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There was a question regarding the closest distance regular people are allowed to be near a rocket. At that distance (Baikonur I, 1.1 miles, 1.8 km), even if a rocket explodes, there is a small chance someone is getting killed.

What I ask you is how close can I be near a rocket assuming a successful launch and survive? And what thing will be most likely to kill me? Heat? Rocket debris? Little rocks flying away? Toxic gas?

Of course, the specific answer highly depends on a variety of factors, perhaps most importantly the type of a rocket, therefor I only want an approximate answer (like 100m, 50m, 10m, most probably heat).

NO MEANS OF PROTECTION ALLOWED!

Fred
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    I know folks who have bribed the security guards at Baikonur and been about 1/4 mile (400 m) away from a Soyuz launch. I'm told it's very impressive. Hearing protection is a must. And if it explodes, there's a good chance of being killed. – pericynthion Dec 25 '14 at 19:58
  • No protection at all, not even ear defenders? – pericynthion Dec 25 '14 at 19:59
  • Yes. Deaf or blind, he/she is still alive. –  Dec 25 '14 at 20:01
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    The closest you can get is to sit on top of the rocket. – Mark Adler Apr 12 '15 at 05:39
  • The proviso "assuming successful launch" is perhaps a bit silly, in my view. Consider that the N1 launch failure was the largest non-nuclear explosion on earth at 7Kt (29 TJ) I would want to be a long long long way away ..... – Brian Tompsett - 汤莱恩 Apr 12 '15 at 11:19
  • @pericynthion by my calculations the soyuz rocket would be 7 times as bright as the sun at 400m distance. See my answer. My assumption of emissivity of 1 is admittedly conservatively high. However I'm curious as to the heat protection and eye protection used at that distance. Do you have any more info? – Level River St Apr 12 '15 at 13:42
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    I think they had ear defenders. Your calculations are out of whack; it would be tremendously wasteful for the rocket to throw out that much heat and light. Also, surface brightness of any object is constant with distance (neglecting atmospheric absorption) until the observer is too far away to resolve it. If you were correct, it would be unsafe to look at rocket launches even from many miles away, which is clearly untrue. – pericynthion Apr 13 '15 at 06:13
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    @pericynthion as a single point of reference, wikipedia puts the efficiency of a rocket engine at 60%. Regarding the distance, it's an inverse square law, so the perceived intensity (not the actual brightness, maybe I used the wrong word) falls to a quarter when we double the distance from the launch, so no problem viewing a launch at distance. The only thing out of whack in my calculations is the emissivity, which is always a tremendous fudge in any kind of combustion process. I was hoping your knowledge of Baikonur could shed some light on the thermal radiation at 400m from the Soyuz launch. – Level River St Apr 13 '15 at 10:39
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    I've always been told that there's a distance (100 feet or so) within which your internal organs would basically liquify/be vibrated apart by the acoustic shock of launch -- this was for a Shuttle launch. –  Oct 17 '16 at 14:46
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    @BrianTompsett-汤莱恩 That list you linked says the N1 was 1kT, not 7, and lists 7 or 8 documented non-nuclear explosions with higher yields. – TylerH Oct 17 '16 at 19:50
  • @SpaceGeek See Dudely's answer below from September of last year. – TylerH Oct 17 '16 at 19:51
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    38 meters. If you're on top of it. That's how tall the Vostok launcher was that put Yuri Gagarin in orbit, safely. That's however not for "regular people", but for crazy test pilots. I'm pretty sure 38 meters is the record for crazy people. – LocalFluff Apr 13 '17 at 11:52
  • 5 meters, I correct myself. That's how sure I just were above. In 1945 another crazy test pilot was launched on a suborbital rocket. On the Bachem Natter. The launch seems to have been safe. Just not the landing very much. – LocalFluff Apr 13 '17 at 12:07
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    Ah, wait... So, third degree burns, deaf, blind, wishing to die, will never walk or use hands, but alive - that still counts? – SF. Apr 13 '17 at 14:31

6 Answers6

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The American Petroleum Institute, in its standard 521, outlines limits for exposure of personnel to heat radiation from flares. As hydrocarbons and hydrogen are commonly flared, and also commonly used as rocket fuel, the data is relevant. This publication is used throughout the oil industry worldwide (and therefore is in far wider use than anything produced by any space agency.)

