I am an evil genius owning a space station in a 400km orbit around Earth. As part of my plan for world domination I would like to equip this space station with a ground-attack laser cannon. Approximately how powerful (in watt) would an orbital laser need to be so that a one-second burst could destroy:

  • an unprotected person
  • an unarmored car
  • a one-family house
  • a battle tank
  • an office building
  • a small town

Assume that the target is on normal-null altitude, directly below the orbit of my station and that the weather conditions are as favorable as can be reasonably expected.

  • $\begingroup$ This (youtu.be/xOK-s97c6Ko) could be relevant to you it's a talk about a system that people actually are trying to set up to deflect or even destroy asteroids and such. $\endgroup$ Commented Sep 4, 2015 at 7:50
  • $\begingroup$ I have seen an estimate that said hundreds of terawatts would be needed to give someone a suntan over the course of an hour or so because of atmospheric scattering but I've taken note of the question and I'll run the math for some different frequencies and see what I can work out, be a few days though. $\endgroup$
    – Ash
    Commented Aug 9, 2017 at 18:25

4 Answers 4


So I'm no rocket scientist, but here goes. Much of this is shamelessly lifted from Project Rho.

First, lets ignore the diffraction caused by passing through the earths atmosphere for the following. I'm also assuming you're talking an actual laser, not a particle weapon, blaster, or some other variant.

A general idea to start with is for the US military, the minimum threshold for a tactical weapons-grade laser is 100 kilowatts.

Maximum range will be a few hundred thousand kilometers, otherwise almost every shot will miss due to light-speed lag, Unless we are shooting at targets that don't move.

First we calculate the beam divergence angle θ (θ = 1.22 L/RL)

$θ = beam divergence angle (radians)$
$L = wavelength of laser beam (m, see table above)$
$RL = radius of laser lens or reflector (m)$

Next is beam power BP, then calculate the beam intensity at the target (the beam "brightness"):

$$BPT = BP/(π * (D * tan(θ/2))2)$$

$BPT = Beam intensity at target (megawatts per square meter)$
$BP = Beam Power at laser aperture (megawatts)$
$D = range to target (meters)$
$θ = Theta = Beam divergence angle (radians or degrees depending on your Tan() function)$
$π = Pi = 3.14159...$

The following equation will be of use for determining the diffraction:

$$R_T = \frac{0.61 D L}{R_L}$$


$R_T$ = beam radius at target (m)
$D$ = distance from laser emitter to target (m)
$L$ = wavelength of laser beam (m, see table below)
$R_L$ = radius of laser lens or reflector (m)

As for laser wavelengths:

  • Near Infrared 2.5$\times$10-6 to 7.5$\times$10-7 m (2,500 to 750 nanometers)
  • Red 7.5$\times$10-7 to 6.2$\times$10-7 m (750 to 620 nanometers)
  • Orange 6.2$\times$10-7 to 5.9$\times$10-7 m (620 to 590 nanometers)
  • Yellow 5.9$\times$10-7 to 5.7$\times$10-7 m (590 to 570 nanometers)
  • Green 5.7$\times$10-7 to 4.95$\times$10-7 m (570 to 495 nanometers)
  • Blue 4.95$\times$10-7 to 4.5$\times$10-7 m (495 to 450 nanometers)
  • Indigo 4.5$\times$10-7 to 4.2$\times$10-7 m (450 to 420 nanometers)
  • Violet 4.2$\times$10-7 to 3.8$\times$10-7 m (420 to 380 nanometers)
  • Ultraviolet A 4$\times$10-7 to 3.15$\times$10-7 m (400 to 315 nanometers)
  • Ultraviolet B 3.15$\times$10-7 to 2.8$\times$10-7 m (315 to 280 nanometers)
  • Extreme Ultraviolet 1$\times$10-7 to 1$\times$10-8 m (100 to 10 nanometers)

Below Extreme Ultraviolet, you can't use the laser outside of the vacuum of space as the atmosphere would absorb it, so we can ignore those.

