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When I look at the full Moon through my 10cm telescope, it is so bright that it hurts. Can large scientific telescopes observe the moon at all? Does that require special protection equipment? Or dedicated telescopes (or none at all)?

In particular, the E-ELT (European Extremely Large Telescope) will have a mirror with a 39m diameter. If it were pointed at the full moon, would that damage the science instruments? Would that generate a significant temperature at the focal point?

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    $\begingroup$ Obligatory xkcd reference: what-if.xkcd.com/145 $\endgroup$
    – vsz
    Aug 30, 2021 at 16:56
  • $\begingroup$ Obligatory and very frustrating, @vsz! $\endgroup$
    – AnoE
    Sep 1, 2021 at 8:57

3 Answers 3

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No chance of damage (the giantness of the telescope makes little difference!), but these cameras can not shoot at a shutter speed of 1/1000 second, so the lit part of the Moon is out of reach due to overexposure.

Can large scientific telescopes observe the moon...

The bright part will probably be too bright to easily image with a deep field camera because it's designed for integration times of seconds to minutes.

The hardware won't be able to provide a 1/1000 second exposure, so only objects in shadows (or the unlit side of the Moon) have a chance of being exposed.

When I look at the full moon through my 10cm telescope, it is so bright that it hurts

Because of conservation of etendue (see below) the Moon has the same surface brightness when seen through any telescope or binocular. It's just that it's bigger and so is spread over a larger area of your retina.

It's just like looking at 100 full Moons in the sky, but each Moon is no brighter than the one we see now.

Put in less than precise but simple wording, magnification increases the size, but not the apparent brightness per unit area of extended objects.

...without being damaged?

There's no chance of damage.

This answer to Can a telescope ever increase the apparent luminance of an extended object? says No and explains that this is the result of conservation of etendue

In big telescopes, the focal planes are also pretty huge.

(units: mm)                aperture focal length   f/no.
Human eye                        6         17      2.8      
Vera C. Reuben telescope     8,360     10,310      1.23

So per square micron, the image of the moon will be $(2.8/1.23)^2 \approx 5$ times brighter on the worst case1 telescope's focal plane than on our retina (seen through a telescope or by eye), that's not going to hurt the silicon.

After all we often take outdoor photos with the Sun in the field of view and that doesn't even melt the polymer coatings and color filters on top of the CCD!

1lowest f/no. big telescope so brightest per unit area on the sensor.

LSST Focal Plane

Source

Suzanne Jacoby with the LSST focal plane array scale model. The array's diameter is 64 cm. This mosaic will provide over 3 gigapixels per image. The image of the moon (30 arcminutes) is present to show the scale of the field of view.

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    $\begingroup$ "that doesn't even melt the polymer coatings and color filters on top of the CCD" ... perhaps because of very short shutter speeds, which you have established are not available to scientific instruments. See also photo.stackexchange.com/questions/4016/… $\endgroup$
    – Eric Towers
    Aug 30, 2021 at 12:18
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    $\begingroup$ A link on the photo.stackexchange page Eric Towers referenced, showing the damage done to rental lenses by people taking photos of a solar eclipse: lensrentals.com/blog/2017/09/… $\endgroup$
    – Peter Erwin
    Aug 30, 2021 at 12:27
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    $\begingroup$ And this video, showing how easy it is to damage a DSLR by attaching it to a large telephoto lens pointing it at the Sun (even with the shutter closed, you can totally destroy the sensor): youtube.com/watch?v=2TO_yZDxryQ&t=109s $\endgroup$
    – Peter Erwin
    Aug 30, 2021 at 12:30
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    $\begingroup$ @uhoh: No, the reason our eyes don't start on fire (besides being made of soggy wet stuff) is that they don't have a large enough area to collect sufficient solar energy and concentrate it. Increase the area substantially, and you can set almost anything on fire (or melt/vaporize it): en.wikipedia.org/wiki/Odeillo_solar_furnace And I don't express this as a question, because I don't HAVE a question. I was just trying to get you to think about why your answer just might be wrong. $\endgroup$
    – jamesqf
    Aug 30, 2021 at 21:54
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    $\begingroup$ Wow! A GIGApixel camera is awesomely big :D $\endgroup$
    – The_Sympathizer
    Sep 1, 2021 at 0:23
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It could damage the instruments, but it can't generate very significant temperature. You can't use optics (mirrors, lenses) to heat something more than the temperature of the thing itself! If you could you would be transferring heat from something cold to something hot, without doing any work, and that breaks the second law of thermodynamics.