Here are the limits from the 1997 edition (a bit easier to interpret for the purpose of this question than the latest edition.) The odd numbers are a result of conversion from round numbers of $\frac{BTU}{hft^2}$. For comparison, solar radiation is about 1 $\frac{kW}{m^2}$.

9.45 $\frac{kW}{m^2}$ - Exposure must be limited to a few (approx. six) seconds, sufficient for escape only. May consider tower or structure provide some degree of shielding.

6.31 $\frac{kW}{m^2}$ - Emergency actions lasting up to 1 minutes without shielding but with appropriate clothing.

4.73 $\frac{kW}{m^2}$ - Emergency actions lasting up to several minutes without shielding but with appropriate clothing.

1.58 $\frac{kW}{m^2}$ - Personnel with appropriate clothing can be continuously exposed

The latest edition reduces the times for 4.73 and 6.31 to 2-3 minutes and 30 seconds respectively, and rather unhelpfully for the point of view of this question, does not specify any time for $9.45\frac{kW}{m^2}$.

Let's take an example with a popular engine. According to Wikipedia, a SpaceX Merlin 1-C engine has a thrust of $420000 N$ and a nozzle velocity of $2600 \frac{m}{s}$ at sea level, which means a propellant consumption of $\frac{420000}{2600}=161\frac{kg}{s}$, about two thirds (by mass) of which is oxygen. The rest (say $50\frac{kg}{s}$) is kerosene. The Lower Heating Value (i.e. not considering heat recoverable by condensation of water produced in combustion) of kerosene is about $43\frac{MJ}{kg}$ so the power of a merlin 1-C is about $43 \times 50 = 2150\ MW$ or $2150000\ kW$.

Let's assume we want to at the $6.31 \frac{kW}{m^2}$ distance and assume (as the API 521 standard does) that the radiation of a combustion source is identical in all directions. To keep the calculation simple, we will assume (for now) that the emissivity of the combustion source is 1: that is, perfect radiation.

We now need to calculate the radius of a sphere such that $6.31 \frac{kW}{m^2}$ radiation will be experienced from a point source of $2150000\ kW$. Such a sphere will have an area of $\frac{2150000}{6.31} = 340729\ m^2$. As the area of a sphere is $4\times \pi\times r^2$, this works out as a distance of 165 m.

Two more things to consider: First, a Falcon 9 launch vehicle has 9 engines, not one. to factor this in, we need to multiply by $\sqrt{9}=3$ so we need to be at $165 \times 3=495 m$ distance. (Say, 500m.)

Secondly, the emissivity may be quite a bit less than 1 (values for combustion with oxygen are difficult to come by) but because of the square law it won't make much difference. Opaque smoke can make quite a difference to emissivity, but most rockets burn cleanly once they are clear of the launch pad. A low value for a smokeless flare burning heavy hydrocarbon would be 0.25 ($\frac{1}{4}$) so if this is was applicable to a rocket the distance would be halved to $250 m.$

I reckon you would survive witnessing a Falcon 9 launch at a maximum radiation of $6.31 \frac{kW}{m^2}$, though quite possibly with significant burns. It's a fairly short time before the rocket is well clear of the Earth, but it would be hot and uncomfortable (painful) with 6.31 times the solar radiation in your face. I wouldn't be surprised if you turned and ran.

Most propellants are not that toxic. Perhaps the worst exhaust fumes would be from the Space Shuttle solid rocket boosters, which produced aluminium oxide in a fine white powder form which would be very bad for your lungs. I'm pretty sure the heat radiation would still be the limiting factor though.