An example of how this works is as follows:

Say you have an ultraviolet (20 nanometer) laser cannon with a 3.2 meter lens. Your hapless target is at a range of 12,900 kilometers (12,900,000 meters). The Beam Radius equation says that the beam radius at the target will be about 4 centimeters (0.04 meters), so the beam will be irradiating about 50 cm2 of the target's skin (area of circle with radius of 4 centimeters). If the hapless target had a hull of steel armor, the armor has a heat of vaporization of about 60 kiloJoules/cm3. Say the armor is 12.5 cm thick. So for the laser cannon to punch a hole in the armor it will have to remove about 625 cm3 of steel (volume of cylinder with radius of 4 cm and height of 12.5 cm). 625 * 60 = 37,500 kiloJoules. If the laser pulse is one second, this means the beam requires a power level of 37,500 watts or 38 megawatts at the target.

A note: using a pulsed laser rather than a single focused beam would require less power, in effect drilling into the target rather than trying to vaporize it.

Now aside from all of that, you're generating a massive amount of waste heat that will require dissipation, but that's a whole extra problem.

There's also a neat laser maker calculator here that I would recommend for those that don't math.

Also to read the source of this information, you can find it here.

  • 1
    $\begingroup$ Thank you for your answer. But I don't think that you can just ignore the atmosphere. $\endgroup$
    – Philipp
    Commented Sep 3, 2015 at 17:29
  • $\begingroup$ Also, would an uv-laser be visible when crossing the atmosphere? Having a visible beam striking down from the sky is essential for the coolness factor of my kill sat. $\endgroup$
    – Philipp
    Commented Sep 3, 2015 at 17:31
  • $\begingroup$ You would have to increase the lasers strength to compensate for the atmosphere, thought as far as I know no data on that is available. Just increase until you feel the interference is gone. The same formula's apply to other wavelengths if you want a different color. $\endgroup$
    – Nonafel
    Commented Sep 3, 2015 at 17:40
  • $\begingroup$ If his kill sat is at 400km orbit and he's shooting at the ground, what target is 12,900 km away? $\endgroup$ Commented Sep 3, 2015 at 19:15
  • $\begingroup$ You need to redo your math. 400 km is much less than 12,900 km (not meters!), so the laser optics can be much smaller. $\endgroup$ Commented Sep 3, 2015 at 19:22

1 watt second is a joule, so I will use J instead of W/sec

To be accurate you need to determine what is a definition of destroy. For example, to burn a person could mean incapacitate by 2nd or 3rd degrees burns of much of the body, or it could mean turn to ash.

Published data from atomic testing suggested that 8-10 cal / cm^2 is enough to cause 3rd degree burns on unprotected flesh. 10 cal / cm^2 is about 420 KJ / m^2. Average adult has about 2 sq meters of skin, but since you can only irradiate 1 side, you have to settle for burning one side of a person, so 400 KJ one the target will incapacitate a human. Without medical treatment 3rd degree burns over 50% of body is often fatal due to shock, fluid loss and infection. To actually guarantee a kill, you need more power on target of course, 10 MJ is probably enough and their clothes will catch on fire too overriding more common forms of protection outdoors.

Setting hardwoods on fire requires about 1 MJ / m^2 this would be sufficient for many homes, but a crushed white rock roof would be mostly unaffected, perhaps 10 MJ / m^2 would be enough -- 200 sq meters would include most homes so you need 2000 MJ or 2GJ on target.

A car again is difficult determine what destroyed requires. Unlike on TV cars to not catch on fire or explode at the least provocation. To temporarily incapacitate a car via space laser, the softest target is probably the tires, could not find direct data, but I suspect 1 MJ / m^2 would ignite the tires most of the time, but since the tires are often shielded by the vehicle they would often not catch on fire until far beyond this point and in any case you could easily fix this. To destroy a car you need to get it hot enough to destroy vital components like electrical wiring or engine belts - though again belts are easily replaced. I believe if you can heat the car to 500 C you will destroy most of the wiring as well as other soft components (seals, etc). So, how much energy to raise a car temp by 500 C (it could be winter). Curb weight on a new Ford Exposition is up to about 6000 lbs, as a first order approximation lets model this as 2720 kg of iron. Iron has a heat capacity of 0.45 J/g/deg. So 2720 kg * 1000 g/kg * 500 deg = 612 MJ on target.

