[mattlondon]

I'll bite.

In those pictures of neptune, what is the KM-per-pixel were looking at?

Is there a minimum focal length on this? Purely hypothetical: Could we basically see astronaut's footprints on the moon with this? What about looking into the window of the ISS?

[eisstrom]

In this narrow-field mode of MUSE, the CCD detector can resolve 0.025 arcseconds per pixel (arcsecond is a weird unit for angles used in astronomy). At the current distance to Neptune (according to wolframalpha: about 30 au = 4.5 bn km), this corresponds to about 500 km/px. Due to observing conditions, I think the real resolution was more like 0.07...0.08 arcseconds, so maybe it was 1000 to 2000 km/px.

I'm not sure if the focal length plays any role here. The resolution is usually limited by the telescope size (true for all telescopes, scales with 1/diameter) and atmospheric conditions (only relevant for ground based ones). At the distance of the moon (300,000 km), the physical resolution is 36 m/px and for the ISS (400 km) it is 5 cm/px.

If you want to play around with it, here's the formula: length_still_resolved = angular_resolution * distance

The angular resolution is 1.2 * 10^-7 (= 0.025 arcseconds converted to radian), distance and length_still_resolved have the same units.

[m-app]

From the FAQ: http://www.eso.org/public/about-eso/faq/faq-vlt-paranal/#18

> Q: Could the VLT take a picture of the Moon-landing sites?

> A: Yes, but the images would not be detailed enough to show the equipment left behind by the astronauts. Using its adaptive optics system, the VLT has already taken one of the sharpest ever images of the lunar surface as seen from Earth: http://www.eso.org/public/news/eso0222/. However, the smallest details visible in this image are still about one hundred metres on the surface of the Moon, while the parts of the lunar modules which are left on the Moon are less than 10 metres in size. A telescope 200 metres in diameter would be needed to show them. [continued]