EDIT 1: The Soyuz launcher has five (quadruple nozzled) engines, of 813 kilonewton thrust and $2.4\frac{km}{s}$ velocity, giving a total propellant consumption of $1694\frac{kg}{s}$. That is marginally more than the $9 \times 160 = 1440\frac{kg}{s}$ used by the Falcon 9. Therefore I find the claim in the comments that the launch can be watched from $400 m$ surprising, though it does not conflict with an emissivity of $0.25$. The emissivity is something of an unknown, and the cloud of debris and steam at the launch pad would shield the observer from the heat radiation until the rocket gained some height. It's still closer than I would like to be to a launch.

EDIT 2 I am receiving comments that my thermal calculations are an overestimate. I've checked the overall energy release and that at least is correct. So let's see what may be wrong:

  1. The spherical radiation model is an oversimplification. In fact, most of the radiation will be downwards, so this would actually increase the thermal energy felt by an observer on the ground.

  2. I took no separate account of the energy converted to thrust. Wikipedia indicates around 60% efficiency, leaving 40% energy available for emission. I checked this with my own expansion calculation:

Chamber pressure $6.77\ MPa$ (Merlin) $5.85\ MPa$ (Soyuz): consider $60\ atm$ (approx. $6\ MPa$) for convenience.

Specific heat ratio: Both CO2 and H2O are around 1.3.

Heat not converted to thrust = $\frac{T2}{T1} = 60^{\frac{1-1.3}{1.3}} = 0.389$

This is surprisingly close to the Wikipedia efficiency value.

Given the general uncertainty of emissivity values, I do not consider a factor of 40% to be particularly significant.

  1. After some thought, it occurred to me that perhaps the most important difference between a flare (which as a combustion engineer in the oil industry I am very familiar with) and a rocket engine (which I am admittedly less familiar with) is the much greater turbulence with ambient air. This may lead to much greater mixing and a consequently lower emissivity.

I'm reluctant to make another guess at emissivity, but if it was as low as $\frac{1}{25}$ (that's just 4% of the heat released being converted to thermal radiation!) my estimate for the minimum non fatal distance from a Falcon 9 would be $\frac{500}{\sqrt{25}}=100\ m$ (at which distance your hearing would be severely damaged.)

It's notable that this is not much different from the radius of the cloud of dust and steam that forms at the launch pad. That debris cloud must be pretty hot (all that heat that doesn't get radiated has to go somewhere) so I think the risk of being killed by flying debris is irrelevant, as the heat would get you anyway.