The surface of the moon is at about 100*C, so thermodynamically it is impossible to use moonlight to heat something above that temperature. In practice, even with a large mirror, you don't get lots of heat. At least not enough to melt the CCD.

However, the instruments are not designed to deal with that much light for a long time. This could be handled by simply using very short exposures.

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    $\begingroup$ The thermodynamic argument only applies to the temperature of the sun, not the moon. I realize that Munroe says otherwise, but as far as I can tell he's simply wrong. The temperature of the moon rocks depends on the light they absorb, while moonlight is the light they don't absorb, and there's no particular connection between the two. $\endgroup$
    – benrg
    Aug 29, 2021 at 23:24
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    $\begingroup$ Exactly, as a thought experiment imagine a giant mirror in space instead of the Moon, it would melt stuff regardless of whether it's cold or not. Moon is just a not-very-good mirror. $\endgroup$
    – JohnEye
    Aug 30, 2021 at 7:47
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    $\begingroup$ @JohnEye but the not-very-good part is important. $\endgroup$
    – user253751
    Aug 30, 2021 at 8:46
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    $\begingroup$ @JohnEye The second half of the what-if article addresses your exact argument. What about it is incorrect? $\endgroup$
    – Spacedog
    Aug 30, 2021 at 16:03
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    $\begingroup$ @Peter - Reinstate Monica: Suppose we have a large number of mirrors at our disposal. Say each mirror is 1 m^2, and so intercepts 1 KW of solar radiation. (We're doing this thought experiment in space.) The energy is reflected to a 1 m^2 spot, so the spot recieves 1 KW/m^2. Add a second mirror, it gets 2 KW/m^2. Increase the number of mirrors to a million, and that spot is getting 1 GW/m^2, no? And (ignoring problems of stable orbits) we could increase the number until we're focusing a large fraction of the sun's output on that 1 square meter. What happens? $\endgroup$
    – jamesqf
    Aug 31, 2021 at 18:46
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It probably depends on the details of the equipped sensor(s).

Let's assume the sensor size in the focal area is around 10cm x 10cm (a typical size, if you look at VLT instrumentation).

The sun is about 400.000 times brighter than the moon at 1370W/m².

So the irradiation at the focal area hence is:

$$ P = \frac{1370W/m²}{400.000} \cdot \frac{\left(\frac{39m}{2}\right)^2\cdot\pi}{0.01m²} = 410W/m² $$

Thus we end up with an irradiation intensity about 1/3 of that of the naked sun - that's something typical CCDs can still cope with - though I'd caution to do so for extended exposures and not attach an objective for visual observation - it would damage your eyes. Depending on the detailed thermal properties of the sensor, its cooling and fixation might sustain damage when some elements become overheated due to prolonged exposure. This is an especially likely scenario if filters are involved as often employed - they are not designed to absorb much energy, they are designed for low light situations and wavelength precision.

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  • $\begingroup$ "an irradiation intensity about 1/3 of that of the naked sun" When I stand outside in the Sun, my skin is slightly warmed by sunlight, but I don't see how 1/3 of that is going to sustain damage to a CCD. You might check how much power is dissipated by the on-chip readout circuitry per square meter and see if it's not actually larger. Light itself is certainly not going to hurt it. $\endgroup$
    – uhoh
    Aug 29, 2021 at 23:20

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