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There was just a question (actually, on Astronomy) about why the data rate from New Horizons is low.

Of course, even with the most directional radio antennas, the spread is enormous.

I guess conceptually, some sort of laser signaling system would have tremendously less spread. (Or zero spread? I don't know).

Since the Earth has crap seeing, I propose

  • On New Horizons, some sort of laser modem

  • On our moon, some sort of laser modem receiving base

What are the numbers here? If (1) New Horizons has a power available of 1 NHPU, how many NHPU would be needed on board for my scheme? Is it overwhelmingly too much, or just "a bit more"? (For that matter, is it much less? Like, "a really good LED" or some such? After all, directional is a fantastic energy saving in the abstract.)

How complex would Moon Base Laser need to be? What I mean is, (2) would it be "surprisingly small" rather than the huge sizes needed with radio antennas. Would it basically be ........... a telescope?? With a $200 nikon camera body glued on, or is "a laser receiver" different somehow than "a telescope with a CCD"?)

So, would Moon Base Laser be a natty device (I'm thinking, say, the size of a car) which we could easily throw on to the moon with current systems, or, would it more be "a large construction". i.e. somewhat like current large Earth telescopes??

I know we already have (tricky) devices which receive lasers bounced off a mirror on the moon ... would one of those be ready-made to do the job, or ...?

enter image description here

With our current awesome radio dishes, they need to get (much) bigger as the spacecraft gets further ... would (3) this be the case with Moon Base Laser or would the issue not be so bad? Would reception / data rate be pretty much the same as the craft moved through our solar system ("since lasers are directional!") or is that all wrong?

Has anyone proposed, or indeed do we (4) already use, laser-style communications in space?

Are there any other narrow-paradigm communication concepts I don't know about, other than "laser"? Which we could use instead of good old dispersing radio??

(I guess proposed multi-craft fleets like LISA indeed do this, right?)

Summary,

  1. how many NHPUs of power would be needed on board for this scheme? what's the order of magnitude?

  2. what size paradigm would Moon Base Receiver be? suitcase-sized? town-sized?

  3. would Moon Base Receiver have to increase in size dramatically w/ distances in the solar system (as do radio dishes)

  4. do we do this already?

PS I'm just assuming this would be worthless from Earth, so, I'm just assuming a Moon receiver (or I guess an orbiting receiver) is a must.

Action update! : the best current answer seems to be "it would offer 100kb/s rather than 1kb/s (using the same power, 12W)"

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  • $\begingroup$ PS I'm already familiar with the "light up the moon" xkcd :) $\endgroup$
    – Fattie
    Jan 3 '19 at 13:11
  • $\begingroup$ Ooops: I misread a page: the photo was taken from a distance of 1.9 million km... $\endgroup$
    – TripeHound
    Jan 3 '19 at 14:40
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    $\begingroup$ Sorry I was massively wrong (I was surprised as well). Googling "distance to Ultima Thule" gave this page which -- to me -- read as the distance to UT from the earth as 1.9 million km. However, this page says it's 6.4 billion km which makes more sense. Which gives a factor of 16,650 times further! $\endgroup$
    – TripeHound
    Jan 3 '19 at 14:49
  • $\begingroup$ sure, I'm often massively wrong :) I'll deleet to clean up $\endgroup$
    – Fattie
    Jan 3 '19 at 14:50
  • $\begingroup$ well, multiplying "2.5" by "about 10,000" we get "about 20 thousand". $\endgroup$
    – Fattie
    Jan 3 '19 at 15:11
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It absolutely could happen, but it would require a more precise pointing than New Horizons has. Lasers of some kind are the best for the high data resolution. The spacecraft to most heavily use lasers in communication is the Lunar Reconnaissance Orbiter. It has also long been talked about as a goal for a Mars communication satellite, which would allow for much more data back from Mars. The problem is the pointing requirements are pretty extreme, you even have to know which site on Earth you are going to target, the laser beam will not cover the entire planet Earth from Mars. For example, MRO has a pointing accuracy requirement of 0.0032 mrad. Pointing requirements for a laser system are in fact similar to this requirement, however, they require stability for much more time. HiRISE only requires it for a few milliseconds, while a laser communication system requires it essentially indefinitely.

In the case of New Horizons, it just isn't needed. Yes, it will take a long time to get the data back, but that isn't a problem, there's lots of time to wait.

As a reference, laser power systems generally require less power then radio based systems, because they broadcast power more directly and thus waste less power.

