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How bright (in Watts, preferably, or lumens) is the plume from a rocket launch (e.g. a SpaceX launch)? Obviously in its direction of travel it isn't particularly bright, but what is its effective isotropic radiative power abeam?

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    $\begingroup$ There is an unclassified McDonnell report on optical signatures of rocket plumes. It has some example data for plume contour and discusses the processes that contribute to emissions. $\endgroup$ – Andreas Dec 8 '16 at 18:11
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    $\begingroup$ There's a similar question also with no answer: space.stackexchange.com/questions/20601/… $\endgroup$ – Aaron M Jul 5 '17 at 0:06
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    $\begingroup$ It's not a real answer to the question, but anecdotally, I found the SRB plume from STS-135 uncomfortably bright from my vantage point at the press site -- about like looking at the sun. $\endgroup$ – Tristan Aug 17 '17 at 13:37
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The chemical energy from burning 1 mole (2 grams) of hydrogen in plentiful oxygen is 242 kJ. Just enough oxygen is half a mole, or 16 grams.

The Space Shuttle external tank contained 629,340 kg of liquid oxygen and 106,261 kg of liquid hydrogen. Note that this isn't enough oxygen to burn all the hydrogen, for rocket science reasons, so we'll figure out the total energy using the oxygen.

629 tonnes of LOx, or 629 million grams, is 39.3 million half-moles. At 242 kJ per half-mole, that's a total output of 9510 GJ.

The SSMEs ran for roughly 510 seconds. Dividing the energy by the time gets us 18.6 GJ/s or GW.

From the ever-useful Atomic Rockets site, I get a total thrust for the three SSMEs of 12.1 GW. Divide the useful energy, or thrust, by the total energy from burning the fuel gets an efficiency of around 65%. I remember that the SSMEs were supposed to be especially efficient, so I might guess at around 40% wasted energy for most rocket launches. Look up the thrust in megawatts for the rocket you're interested in, multiply by 40%, and divide by ten (at a guess) for the amount that goes into visible light.

Alternatively, look up the temperature of the exhaust, and work out the proportion of visible energy from a blackbody curve for that temperature. (Hint, the SSME operated at 3300°C.)

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    $\begingroup$ One complication is that the SRBs burned a fuel containing a lot of aluminum and a fair bit of carbon, which produced a lot of visible light; on the other hand the SSMEs, burning only hydrogen and oxygen, emitted very little visible light (aside from a short blue cone, most of the SSME exhaust plume appeared transparent). $\endgroup$ – Anthony X Oct 21 '18 at 0:40
  • $\begingroup$ This seems like a guess. I don't see why this question should not be answerable, in principle. $\endgroup$ – Organic Marble Oct 21 '18 at 0:48
  • $\begingroup$ @Organic Marble: I agree that it should be answerable in principle. The difficulty is that I don't know how to do it in practice. This Quora answer hints at a solution, but my integration skills aren't up to it, and I broke an online integrating calculator on it. $\endgroup$ – Nelson Cunnington Oct 21 '18 at 18:39
  • $\begingroup$ @Anthony X: And yet, one night during a Shuttle launch at high inclination, I saw it rise above the western UK horizon as a moderately bright, fast-moving star, well after SRB separation but before MECO. The SSME exhaust was only relatively dim in bright sunlight, giving an impression of invisibility. I think it's still quite luminous in absolute terms. You can actually see it in daylight video and photos, after all. $\endgroup$ – Nelson Cunnington Oct 21 '18 at 18:49
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An OV-103 or also called Space Shuttle Discovery has an average lift-off mass of 2,050,000 kg.

2,050,000 kg x 9.81 = 20,110,500 N is the total weight at lift-off

The Space Shuttle Discovery has two big white solid rocket boosters on the sides and three main engines at the tail.

Thrust of both boosters is equal to 2,404,040 kg x 9.81 = 23,583,632 N

Thrust of each main engines equals to 179,097 kg x 9.81 = 1,756,942 N

Thrust of all three equals to 1,756,942 N x 3 = 5,270,826 N

Overall thrust is 23,583,632 N + 5,270,826 N = 28,854,458 N

Resultant force = thrust – weight = 28,854,458 N – 20,110,500 N = 8,743,958 N

Acceleration = resultant force ÷ mass = 8,743,958 N ÷ 2,050,000 kg = 4.27 m/s^2

The spaceship is at rest before launched, so its initial speed is 0. After one second of the launch, its speed is going to be 4.26 m/s which is going to increase as the ship accelerates. We could say that the plume from a rocket launch could be happening at the first second after the launch.

Final Answer: The energy released at such moment is (1/2) x 2,050,000 kg x (4.27 m/s)^2 = 18,688,722.5 J

Note: The amount of joules in that second equals 18,688,722.5 Watts, because J/s = Watt

I believe 18,688,722.5 J is the energy released at such moment, but it would not all convert into visible light because most of the energy is lost as heat.

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    $\begingroup$ So all the energy released is converted into light? I am surprised by that. $\endgroup$ – Organic Marble Aug 16 '17 at 15:11
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    $\begingroup$ This is an interesting attempt. The amount of energy released is much much more than this. In rocket problems we say that the momentum change of the rocket is equal to the momentum of the exhaust (in some frame) but I'm pretty sure most of the power is lost as heat, and so is much more than the rate of change of the rocket's kinetic energy. Therefore the number could be right by sheer coincidence, but this is not the right way to calculate the radiated power by the exhaust. $\endgroup$ – uhoh Aug 16 '17 at 15:41
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    $\begingroup$ Instead, you might try instead is to look at the plume shown in this question choose a "white hot" temperature of maybe 1000C or look up a better number, choose a surface area $2\piRh$ and use $\sigma T^4$ to approximate the blackbody radiation. It's a very rough approximation, but it is better based in physics than the kinetic energy of the rocket. $\endgroup$ – uhoh Aug 16 '17 at 15:46
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    $\begingroup$ I see what you're doing here, but you might be better served to find the total chemical energy released over that time. Then you would subtract this number you calculated above as "work extracted" to find the waste heat produced. That would be a closer representation. It still wouldn't be right, because not all of the waste heat would be converted into visible spectrum radiance, but it would be closer. $\endgroup$ – Tristan Aug 17 '17 at 13:40

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