# Seeing Apollo

Why you can’t see the Apollo lunar landing sites from Earth.

Something that Apollo denialists^{1} sometimes say is: if the Apollo programme put people on the Moon, why can’t we see the landing sites?

Well, we can. In 2009, the Lunar Reconnaissance Orbiter (LRO), in orbit around the Moon, took pictures of some of the Apollo landing sites, including an astonishing picture of the Apollo 12 landing site in which you can see Surveyor 3 as well as clear signs of tracks made by the astronauts as they walked on the Moon. It is even possible to work out what happened to some of the flags & that the flag planted by the Apollo 12 astronauts is still standing. In 2011 LRO took some even better pictures: in these images it is very easy to see the tracks left by the astronauts. A summary page points to images of the landing sites of Apollo 11, 12, 14, 15 & 16.

‘But’, they say, ‘you can’t see the sites *from Earth*, and we don’t believe that LRO actually exists: it’s just part of the giant Apollo conspiracy. If people landed on the moon you would be able to see them with Earth-based telescopes^{2}’.

Here’s why you can’t.

The angular resolution of a telescope with diameter \(D\) working at a wavelength of \(\lambda\) is given by:

\[\Delta\theta = 1.220 \frac{\lambda}{D}\]

This is the smallest angle it can resolve, even in theory. The \(1.220\) comes from the properties of the Airy discs that correspond to the diffraction patterns of light — in fact it’s the position of the first zero of intensity in the pattern.

For a telescope where \(\Delta\theta \ll 1\) (which is true for anything worth being called a telescope) then we can translate this into an absolute resolution, \(\Delta l\), at a distance \(r\), using \(\sin\theta \approx \theta\) if \(\theta \ll 1\).

\[\Delta l = 1.220 \frac{\lambda r}{D}\]

And we can rearrange this to tell us what \(D\) needs to be to resolve an object of size \(\Delta l\):

\[D = 1.220 \frac{\lambda r}{\Delta l}\]

So, for the LEMs on the Moon:

- \(\Delta l \approx 9\,\mathrm{m}\) (size of LEM);
- \(r \approx 4 \times 10^8\,\mathrm{m}\) (distance to Moon);
- \(\lambda \approx 5.6 \times 10^{-7}\,\mathrm{m}\) (green light).

And this gives us \(D \approx 30\,\mathrm{m}\).

This means that to even be capable of resolving the LEM on the moon we would need a telescope with a diameter of thirty metres. This is about three times larger than the largest optical telescopes in the world. Telescopes this large are planned (and even larger ones too), but they do not exist yet.

And this size is the absolute minimum: to see any kind of detail you would need a truly enormous telescope: perhaps something with \(D\approx 100\,\mathrm{m}\): nobody is building anything like that soon. The LRO can see the Apollo sites because it is in an orbit around the Moon which gets as low as \(20\,\mathrm{km}\): twenty thousand times closer than the Earth is, which means it needs a telescope with a diameter twenty thousand times smaller.