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  • $\begingroup$ Very interesting, thanks. I'm particularly interested in Q.1 as well ! It's quite an insight that it's "actually not really needed" in the example at hand! Happy new year. $\endgroup$
    – Fattie
    Jan 3 '19 at 13:51
  • $\begingroup$ We do see to be incredibly good at pointing - consider GAIA for instance ....... ?? No? Or is that perhaps a much much bigger system and I just don't know that? $\endgroup$
    – Fattie
    Jan 3 '19 at 13:53
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    $\begingroup$ Added more details. It can certainly be done, but only the highest end imaging satellites have the ability to point that accurately, and many of them for not as long. $\endgroup$
    – PearsonArtPhoto ♦
    Jan 3 '19 at 14:54
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    $\begingroup$ I disagree with your statement about power usage. Lasers, even laser diodes, are not particularly efficient. What they provide is the possiblity of high data rates, but that would require high-data-rate electronics behind the transmitter, which is another power-hungry operation. NH is " running on fumes" and probably couldn't even generate a GHz message even if the laser part was for free. $\endgroup$
    – Carl Witthoft
    Jan 3 '19 at 16:36
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    $\begingroup$ @Fattie The difference between this problem is GAIA is the difference between "knowing where you were pointed" with huge accuracy after a lot of data analysis and "controlling where you are pointing now" with the same accuracy. It is a bit chalk and cheese. $\endgroup$
    – Steve Linton
    Jan 3 '19 at 16:39
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Let's try and do some numbers. We will need to make a few assumptions.I'm going to choose ones which make the calculations easy, varying might produce variation of a factor of 10 or 100 in the answer.

  1. A near IR laser with a wavelength of $1 \mu m$
  2. NH transmitting using the LORRI telescope with an aperture of about 20cm. We know NH can point accurately enough to keep this telescope on target.
  3. A detector in Earth orbit (easier than on the Moon, I think) identical to the JWST with a 6m mirror.
  4. A goal of 10 photons per bit hitting the detector to be reasonably sure of picking the signal from the noise.

So now we get a beam width of $1 \mu m/ 20cm = 5 \mu Rad$

At six billion km this is a beam width near Earth of $30 000 km$.

So our detector picks up $\left(6m/30000km\right)^2 = 4\times 10^{-14}$ of the beam.

So we need transmit about $2.5 \times 10^{14}$ photons per bit.

At this wavelength, one photon has about $2\times 10^{-19}$ Joules of energy, so we must transmit $5\times 10^{-5}$ Joules per bit of laser light.

So now we have a tradeoff of power against data rate. Using the same power (12W) as the current radio transmitter, we could manage about $240 kb/s$ assuming a perfectly efficient laser, probably 10 to 50 % of that in practice. Using 100% of the power from the RTG (190W), we might get 3 Mb/s or so and so on.

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    $\begingroup$ @Fattie I'd say 10-50% of 240 Kbps is the headline. There's a reason NH only uses 12W for transmissions. You need power for computers, heaters, etc. as well as transmission. Also note you need a 10 billion dollar space telescope as the receiver. $\endgroup$
    – Steve Linton
    Jan 3 '19 at 15:14
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    $\begingroup$ @Fattie, but yes, they do wany to shift to laser comms for deep space missions in time. It's just a lot of engineering to get right, and if you mess up you lose a billion dollar space probe. For NH there is no real hurry at all, as someone has already said. $\endgroup$
    – Steve Linton
    Jan 3 '19 at 15:16
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    $\begingroup$ @jpa, also we have a 70m radio dish at this end, and I'm somewhat optimistically assuming a 6m IR dish. $\endgroup$
    – Steve Linton
    Jan 3 '19 at 15:17
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    $\begingroup$ @Fattie Due to scattered light, I wouldn't be surprised if it infrared- and visible light communication at that power level and distance was altogether impossible from earth surface. $\endgroup$
    – jpa
    Jan 3 '19 at 15:30
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    $\begingroup$ For IR it makes a huge difference how high you are. Your main enemy is water vapour. From a typical astronomy site in a high desert it won't make too much difference. On the other hand a 6m ground based telescope is also quite expensive, and you need NH to be visible. $\endgroup$
    – Steve Linton
    Jan 3 '19 at 15:48
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Steve Linton's answer is excellent, although possibly a bit conservative. Information has been transmitted in the lab via laser at a rate of 1 bit per photon. For proposed uses, Error Detection and Correction codes are definitely indicated.

Has anyone proposed, or indeed do we (4) already use, laser-style communications in space?

Yes to both, although just barely. In 2013, the Lunar Laser Communication Demonstration successfully operated in a LADEE mission. Lead times for exploiting technology in deep space missions are very long, at least a decade, so it will be a while before actual missions using the technology are launched. Nonetheless, astronomers are salivating at the thought of high-resolution, high frame rate imagery.

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  • $\begingroup$ that's a great link! thanks. I'm pleased to see we have a LASERCOMM ! :) $\endgroup$
    – Fattie
    Jan 3 '19 at 20:09
  • $\begingroup$ @Fattie - Actually, we've had LASERCOMM for a coupe of decades now. Military comm satellites relay data via laser. $\endgroup$
    – WhatRoughBeast
    Jan 4 '19 at 17:34
  • $\begingroup$ awesome! we need a MOONLASERCOMM $\endgroup$
    – Fattie
    Jan 4 '19 at 18:05
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This is closely related to the concept of Antenna gain, which for radio transmission measures how narrow a beam the antenna can focus. The narrower the beam, the more accurately you need to point it.