Tanks are a lot like cars, so I will use the similar assumptions but I will require 800 deg and 63,000 kg (Leopold 2) and I will need about 22.7 GJ on target.

Office building. Likely brick, stone, steel in construction. Again how much is enough to consider it destroyed? If you really want to be sure, you need to err on the high side. The Empire State building used 60,000 tons of steel, 200,000 cubic feet of limestone and granite, 10 million bricks and 730 tons of aluminum and stainless steel. To simplify, I will assume 200,000 cu ft of limestone and ignore the bricks and aluminum. How much does granite weigh about 2.75 g/cm^3 and a heat capacity of 0.19 J/g/deg. Total granite mass 15,600 long tones. For a 500 deg temperature rise I need 13,500 GJ for the steel plus 1,482 GJ for the granite or about 15 TJ to take out the empire state building. In actually fact, you don't heat up the lower floors in this case as the upper floors absorb nearly all of the heat. A 1 second pulse would need to essentially vaporize the upper part of the building in order to destroy. But that is not even enough as the vapor cloud itself would absorb and scatter a large percentage of the laser blast. A thousand times might be enough though.

A small town, again what is threshold of destruction, and just how big is this. I was raised in Columbus Indiana, so seems like a good model to me. Columbus IN, population 44,061 (2010 census). Area 72.23 sq km. 500 deg temperature rise and steel & concrete destruction (to take out all of the buildings). I will simply revert to the energy per unit assuming the 100 MJ per sq. M is enough to take out everything (100 times the energy to ignite hardwood). 100,000,000 MJ / m^2 * 72,230,000 m^2 = 7.223 PJ -- petajoules on target.

Needless to say, there are no experimental confirmations of the beam attenuation that would be encountered firing from orbit to surface targets, certainly not anything approaching these energy levels. However, I did find Laser Atmospheric Attenuation Tables for LTAS. Without going through the justification at this I think you would need about 20-50 times as much light at the source to reliably punch through the atmosphere as you have at the target. I have not yet performed more detailed calculations to confirm this. So, take all of the amounts above and multiply by 50 if you want to be able to destroy targets with reasonable certainty. This is based on bad weather conditions. For best case but still realistic weather conditions it would appear that you only need 5-10 times the on-target energy at the source.

The higher energy levels deserve better accuracy as their are many non-linear and probably classified effects that doubtless come into play.

I am also understating the actual power required because I generally ignored that fact that for everything except the unprotected human case, the target will ablate. The surface layers will cook off turning into gas and taking much of the incoming heat with them, and continuing to absorb and scatter the incoming laser light. I also generally ignore the fact that the target will reflect some of the laser light, when in can reflect at least some of the incoming light. At most of these energy level, a mirror will not provide protection, at least not for long as even a good mirror will absorb enough energy to quickly be destroyed.

These power levels are beyond ridiculous. For everything larger than a car, the power at the target exceeds that of even the largest electrical generation stations. The entire United States only has a electrical capacity of 1060 GW.

For our simple 3rd degree burns on a single human 1MJ on target requires a minimum of 5 times that at the source and 5 times again for the loss of electrical conversion to laser energy, i.e., 25 MJ. A common comparison is that 1 MW will power 1000 homes, so 25 MW is over half of the power needed to supply Columbus Indiana just to incapacitate 1 person.

Using capacitors to store up for a bolt from the sky may sound like a good idea, but your requirements dwarf standard capacitors. The largest capacitor bank in the world is only 50 MJ, built at a cost to 10 million Euro

Somewhere during your construction phase people are going to wonder what you are building and will get motivated to put a stop to your evil plan.


Most of the numeric factors have been accounted for, but there are a few other things you need to consider in your orbital fortress.