No Nonsense
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Level River St
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    A rocket engine is not a petroleum flare. The combustion occurs inside the thrust chamber and much of the exhaust products' thermal energy is transformed into velocity by the nozzle. – pericynthion Apr 13 '15 at 06:09
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    @pericynthion sure, I know it is not a perfect analogy. Adiabatic flame temperature for petroleum combustion in air is around 2000C (a lot of energy goes into heating nitrogen.) http://en.wikipedia.org/wiki/Rocket_engine gives an efficiency of 60% which would leave 40% thermal energy available for emission. Because a rocket engine breathes pure oxygen, the combustion temperature prior to expansion would be higher than in air. I haven't found data on final exhaust temperatures, but I guess it's not far off 2000C. All these factors I would lump together into the uncertainty of emissivity – Level River St Apr 13 '15 at 09:29
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    So you're right that emissivity of 1 (Soyuz 7 times brighter than the sun at 500m) is not correct. At the lower end of my range of estimation, emissivity of 0.25 would put the Soyuz at 1.75 times brighter than the sun at 500m, which doesnt conflict with your experience. Reducing the distance by half to 250m would quadruple it back to 7 times brighter than the sun. When estimating distance, the uncertainty in emissivity is square rooted, so my imperfect analogy should give reasonable ballpark results. I just wondered if you could shed more light on the thermal radiation at 400m at Baikonur. – Level River St Apr 13 '15 at 09:42
  • In the extreme case, exhaust can be below 0 Celsius: https://www.youtube.com/watch?v=4eM1mNNdguA – pericynthion Apr 13 '15 at 15:36
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    From my personal experience, I think the noise will kill you first, or at least make you deaf. Your heat calculation is way off. You could get a decent estimate by making the following assumption: The outer layer of the flow must not melt the nozzle, so for an upper estimate, take 1000°C. At the exhaust it's probably less. Use CEA to get an idea, but it doesn't factor in cooling. As for sticking your fingers in your ears... Oh sweet summer child, you never stood too close to a rocket engine. – Rikki-Tikki-Tavi Apr 13 '15 at 16:27
  • @Rikki-Tikki-Tavi Sticking fingers in ears wont help much, but OP says going deaf is OK, so long as you don't die. Emissivity's a difficult thing to get a handle on. I know it can't be as high as 1 for efficiency reasons, I'm surprised my 1/4 is that far off. If it was 1/25 then my calc would be 495m/5=99m from the Falcon. I know my answer would be greatly improved by a single data point of radiation, power and distance. If a nozzle skin is cooled, bulk flow can be hotter, and on a large engine nozzle skin cooling will make little difference to the flow. Will try an expansion calculation later – Level River St Apr 13 '15 at 17:40
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    You should look into black-body radiation again, how you applied it is not how emissivity works. You should also look into equilibrium chemistry, and how it relates to rocket engines as well as into gas dynamics, to learn about the temperature of exhaust gases. – Rikki-Tikki-Tavi Apr 13 '15 at 23:26
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    If the heat kills you, my money is on convection, not radiation as a method of transfer. – Rikki-Tikki-Tavi Apr 13 '15 at 23:26
  • @Rikki-Tikki-Tavi Convection: That's basically the conclusion I reached (I was editing as you commented.) With a rocket, mixing with ambient air occurs after combustion (not before as in a flare) and is therefore much more intense. As deafness is disqualified by the OP, the possible causes of death are: toxicity/asphixa, debris, heat. Any cloud that kills you by asphixia or debris abrasion is going to be hot enough to kill you by heat alone. It's a multi-gigawatt combustion process. All that heat has to go somewhere. – Level River St Apr 13 '15 at 23:41
  • BTW I have 7 years experience as a combustion engineer, and the way I applied emissivity is the fudged way they apply it in API521. Combustion gases cannot be treated as black bodies. Nitrogen is a very poor infrared emitter as it has no dipole moment. CO2 and H2O are rather better. But when you have combustion, emission, mixing/convection (and in the case of a rocket engine, expansion) all occuring at the same time, some kind of empirical fudge factor is the best you can do. – Level River St Apr 13 '15 at 23:48
  • Well then you shouldn't have any problem using CEA. It's a lot of fun to mess around with, and you can use it to make a lot better assumptions than you did. I'm guessing your work was not with rockets? – Rikki-Tikki-Tavi Apr 14 '15 at 00:11
  • http://www.grc.nasa.gov/WWW/CEAWeb/ceaHome.htm – Rikki-Tikki-Tavi Apr 14 '15 at 00:12
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    Rocket's don't use stoichiometric combustion so maybe you would die from the spray of unburnt fuel. – Rikki-Tikki-Tavi Apr 14 '15 at 00:20
  • Let us continue this discussion in chat. – Level River St Apr 14 '15 at 00:26
  • Another factor. You were setting the exposure limit to the 1-minute value but 60 seconds after the rocket lit off it's a long ways away and no threat. The actual exposure duration is a lot shorter. – Loren Pechtel Sep 23 '15 at 02:16
  • @LorenPechtel that may be relevant, but the difference between 6.31kw/m2 (1 minute) and 9.45kw/m2 (6 seconds) is not that much, especially when square rooted. Videos of Falcon 9 and Soyuz rockets seem to show about 5-10 seconds of combustion before the rocket begins to move. The rockets seem to gain a height equivalent to their own length (about 50m in both cases) in about another 5 seconds. – Level River St Sep 23 '15 at 09:07
  • @steveverrill Much of that initial energy on the pad is going to be absorbed by the water systems. It's the energy emitted after liftoff that's the real threat. – Loren Pechtel Sep 24 '15 at 03:34
  • I suspect that the sound suppression system greatly change those numbers. – JFL Dec 08 '22 at 15:10
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Keep in mind that a rocket engine is simply a controlled explosion. The explosion is directed at the ground, so most of the heat/flames/exhaust won't reach you unless you were very close. It would be simple to just stand on the side opposite to where it was being directed on launch.