However, any laser or radio beam will still be divergent - it will spread out the further away it goes. And it will spread out at the same rate, with the beam diameter doubling every time the distance doubles. This leads to area of the beam increasing by the square of the distance, giving the Inverse-square law.

A narrower beam only improves the initial signal power - e.g. 10x smaller beam area will give 10x transmission power. When you get to the Kuiper belt, signal will be only 1/400 000 000th as strong as it would be in lunar orbit. A narrow-beam improvement of 10x or even 100x pales in comparison to that.

Or to put it more simply: a narrow beam is faster, but when you go that far, it will be slow anyway.

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  • $\begingroup$ ? but lasers suffer millions or billions of times less spread than the best directional radio. $\endgroup$
    – Fattie
    Jan 3 '19 at 14:45
  • $\begingroup$ wait - lasers do not spread out with the inverse square law. Right? $\endgroup$
    – Fattie
    Jan 3 '19 at 14:46
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    $\begingroup$ "A narrow-beam improvement of 10x or even 100x pales in comparison to that." .... wait .. so you're saying the suggested scheme would "only" (what?!) offer a 100x improvement ?? I'm sorry, I'm not really getting with this answer because it doesn't really deal with the issue: what is the potential order of magnitude improvement, how much (energy, etc) would it cost. $\endgroup$
    – Fattie
    Jan 3 '19 at 14:49
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    $\begingroup$ @Fattie they do, It's just that a small laser (like a laser pointer) can create a beam that only spreads noticeably over a long distance (miles), whereas you need quite a big radio antenna to do the same job. The number that matters is the ration of the wavelength of the radiation to the diameter of the telescope/antenna. For a green laser pointer with a 1mm lens, that ration is about 2000. To achieve the same ration for 2.4GHz (WiFi) you'd need a dish 200 meters across (one of the world's biggest radio telescopes). $\endgroup$
    – Steve Linton
    Jan 3 '19 at 14:51
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    $\begingroup$ @Fattie It's still inverse square -- go twice as many miles and it spreads out over four times as many (square) meters. The power law is the same, the constant is just lower $\endgroup$
    – Steve Linton
    Jan 3 '19 at 15:11
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If you want to point a laser exactly at a point at earth, then you would need to know the exact transverse momentum of each photon. The uncertainty principle says that for that you must have infinite uncertainty about its transverse position, which means that you need an infinitely large transmitter.

As a rough, order of magnitude estimate of the spread of the beam, you can take (distance beam travels)*(wavelength of light)/(width of transmitter). New Horizons is 6*10^12 m away, and about 2 m wide. Visible light has a wavelength of about 5*10^-7. So a laser in the visible light spectrum has a lower bound of spread of about 3000 km. Thus, to answer your second question, even if you could build a laser with very low spread, and target a point on earth precisely, it would violate basic physics to hit a receiver exactly, unless the receiver is about the same size as the moon, and to answer your third question, as New Horizons gets further away the size would increase. And the smaller the wavelength, the more power it takes to create each photon.

So the answer to your third question is yes, even a laser would require

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    $\begingroup$ @JMac You don't know the exact position of the photons, no. You can't aim a laser exactly at a point. That's the point of my answer. $\endgroup$
    – Acccumulation
    Jan 3 '19 at 18:09
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    $\begingroup$ @Acccumulation But how does pointing it "exactly at a point on Earth" give you the "exact momentum of each photon"? Wouldn't that, by definition, be giving you an exact position of each photon? It seems like a conflicting statement. $\endgroup$
    – JMac
    Jan 3 '19 at 18:59
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    $\begingroup$ @JMac Emitting a photon pointed exactly at a point on earth means that, at the time of emission, you know the momentum of that photon exactly. $\endgroup$
    – Acccumulation
    Jan 3 '19 at 19:06
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    $\begingroup$ @Acccumulation Wouldn't you also need to know it's position exactly to accomplish that? I'm not arguing that it's possible, just that the wording seems really poor compared to what you are trying to get across. $\endgroup$
    – JMac
    Jan 3 '19 at 19:30
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    $\begingroup$ This answer seems to assume that lasers would be completely pointless unless essentially all of the beam was received, and uses a rough approximation of diffraction-limited spot size to demonstrate how impractical it would be to receive the whole beam. But since the usual radio communications receive only a miniscule fraction of the transmitted power, even receiving 0.01% of the laser beam (say, with a 5-meter reflector somewhere in the 500-meter spot) would be a huge improvement. $\endgroup$
    – Nathan Tuggy
    Jan 4 '19 at 10:24

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