8
$\begingroup$

Given that a laser beam is made up of coherent light waves running in parallel in the same direction, and that space is not a complete vacuum (dust, radiation, electromagnetic forces etc.), what kind of effective range could a laser weapon have in space?

And what form would a laser take, once technology has progressed to the point where a laser is reliably weaponisable? Power requirements and wavelength?

Additionally, what form of defense could a spacecraft use against lasers? A mirrored hull? A thick ice shield?

I'm assuming an initial contact distance of several hundred kilometers (call it 150 miles if you like), closing as the two combatants approach each other. Is this a realistic expectation?

The environment in question is in the asteroid belt and the ship is powered by a nuclear power source.

$\endgroup$
7
  • 3
    $\begingroup$ What kind of space? LEO has a great deal more material than interstellar space does. $\endgroup$
    – Cort Ammon
    Jan 3, 2016 at 16:32
  • 1
    $\begingroup$ For defense, a "time reversed mirror" based on superconductors so it doesn't have any problems with the high power: the beam is reflected back along the incoming path, where it destroys the laser! Given the vast range of space, the ship may have moved a little so it hits the attacking ship near the gunport rather than messing up the laser amplification stage. $\endgroup$
    – JDługosz
    Jan 3, 2016 at 18:30
  • 1
    $\begingroup$ There is no mechanical movement. It's simply a material that reacts to the incoming E field of the light, just like a common metal does to be a mirror. It does not tilt mechanically; it just reflects, but differently than a (regular) mirror. $\endgroup$
    – JDługosz
    Jan 3, 2016 at 18:58
  • 1
  • 1
    $\begingroup$ It occurs to me that if you use metamaterials to implement time-reveral reflection, that's the same kind of tech you would use to make something invisible. A laser won't damage an invisible hull, right? So more generally, use metamaterials containing suoerconductive elements to absorb and "get rid of" the E field of any incoming radiation, no matter how intense. $\endgroup$
    – JDługosz
    Jan 3, 2016 at 19:06

2 Answers 2

9
$\begingroup$

The equation for the power emitted by a laser beam at a distance in a circle of radius $r$ at a distance $z$ where the beam diameter is $w(z)$ is $$P(r,z)=\frac{1}{2}\pi I_0w_0^2\left(1-e^{-2r^2/w^2(z)}\right)$$ where $I_0$ is the initial intensity and $w_0$ is the initial beam diameter (see these course notes). If we assume that the entire diameter of the beam hits the target, then we can set $r=w(z)$ and get $$P=\frac{1}{2}\pi I_0w_0^2\left(1-e^{-2}\right)$$ We could write $I_0$ as a function of the electric field amplitude $E$ and the characteristic impedance $\eta$ (see this presentation), but it might be better to just work off of numbers. At any rate, this equation assumes that all of the beam hits the target, which isn't necessarily the case. Were we to use the world's most powerful laser, we could get around 1.3$\times$1015 watts of power - for half of a trillionth of a second.

Let's think about beam divergence, and the Rayleigh length. This is the value of $z$ for which $w(z)=2w_0$. It is given by $$z_R=\frac{\pi w_0^2}{\lambda}$$ The NIF laser operates at a wavelength of 351 nm (Haynam et al. (2007)), with all 192 beams being focused through a hole less than 1 mm in diameter. Assuming that our lasers have beam widths of about this much, then we have a $z_R$ of ~22 meters. That's not good.

However, this is because the beams must be so small. The Boeing YAL-1 could have been effective at up to 300 kilometers (see a summary of a report). So if we up the power, then we could in theory get results more like the ones seen at NIF - really, really explosive. A couple hundred kilometers should be achievable.

Laser shielding is a whole different problem. At these energies, there's not a whole lot that can stop these lasers. Most things will catch fire or blow up (or both). Heck, that's why the NIF uses them!

One option is to use a shield of "trash" - basically, laser cannon fodder. It gradually gets eaten away by laser attacks.

Problems:

  1. It blocks the vessel from doing anything (seeing the opponent, launching missiles, etc.).
  2. It's temporary, must be replaced, and may not last long.

A second solution might be to use a shielding gas. This is commonly used in industrial welding to absorb some heat from welding lasers. I have absolutely no idea if it could work. You would likely need to rig up a magnetic field to contain it (if possible), and it would obscure visible light. But it might be better than nothing.

