I answered this question for someone the other day, whose work colleagues wanted to know if he could see the flag left behind on the moon from the first lunar landing using his new 300mm telescope and thought I’d write the answer up properly.

The answer is no, you can’t. Not even close. And here’s why. The explanation involves a small amount of maths, but it’s really not that scary.

The scientist John Strutt, Baron Rayleigh, showed in the late 1800s that the angular resolution of a telescope can be calculated as:

R = λ / d

where R is the resolution in radians, λ is the wavelength of light in metres and d is the diameter of the telescope, also in metres. Simple enough so far. Now it has to temporarily get a bit more complicated, but we’ll sweep some of that under the carpet in a moment. Ignore the fact that we’re measuring angles in radians too, as we’ll treat that similarly when the time arises.

Taking the flag example, imagine the light from opposite corners of the flag which, for the sake of argument we’ll assume are a metre apart, travels all the way from the moon down to the earth and into your telescope. To be able to resolve the flag from the rest of the lunar landscape the angle separating those two beams of light must be at least the angular resolution of the telescope. If we calculate the angle between the beams of light, which we can do as we know the distance to the moon and we’ve set the size of the flag, then from Rayleigh’s equation we can work out what size telescope we’d need.

A purist might suggest at this point that we should assume the distance to the moon is the line from your eye behind the telescope to the middle of the flag and that to calculate the angle we’re interested in we should construct a notional right-angled triangle with the right-angled corner at the centre of the flag, one remaining corner at one corner of the flag and the third at your eye and calculate the angle at the eye end, doubling it to account for the angle between the corners of the flag. In practice, the angle is going to be so small that it really won’t matter. In terms of the maths,

tan ( R/2 ) = w / ( 2 x D )

where w is the width of the flag and D is the distance to the moon, but where D is far bigger than w we can closely approximate this as

tan R = w / D

We can also make another approximation, as when we’re dealing with very small angles (again when D is very much larger than w), tan R = R, so we get:

R = w / D

and substituting R from the Rayleigh equation we get

λ / d = w / D

and now all the awkward maths stuff has disappeared and we’re left just with multiplication and division.

We’re going to need to come up with a value for the wavelength of light at some point. Humans can see light of around 400nm to 750nm, so let’s take an average value of 575nm, or 5.75 x 10^{-7}m. Rearranging the above equation to calculate the size of telescope we’d need to see the flag we get:

d = λD / w

We’ve said w is 1m and D, the distance to the moon, is 356,400km or 3.564 x 10^{8}m at its closest. So that gives us:

d = 5.75 x 10^{-7} x 3.564 x 10^{8} / 1

which is 204.93m. So, to be able to see the flag (assuming it was 1m wide) from the first moon landing from Earth you’d need a telescope more than 200m in diameter.

If we set ourselves an easier target, say, to see the bottom of the lander module which was about 9.5m diameter, we’d have:

d = 5.75 x 10^{-7} x 3.564 x 10^{8} / 9.5

which is just over 21.5m, so even to see the lander module you’d still need a 21m+ telescope.

Let’s turn things around. Given a 300mm (0.3m) telescope, what is the biggest thing that you **could** resolve on the moon. This is given by:

w = λD / d

or

w = 5.75 x 10^{-7} x 3.564 x 10^{8} / 0.3

which works out as 683.1 metres. If you have a 100mm refractor then it’s closer to 2km.

So what about the Hubble Space Telescope (HST)? That can see some amazing stuff, can’t it? Well, yes it can. But not stuff “we” left on the moon. The HST isn’t that big. It had to fit in a Shuttle payload bay. In fact the primary optical element is only 2.4m in diameter. The HST orbits at a height of about 560km, but obviously that sometimes takes it closer to the moon and sometimes further away. At it’s closest point it’s going to be 3.5584 x 10^{8}m from the moon, so the smallest thing it can resolve is:

w = 5.75 x 10^{-7} x 3.5584 x 10^{8} / 2.4

which is about 85.25m, so still not even close to resolving either the flag or the lander module.

So, the only way we can see things left behind by the moon missions are through photographs taken from the lunar orbiter missions which are far closer. In those it’s possible to see which way the wheels on the abandoned lunar rover are turned and even people’s footprints.

On other thing that has come out of the maths for this problem is that as a rough first approximation, the smallest thing that can be resolved on the moon is given by:

w = 200 / d

Unfortunately that isn’t the end of the story though, as atmospheric conditions will make life even harder by causing distortion of the view. Life is never easy.