How large must an object be to be seen through a telescope?

A few clarifications.

Telescopes in general operate at very large distances - "at infinity" is the term used in optics parlance.

A bright enough object can be seen from any distance, no matter what its size is. All that matters is that:

1. It's bright enough to produce an impression on whatever sensor
   you're using (or your eye)




2. The background is dark enough to produce sufficient contrast

But then it would be just a bright but tiny spot.

I believe what you're really asking for is: what is the combination of factors that shows the object as bigger than a simple dot? In that case, it's two factors: aperture of the telescope, and angular size of the object.

Assuming a flawless telescope, its aperture is what determines its resolving power. The resolving power is the angle at which two dots can be separated by the telescope. The formula is:

resolving power = 1 / (10 * aperture)

where resolving power is in arcseconds, and aperture is in meters. Examples:

aperture         resolving power
10 cm            1 arcsec
20 cm            0.5 arcsec
1 m              0.1 arcsec

As long as the object's angular size is bigger than the resolving power, it will appear bigger than a dot.

That's all. In astronomy, we don't speak of an object's absolute size, we only speak of the angular size. But that should be enough. As soon as you have the angular size, and say the distance, then you could deduce the absolute size, it's a simple matter of trigonometry.

absolute size = distance * tangent(angular size)

E.g., this is the size of an object of 1 arcsec angular size, situated at 384,000 km (the orbit of the Moon):

http://www.wolframalpha.com/input/?i=384000+km+*+tangent%281+arcsec%29

It's 1.8 km (in case the link above is unavailable).

In other words, that's the minimum absolute distance resolved by a telescope 10 cm in aperture, for objects on the Moon. Any two dots closer together than 1.8 km, placed on the Moon, are seen as one dot in a 10 cm telescope.