Punching all that energy through the atmosphere, even in "window" wavelengths will cause a huge amount of energy to be deposited in the atmosphere. To a certain extent the initial thermal blooming and other distortion effects can be accounted for using "adaptive optics" (AKA a "rubber mirror") which flexes to compensate for atmospheric effects. At higher energy levels, like those needed for burning hardened targets like tanks, the amount of energy could easily cause air molecules to dissociate into plasma, blocking the beam entirely as the plasma absorbs the energy and radiates it away as incoherent light and heat. Luckily, since you are in orbit, you will miss the worst effect of the plasma running back "up" the beam to the emitter and depositing the energy on the mirror.

The second consideration is that lasers in general have very low conversion efficiencies, so if you need as much as 5X the actual beam power as input energy for your laser system. This also means that you are join to be dealing with incredible amounts of waste heat, so your space station will be dominated by the energy system and vast arrays of radiators. Even at 400km altitude, the sheer size of all this could cause enough atmospheric drag to bring you out of orbit. The mass of all the stuff will also make your space station virtually immobile in terms of orbital manoeuvres, so people who object to being lasered from orbit won't have much difficulty in sending clouds of ball bearings or even sand into the orbital path of the space station, meaning that even with the impressive laser weapon, it could be destroyed.

Now the most impressive laser weapon that I have ever seen described is also on the Atomic Rockets website (http://www.projectrho.com/public_html/rocket/spacegunconvent.php); the Ravening Beam of Death (RBoD), which uses a kilometre diameter accelerator ring to power a Free Electron Laser (FEL) capable of vaporizing metal, ceramic and carbon fibre a light second away in milliseconds, and I don't think that has the energy levels described to vaporize entire towns in a single shot. But it does provide an alternative, since a light second is almost the distance from the Earth to the Moon (and the beam itself is dangerous far beyond, a light second is arbitrarily chosen to make targeting responsive, since no target can move very far in a single second or the two seconds needed to see the effect of the shot and correct).

So set up your fortress on the Moon, fire short pulses of laser energy rather than long beams, and remember to allow for the energy dumped in the atmosphere to dissipate between shots.


The biggest problem for your laser of death is going to be the atmosphere. As Thucydides noted, even at wavelengths specifically chosen to not interact with the atmosphere, a considerable portion of your power is going to be transferred into the air. If your laser is powerful enough to punch through the armour of a tank (as worked out by Gary Walker) then the beam has to be about 22 GJ *20 (minimum attenuation) or 440 GJ. Lets assume that 40 GJ of energy gets through and the rest is lost to atmosphere, so 400 GJ pumped into the atmosphere. A lightning bolt delivers 1-10 GJ and turns the atmosphere to superheated plasma in a fraction of a second. I'm not sure what the energy flow rate is going to be like for your laser, given the atmospheric pressure gradient, distortion, diffraction and other effects, but I think it's fairly likely that with even a conservative estimate of the energies involved you're going to literally set the air on fire.

You're looking at energy densities that can break water molecules apart and then set them on fire, ionise the air and generally do all sorts of bad things to the medium you're trying to push the laser beam through. If the laser makes it harder to push the beam through to the target you'll have to pump in more energy to hit, which will cause even worse things to happen, all the way up to turning the air into plasma and literally forcing it out of the way in order to make a path for the laser beam to travel through. This sets fire to an awful lot of other things too.

In essence: this weapon will not be precise, will require an ungodly power source, and is generally hideous to try and build without anyone noticing.

Or... You could just drop ceramic coated lumps of tungsten with some guidance packages on them... From your altitude you could knock out a city with something the size of a car, a tank with a computer tower, and a person (and his mates) with a tangerine...

  • 1
    $\begingroup$ I wonder if that "literally setting the air on fire" part might in fact be beneficial when using my laser cannon as a weapon of mass destruction instead of a precision assassination device. $\endgroup$
    – Philipp
    Commented Sep 4, 2015 at 12:12
  • $\begingroup$ It depends on exactly how much destruction you're after. You could accidentally set the whole atmosphere ablaze if you pump enough energy in... Look up the XKCD What if on lighting up the moon with a laser if you want a fun and amusing fifteen minutes on why high power lasers can be hilariously bad. $\endgroup$
    – Joe Bloggs
    Commented Sep 7, 2015 at 8:30

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