Would a free floating flashlight emitting light in space accelerate due to the photons emitted?
Long answer: Let’s go an adventure. For a normal pocket flashlight powered by a single 1.5 Volt AA lithium battery, we can do an easy calculation. That battery has a mass of 15 g, and contains a total charge of 2700–3400 mAh (that’s milliAmp hours) which gives 5.10 Wh(watt hours) of energy.
Let’s assume that this thing operates at 100% efficiency. This actually isn’t a bad estimate. Not much energy gets lost to heat in the circuit if we have a good LED, and the bit that does get lost to heat will probably find an easier time radiating from the front, because the forward facing part dominates the angular part of the space that it could be radiating into and the material probably isn’t as thick. Anyway, it puts out:
E = 5.10 Watt-hours = 18360 Joules
This is comparable to the detonation of ~4 gram of TNT. I did some googling to try and find the explosive power of various fireworks to give a sense of how much this is, but that information doesn’t seem readily available. I only seem to have found myself on a government watchlist now for lots of google searches about the explosive power of fireworks. Fortunately, this is America, so information about gunpowder is readily available, which has about 70% the energy density of TNT. Some useful conversation from grains to grams tells us that 4 grams of TNT contains about as much energy as sixteen 9mm cartridges worth of gunpowder. It’s a good thing our flashlight isn’t releasing all this energy at once. Into a bullet. Actually, that’s a shame, because that sounds really cool.
Anyyywaaaay, if we calculate the momentum of the photons with this energy, we find
p = E/c = 18360 J/(3.0x10^8 m/s) = 6.12x10^(-5) kg m/s
which is the total momentum of an entire AA battery’s worth of pocket flashlight light. So now, your pocket flashlight has that much momentum going backwards. If your flashlight weighs 30 g (so half the mass is in the casing and elecrotonics, and the other half is in the battery), then:
v = p / m = 6.12x10^(-5) kg m/s / (0.030 kg) = 0.002 m/s
That’s slow. Coincidentally, it’s also almost exactly equal to the World Record for the fastest snail in the Congham, UK. No, I don’t know if the record for the fastest snail in the world was set in Congham, or if this is a local Congham record.
So to find your acceleration, we need to know how long it took for the battery to run dead. I know my flashlight dies if I leave it on in my pocket over night, but it certainly can get me through a weekend of camping on one battery (provided I don’t leave it on in my pocket the first night), so let’s say it has 8 hours of juice.
In that case, the acceleration can be found by:
a = dv/dt = (0.002 m/s) / (8x3600 s) = 6.94x10^(-8) m/s^2
That’s also small. Very small. It’s so small that the Wikipedia page doesn’t even give us a sense of how small it is. If we multiply our 30 grams back in, we find the force, which is about 10-9 Newtons, which is about the amount of force needed to break a covalent bond. If I haven’t hit this point home hard enough yet, that’s small. Really really small. So maybe, if you hitch your flashlight like a tugboat to one of those lithium ions from that battery, you might just be able to pull it apart.