However, other than a suppression system consisting of trenches and/or water, nothing will protect you from the sound.

Sounds are just pressure waves traveling through the air. The sound of the shuttle launching was 215 dB. Saturn V was 220 dB, and was capable of melting concrete through sound alone. For reference the nuclear bombs they dropped on Japan were 248 dB. Studies show that a sound of 210 dB will cause damage to internal organs, probably resulting in death from internal bleeding. How close would you have to be to the Saturn V to get hit with 210 db? Probably a few hundred feet, which would put you beyond most of the smoke and heat, especially if you stood at the right spot.

So I'm voting that the sound kills you first.

Ref:http://www.makeitlouder.com/Decibel%20Level%20Chart.txt

Fred
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Dudely
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  • This looks good, but do you have references for the dB levels? They are perfect examples of where a good link would seal the argument and also allow a path to learning more. - Welcome to Space Exploration. – kim holder Sep 22 '15 at 14:17
  • Yes, I do :) http://www.makeitlouder.com/Decibel%20Level%20Chart.txt – Dudely Sep 22 '15 at 14:18
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    That's a very cool list. I don't see mention of distances. Those measures must be the energy at some distance. – kim holder Sep 22 '15 at 14:23
  • Ah, found it near the end. Normalized to a distance of one meter. But if you were one meter from a firing rocket you'd be dead for other reasons too. So i don't think this settles the minimum distance question, which is really the heart of the matter. – kim holder Sep 22 '15 at 14:29
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    dB goes down with distance. At 1/2 mile 220 db is 170 db. So I'm guessing at a few hundred feet the sound is still loud enough to kill you, but the heat would not be hot enough to kill you. – Dudely Sep 22 '15 at 14:38
  • Alright. Then i vote for a second reference to the falloff rate of sound energy, especially since it has some wrinkles at high energies. – kim holder Sep 22 '15 at 14:44
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    I'll see if I can find a reference for that. Most likely I'll have to calculate it myself, which would take some time. I'll do it after work, maybe :). – Dudely Sep 22 '15 at 14:48
  • I might look later as well. Cheers. – kim holder Sep 22 '15 at 14:55
  • Having made the mistake of standing too close to a rocket test once, I can see how the sound may well be the first thing to kill you. – Rikki-Tikki-Tavi Oct 17 '16 at 15:19
  • Once you pass the 194 dB threshold, it is a bit meaningless to talk about sound because at that point the sound waves become shockwaves. At 194 dB the pressure difference between the sound wave valleys and the ambient pressure is going to be 101.325 kPa, which is the atmospheric pressure; in other words, the pressure in the wave's valleys is going to be zero, and it cannot drop anymore than that. Hence, shockwaves behave differently than sound waves. It is even in your reference. Consider including that information. Cheers. – No Nonsense May 24 '22 at 01:17
  • I am skeptical of the "melting concrete" claim for the Saturn V. Sound doesn't generally melt things, and concrete is very brittle and I would expect it to crack or shatter. I have searched that claim on the internet and not been able to trace it to a source that actually gave an explanation. If someone could find a good source (or debunking) of that it would be valuable. – Mark Foskey Aug 22 '22 at 16:50
  • What about the impact of the sound suppression system? – JFL Dec 08 '22 at 15:11
  • I'll pile on here in that it's not enough to just survive the sound levels. Animal organs are sensitive to vibration and sufficiently intense acoustic waves can cause connective tissues to break without causing immediate death. Thus, one may be permanently injured without immediate external evidence of it. As for melting concrete, I've done this calculation. It isn't melting, it's ablation from acoustic and turbulent flow damage causing erosion. It's more efficient that mere melting, which would take much more heat and concrete itself doesn't melt, it disintegrates. – Chris Ison Dec 10 '22 at 15:58
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That close.....

