Assuming that you had:

  • some sort of amazing power plant on the moon that can create an unlimited amount of power
  • a similarly amazing solar panel on earth that can receive an unlimited amount of power
  • and an even more amazing laser on the moon that could accurately send the power to the solar panel on earth

How much power could it send before something catastrophic happened like igniting the earth's atmosphere?

more assumptions:

  • it's a nice, cloud-free day
  • the moon is in perfect alignment with the earth for this shot
  • the laser beam is limited to a circular area with a radius of 1 meter.
  • $\begingroup$ Are you imagining a visible light laser? Or is microwave alright? $\endgroup$ – Samuel Dec 18 '14 at 17:55
  • $\begingroup$ @Samuel preferably the one which allows for more energy to be transferred, microwave would be OK $\endgroup$ – user2813274 Dec 18 '14 at 18:00
  • $\begingroup$ related question, though I like the expansion on it worldbuilding.stackexchange.com/questions/3372/… $\endgroup$ – Twelfth Dec 18 '14 at 19:25
  • $\begingroup$ This seemed surprisingly relevant: what-if.xkcd.com/119 $\endgroup$ – Cort Ammon Dec 19 '14 at 0:39

It depends on two things. The power density and the opacity of the chosen wavelength.

enter image description here

You want to choose a frequency with the minimum opacity, that means that frequency passes through the atmosphere with minimum absorption.

As you can see in the graph, radio waves have very low opacity. Microwaves are the first half of that gap in opacity.

According to this research, the maximum density for microwave power transmission is $1.5 \frac{MW}{cm^2}$

So, for your $1m$ radius receiver, you can get $47.12\,gigawatts$ of power before the air catches on fire.

An aside:

Just to point this out, I'm an electrical engineer, so it bugs me when people to the terms energy and power interchangeably. Energy and power are not the same thing. Think of it is this way:

Energy is how much money you have, power is how fast you spend it.

That is you have an amount of energy, power is the rate at which you use it. Please use the terms correctly.

  • $\begingroup$ So.. for microwaves, plugging in some numbers I get 47100 MW - but I think the power transmission for visible light is higher than 1.5 MW/cm^2, when compared to something like bella (but I don't see it giving how concentrated the beam was.. so I am not sure). $\endgroup$ – user2813274 Dec 18 '14 at 18:46
  • $\begingroup$ @user2813274 I don't know much about BELLA, but I would bet they are firing it in a vacuum and/or that it is not a visible laser. $\endgroup$ – Samuel Dec 18 '14 at 18:59
  • $\begingroup$ @user2813274 Yeah, BELLA is 107.5 um wavelength, far below the visible spectrum. It's also designed specifically to ionize gas. You can't use it to transfer energy through the atmosphere. $\endgroup$ – Samuel Dec 18 '14 at 19:05
  • $\begingroup$ I thought I used the terms correctly (albeit the energy tag is throwing off the title a bit..), but if not feel free to suggest an edit. Also, how did you end up with 1.5 gigawatts of power? $\endgroup$ – user2813274 Dec 18 '14 at 19:20
  • 1
    $\begingroup$ @user2813274 For the most part you used them correctly. Except receiving an unlimited amount of energy is not an interesting metric unless there is a time associated with it, which then becomes power. The 1.5 gigawatts is assuming a 1 m^2 receiver, your calculation for a 1m radius receiver is correct. Just multiply the area, in cm^2, by 1.5 MW/cm^2. $\endgroup$ – Samuel Dec 18 '14 at 19:23

The amount of power transmitted is not the issue if we focus (pun intended) on catastrophic effects only. Linear absorption (which will happen for laser beams in the visual range, as pointed out by another answer) as well as scattering of the laser light in the atmosphere will be an unfortunate loss in your transmission but it's not going to ignite the atmosphere. There are also effects of self-focusing of a high power laser pulses due to non-linear effects of the refractive index, but again keeping the intensity low enough keeps those issues away.

What's problematic is the intensity, that is: the power per unit area - in optics called irradiance. Air would be ionized and turned into a plasma at a certain irradiance. As long as you keep the intensity below that point you're pretty safe. The good thing is, simply be increasing the diameter of the laser beam it is possible to decrease intensity for the same given ammount of power - bonus: the area is of relevance here giving us the diameter squared.

From a quick search (see here) we learn that a laser induced plasma in air is produced by 6 ns, 532 nm Nd:YAG pulses with 25 mJ energy. Peak power is about 4.1 MW in this case. Assuming an focus diameter of 10 µm this leads to an irradiance of about $5*10^{12} \text{ W}/\text{cm}^2$. That's pretty close to the $10^{13} W/cm^2$ I've heard about some time ago.

Calculating the maximum power for a beam diameter of 2 m ($A=31400 \text{ cm}^2$) (not taking divergence into account) with - lets assume $10^{10} \text{ W}/\text{cm}^2$ is a safe value - leads to $314*10^{12} \text{ W}$ that is $314 \text{ Terrawatt}$. Meaning one could transmit the worlds yearly energy consumption in about 445 hours (see Wikipedia, 140 PWh/a, 2008)... or if we push it to the limit and go for $10^{12} \text{ W}/\text{cm}^2$ that's 31 Petawatt and just 4 hours.

  • $\begingroup$ So.. lets assume a circular laser with a radius of 1 meter - any idea what sort of magnitude would be possible to transfer? $\endgroup$ – user2813274 Dec 18 '14 at 17:59
  • $\begingroup$ I can't visit your link but it looks like you're basing your calculations on a laser that would, in fact, ignite the atmosphere it's passing through. Is that right? $\endgroup$ – Samuel Dec 19 '14 at 0:13
  • $\begingroup$ The values given are indeed refering to a laser power that would "ignite" the atmosphere - locally only of course. But that would be a major drawback on the intended idea of optically transmitting power. From those values I guessed (or claimed) that a certain irradiance value would be safe (aka not inducing breakdown of air). This value of course is debatable (and significantly of relevance for the final result). $\endgroup$ – Ghanima Dec 19 '14 at 7:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.