Nukes on the Moon - visible from Earth?

[StrayCatalyst]

From Earth with the naked eye, the Moon is pretty small. If a nuclear weapon were detonated on the Moon, would it be visible from Earth? Would it make a difference if the device were detonated during local daytime (in full sunlight) as far as visibility is concerned?

If it makes a difference, assume cloudless night (for observer, on surface of Earth) and W88 warhead (475 kiloton) detonating at whatever altitude is traditional. (Calling it "airburst" would seem inaccurate!)

 

[mfb]

(Calling it "airburst" would seem inaccurate!)

That's the problem. Basically everything you see from a nuclear explosion comes from the heated atmosphere.

Apart from that, the moon is far away, and very big. The full moon receives the total energy released in such an explosion via sunlight 6 times per second (calculation).For atmospheric explosions, this publication gives a maximal visible light output of ~30TW over a timescale of ~100ms for a 19kt-device, long enough to be visible as "on+off" (so I don't need light integration times I found).
This is roughly 0.3% of the light the moon receives from the sun, or 2% of the light the moon radiates back into space (1.5PW) (but concentrated on a single spot). This would be visible (if you know when to look where) - if the moon had an atmosphere to produce that light!

Without air, the light yield should be orders of magnitude lower, and I don't think it would be visible to the naked eye.

 

[snorkack]

Half the energy travels toward Moon surface and heats it.

Assuming best view conditions - fairly narrow crescent, and explosion on the night/ashen light side - would it be visible?

 

[mfb]

This is more guesswork:
Assume that the whole energy directed towards the Moon (so gamma and x-rays together with fast atoms and bomb fragments) gets absorbed uniformly in the first 10cm of the surface. To get the best light output per energy, heating the surface to ~6000K is good. With half of the 2PJ, how large can the area be?

It looks like lunar regolith can be approximated by SiO2 but with a lower density of 1660kg/m^3.
How much energy do we need to break the bonds of SiO2? Afterwards, we have 3/2 R at 20 g/mol -> 0.6J/g. I'll use this as lower limit for the energy we need to heat the material.

Therefore, we can heat 1PJ/(600J/(kg*K)*1660kg/m^3*10cm*6000K) = 1.7km^2. Let's assume that we manage to heat exactly this surface and nothing else. This will radiate with a power of 73MW/m^2 or 73TW/km^2 (WA), 40% of this is visible light (from here). 50TW... but how long does it last? Typical particle velocities of silicon atoms at this temperature are E_kin~kT => ~2km/s (oxygen is a bit faster). After 1/2 millisecond, the hot layer will have expanded by a factor of ~10 (order of magnitude), and cooled so much that its power is negligible. This would lead to ~25GJ of visible light. Using the eye light integration time from above (now it is useful :D), we get an effective power of 250GW.

1/6000 of a full moon.
10 units of apparent magnitude, which puts our nuclear explosion at -3. Close to the maximal brightness of Mars and Jupiter, so definitely visible (but not very spectacular, given the short time).

I think this is more like an upper estimate of the brightness. For an event so short, I guess an apparent magnitude of +4 is a reasonable limit for the visibility, so my estimate is a factor of 200 above that limit.

Hmm... could be possible.

 

[willem2]

I believe the idea of nuking the moon was that the nuke would be detonated on the night side, but close to the terminator, so the resulting dust cloud would be lit up by the sun.

 

[mfb]

How do you distinguish dust from the surface of the moon? If there is no such surface visible, it is new moon, daytime and you try to observe something directly against the sun.

 

[Borek]

I guess willem means if the explosion is close to the terminator, dust cloud can raise high enough to be light by the Sun, and you have a possible bright spot on the dark background.

No idea if it would be visible.