http://www.parabolicarc.com/2017/04/12/close-video-chinese-rocket-launch/

This is pretty insane, the heat must have been immense.

enter image description here

camping
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    Welcome to Space Exploration SE, @camping. Most of the SE network opposes link-only answers, as they can become useless if the host moves or changes their content. You could improve this answer by taking a representative still from the video and adding it to your post (I believe SE usually partners with imgur for this). – Bear Apr 13 '17 at 13:06
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    Judging by the audio delay, that was from 300 m away. Xichang launch center is crazily close to populated areas too, '...crashed 1200 meters away from the launch pad in a nearby mountain village' (https://en.wikipedia.org/wiki/Xichang_Satellite_Launch_Center) – Hobbes Apr 13 '17 at 18:09
  • Voting down for low quality and short length, this post consists just of a link and a picture. Not helpful. – No Nonsense May 24 '22 at 01:01
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It would vary for each type of rocket. The more powerful a rocket the greater the safe stand-off distance. A rocket like the Saturn V would require a larger stand-off distance than a Proton rocket.

It would also depend on what exhaust directing structures exist below the rocket at launch and where the person wanted to be in relation to that. A stand-off distance perpendicular to the exhaust vents would be less than a stand-off distance in the path of the exhaust from the vents.

Fred
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    Definitely varies by kind of rocket. Presumably the OP means ones that go to space. One can be a meter or so away from the smallest of rocket launches with no ill effect. – geometrian Feb 09 '15 at 01:10
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Safety perimeters for rockets and explosives are based on at least three components:

  1. Tolerable acoustic pressure levels (not just fatal but injurious levels).
  2. Absorbed heat by the most sensitive materials nearby (usually bare skin or eyes).
  3. Farthest distance traveled by some fraction of fragments in worst case explosion (usually detonation of all propellants).

A fourth factor also considers local weather and terrain, so perimeters may not be circular around the center of the site.

These perimeters are set for worst case scenarios, not just based on the simple science. Perimeters are never the same except for identical launch sites (like missile silos or some military installations).

Chris Ison
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...one would DEFINITELY be killed from the sound first; the force of the sound alone would liquefy your entire body in a matter of fractions of a second; this is why you don't see anyone around the launch pad starting at T-9 minutes and down. You would probably be killed if you were standing on the LC-39 observation platform less than 1/2 mile from launch pad 39-A. BTW, SLS has 8.8m pounds thrust, as compared to Saturn V's 7.7m at max, so the kill zone is larger, but they are launching SLS from pad 39-B on 8/29, so the distance is closer to a mile from the observation deck.

Bert Peters
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    Do you have a reference for body liquefaction at high sound levels? – Fred Aug 22 '22 at 06:38
  • https://decibelpro.app/blog/can-sound-kill-you/#:~:text=If%20we're%20talking%20about,inner%20organs%20and%20cause%20death. – Bert Peters Aug 23 '22 at 10:23
  • That reference describes 240 dB sound, but in air at sea level the loudest possible is 194 dB. It's not authoritative, it's more like an infomercial. – Camille Goudeseune Dec 09 '22 at 20:23