$\endgroup$
5
  • $\begingroup$ Sterling effort. Unfortunately my mathematical abilities these days don't extend much past time/speed/distance calculations, and a shallow misunderstanding of the coefficient of lift. If I've got this straight, the petawatt laser you describe is immensely powerful, but suffers from too much divergence and therefore a lack of range? And the YAL-1 potentially has the range, but needs more power to be effective for a space battle? What is holding the YAL-1 from being upgraded from a megawatt level to a petawatt level? Fuel, heat? $\endgroup$
    – Smoj
    Jan 3, 2016 at 18:33
  • 1
    $\begingroup$ @Smoj The reason for a short range is that the NIF lasers (all 192 of them) need to be focused on a small target, so they have to have a small initial (and final) beam diameter. Therefore, they'll have a smaller Rayleigh length even though divergence is small. The COIL laser used by YAL was never made more powerful because the program was cancelled, but weight would have been an issue. The NIF laser achieves the petawatt level by using a huge number of lasers. The YAL only used one (powerful) laser. $\endgroup$
    – HDE 226868
    Jan 3, 2016 at 18:36
  • $\begingroup$ That makes sense... Could the size of the COIL be reduced at all? I'm guessing the bulk of it's size was in order to store and accelerate the gasses used, so probably not. A more powerful laser would need much more gas as fuel. Some economy of scale, but not much I assume... $\endgroup$
    – Smoj
    Jan 3, 2016 at 18:47
  • $\begingroup$ @Smoj I don't know much about the specifications of it, so I couldn't say. $\endgroup$
    – HDE 226868
    Jan 3, 2016 at 18:59
  • $\begingroup$ @Smoj I updated the answer to address the second part of the question (shielding). $\endgroup$
    – HDE 226868
    Jan 5, 2016 at 23:37
6
$\begingroup$

Hde's answer can only really be expanded on at the Atomic rockets site (http://www.projectrho.com/public_html/rocket/spacegunconvent.php) which has links to many other sites, calculation tables and worked examples, so I will talk about defense.

The short answer is that most ideas of defense against laser weapons are rubbish. A mirrored hull could reflect 99% of the incoming beam, but the energy impulse will be so short and sharp that the remaining 1% will damage the mirror and allow subsequent shots to burn through the hull. Ablative materials will be rapidly burned away and even rotating the spacecraft will just cause a crazy quilt of burn lines to cover the ship before something fails (and a rotating spacecraft will already be under a great deal of stress).

If you have something which can stop a particular laser, the enemy will be working on a more powerful laser weapon (bigger mirror, shorter wavelengths, the same output packed into shorter beam emissions etc.) with the ultimate weapon being the RBoD (Ravening Beam of Death); a Free Electron Laser with an accelerator a kilometre in diameter capable of delivering a beam of x-rays out to one light second (almost the distance from the Earth to the Moon). A full description is at the end of the laser section at Atomic Rockets (above).

So if he can vaporize you from a light second away, and cause considerable damage out to a light minute, how do you respond?

Defensively, you would have to carry a massive amount of material to absorb the incoming energy. A huge ice shield will absorb much of the incoming energy, and the heat will be dissipated in the mass of ice. The advantage of this is the high energy of the beam will likely vapourize and ionize the water, creating a plasma which will tend to absorb even more energy and travel "up the beam", blocking the laser's punch (although if the RBoD is a light second away, this don't do you much good in the long run). The downside of this is it becomes extremely energy intensive to move your ship at all, much less make combat manoeuvres, so the RBod can gradually cut your ship to pieces. A ship can be built relatively cheaply and easily from water ice (see http://neofuel.com/iceship/index.html), so you might try to overwhelm the RBoD with sheer weight of numbers.

This leads to the other approach, which is active defense, or fighting fire with fire. Rick Robinson's Rocketpunk Manifesto site (http://www.rocketpunk-manifesto.com) often talked about "eyeball frying contests", where lasers attempt to fire down the optical paths of enemy lasers and destroy the optical train. This makes a large number of assumptions which can be questioned, but is a reasonable idea in principle. Counters to that include using external bomb pumped lasers as warheads on drones and missiles, or randomly directing the optical train to different fighting mirrors between shots.

The other approach is to attempt to overwhelm the RBoD with thousands of incoming kinetic energy warheads. These have been given the nickname "Soda Cans of Death" (SCoD), indicating they are about the size of a soft drink can delivered in massive quantities. Since even small objects moving at orbital and interplanetary velocity have massive amounts of kinetic energy, the impact of even one could cause considerable damage, so the fire control system should be programming the RBoD to try to clear the sky of incoming SCoDs, rather than shooting at you. At some point, there are more targets than the RBoD can deal with (cycle times, overheating, repositioning the mirror etc.), so it goes down under a hail of KE strikes. (Too bad they were also shooting busses full of SCoDs back at you at the same time....).

Space war will be very messy, and involve filling the sky with thousands of sensors, weapons and mutually supporting systems (you don't really think there is only one RBoD out there, do you?). I suspect the great expense of all of this will make any real space war more theoretical, much like WWI dreadnoughts spent much of the war in heavily protected bases, glowering at each other across the North Sea because the risk of losing a battle was so great compared to the potential gains of winning.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .