# How close can I get to a rocket launch and not get killed?

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!

• 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. Dec 25, 2014 at 19:58
• The closest you can get is to sit on top of the rocket. Apr 12, 2015 at 5:39
• 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.
– user17310
Oct 17, 2016 at 14:46
• @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. Oct 17, 2016 at 19:50
• 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, 2017 at 14:31

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.

• 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. Apr 13, 2015 at 6:09
• @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.) 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 Apr 13, 2015 at 9:29
• 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. Apr 13, 2015 at 16:27
• 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. Apr 13, 2015 at 23:26
• If the heat kills you, my money is on convection, not radiation as a method of transfer. Apr 13, 2015 at 23:26

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.

• 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. Sep 22, 2015 at 14:17
• Yes, I do :) makeitlouder.com/Decibel%20Level%20Chart.txt Sep 22, 2015 at 14:18
• That's a very cool list. I don't see mention of distances. Those measures must be the energy at some distance. Sep 22, 2015 at 14:23
• 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. Sep 22, 2015 at 14:38
• 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 :). Sep 22, 2015 at 14:48

That close.....

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

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

• 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, 2017 at 13:06
• 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' (en.wikipedia.org/wiki/Xichang_Satellite_Launch_Center) Apr 13, 2017 at 18:09
• Voting down for low quality and short length, this post consists just of a link and a picture. Not helpful. May 24 at 1:01

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.

• 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. Feb 9, 2015 at 1:10

...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.

• Do you have a reference for body liquefaction at high sound levels?
– Fred
Aug 22 at 6:38
• Aug 23 at 10:23