratio approaches.)
This is called the Rayleigh criterion, and it asserts that not only is angular resolution determined by the light waves themselves, but also the size of the telescope. This is intuitive because usually, telescopes are explained as buckets that collect light into a smaller aperture; They make things larger and less dim.
I'll calculate the angular size of the object in kilometers first. (The minimum size telescope we need to resolve the object)
22 light years = roughly 2.0814*10^14 kilometers, object diameter = probably roughly 19,068 km (according to solarsystemquick.com) Side note - according to Wikipedia, this object in particular is currently refuted to exist, but that's not a problem here because the question is much more general.
Arc length formula: Length = Radius * Radians, Radians = Length / Radius, X = 19,068/2.0814E14,
Angular size = 9.1611*10^-11 (Or, 1 / 0.000000000091611 Radians, or a billionth of a degree. It's cool how big space is, right!?)
So, now we can put that into the Rayleigh equation, and I'll put the wavelength at 1 km (Low energy radio) for now, so that we get 9.1611*10^-11 = 1.22 * 1/X, rearrange to X = 1.22/9.1611*10^-11
and so, the diameter, in kilometers of the telescope for light with a wavelength of 1 km is equivalent to 1.1177E12 Km, or over a trillion kilometers, or about 1/10th of a light year.
Ok, that's not as bad as it sounds; for higher energy wavelengths of light (I.E. shorter) the necessary diameter drops proportionaly. We can see this in that, if we held the angular size of the object constant, then to increase wavelength, we HAVE to decrease diameter to maintain the equality.
So, if we go to a very much higher energy light, like one in the 1/1000 km wavelength, (microwave, about) the diameter equivalence drops to "just" a billion kilometers, about less than but on the order of 10 astronomical units. Dropping the wavelength by another 1/1000th brings it to 1 million km, (same order as 10 Lunar distances), and so on.
To answer the other facet of this question, the wavelength obviously changes the situation considerably. Even more than the order of magnitude of telescope sizes, it also matters in the kind of light the object produces, and what kind of stuff is in the way of us and it (for example, radio is preferred because it passes through dust much better than other wavelengths), and even moreso how much interference the light of the star increases. I also haven't even taken into account the want for true detail, which, just to have a pixel equal to about 1 km on the surface requires a telescope 10,000 times stronger.
In essence, unfortunately this question really depends on what kind of astronomy you want. Although, it's not out of the question - I said equivalence a bunch, and that just means that certain kinds of telescopes are more powerful than other kinds, like a reflecting telescope is stronger than a refracting telescope, and interferometry (which was used to image a black hole, of course) is much stronger than the other two. Which, by the way, Sagitarius A*'s angular size is on the same order of magnitude as Gliese 581g.
