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This is almost certainly an oversimplified question born out of ignorance, but I've always wondered why we have to rely on the light of the Sun to find distant,

This is almost certainly an oversimplified question born out of ignorance, but I've always wondered why we have to rely on the light of the Sun to find distant, dark objects. Why would it not be possible to fashion an extremely powerful laser and sweep it across the sky in an area where gravitational clues indicate a dark object may reside? Unless the object was utterly black on its surface, I'd think we would be able to monitor wherever the beam traverses and look for tell-tale spectra coming back in our direction due to the laser being scattered by a solid object's surface. Again, I know I am oversimplifying a complex problem, and I could speculate on factors that might render this impractical, such as atmospheric effects or the necessary power needed for a laser strong enough to travel to such a distant point (it might not be physically possible to create a laser large enough to exceed the light we would already see reflected back from the already formidable output of the Sun), but I don't have the requisite knowledge of physics to really make solid assumptions about such things.

avar 8y ago
Relative to planet #9 we're right next to a giant nuclear fusion powered light 700,000 km across. If it isn't lighting up the surface of planet #9 well enough for us to see it already, then we're not going to be able to construct some laser that makes a difference. That's before you get to the problem that space is ridiculously big, and sweeping it across the entirety of the surface of the imaginary sphere that might contain planet #9 would probably take millions of years (citation needed).
- Rooster61 8y ago
  > That's before you get to the problem that space is ridiculously big, and sweeping it across the entirety of the surface of that imaginary sphere would probably take millions of years. The scale isn't lost on me. I'm aware of how much ground would need to be covered when you get out that far, and I figured that might be a pretty damning factor. That said, you wouldn't need to sweep it across the entire surface, only enough of the surface to rule out any interfering debris between us and the object. Wouldn't take much. We would need VERY accurate models of the orbit based on effects on other bodies, but it's hard to totally rule it out in my head.
  - snissn 8y ago
    I recently listened to an episode of Sean Carroll's podcast with the planet nine astronomer interviewed. Basically they're working on finding it, but their math can only limit the location of the planet to a specific orbit, and they need to check all possible angles of that orbit and it is a lot of the sky to look through. Edit - podcast link https://www.preposterousuniverse.com/podcast/2018/08/27/episode-11-mike-brown-on-killing-pluto-and-replacing-it-with-planet-9/ https://www.preposterousuniverse.com/podcast/2018/08/27/epis...
    - Rooster61 8y ago
      Damn, you just put a hole in my already utterly impractical scheme. CURSES!!! (twirls mustache) :) In all seriousness, that would definitely kill any possibility. Sweeping an entire orbit at the resolution needed to hit something smaller than Uranus/Neptune would take eons.
      - stcredzero 8y ago
        Space is so hugely, mind-bogglingly big, that even paring it down to small fractions using clever orbital analyses still yields hugely mind-bogglingly big regions of space. I mean, you may think it's a long way down the road to the chemist's, but that's just peanuts to space. I think someone should re-imagine all of Star Wars as hard Sci-fi set in the vicinity of a system like Alpha Centauri. That's still a vast setting.
        masonic 8y ago
        "Space is big. You just won't believe how vastly, hugely, mind- bogglingly big it is. I mean, you may think it's a long way down the road to the chemist's, but that's just peanuts to space." Douglas Adams, The Hitchhiker's Guide to the Galaxy
        stcredzero 8y ago
        Even as old as it is, I didn't think it necessary to explain my reference.
- LeifCarrotson 8y ago
  Regarding those two points: A very good laser has a beam divergence of tenths of milliradians, or 0.01 degrees. At the radius of planet 9, this is a very large beam, and we'd have a hard time putting out enough power to compete with the sun. Furthermore, the sky has an area of 40,000 square degrees. Our little laser beam projects a circle with an area of 0.0000785 square degrees. You'd need to aim at 500 million spots to see the whole sky.
  - avar 8y ago
    > You'd need to aim at 500 million spots to see the whole sky. That's smaller than I expected. To convert that into human scale sqrt(500 million) = 22360, in millimeters that's around 22 meters (72 ft). So if we had a square wall 22 meters by 22 meters we could paint a grid of 1 mm by 1 mm squares on it and end up with around 500 million squares. Then we'd need to stand there in front of the wall with a laser pointer and shine the laser at each individual square and check the light bouncing back at us.
    - hughes 8y ago
      Don't forget - due to the speed of light it takes 5-10 hours to get feedback on if the square illuminated, but maybe more because you don't know exactly how long you'll need to wait. And there's uncertainty in whether your measurement is accurate. And there's a background of other illuminated squares that can interfere with your readings. And you're spinning while this is going on. And the target square is continually moving. You might have just missed it.
  - burtonator2011 8y ago
    this has an easy solution. 1. Build a Dyson sphere around the sun. 2. Capture all the mass in the solar system to build a giant freaking laser. 3. Use the energy to from the Dyson sphere to power the laser. 4. Melt your entire civilization.
    - nerfhammer 8y ago
      If it's a super-earth it probably has an iron core. We should be able to use magnets to crash it into the sun. The impact event will be hard to miss. We'll know where it is for sure then. This is a lot less work than building a Dyson sphere.
- thaumaturgy 8y ago
  For those interested in the notion of using lasers to look at things in space, this XKCD what-if might be helpful: https://what-if.xkcd.com/13/ https://what-if.xkcd.com/13/
  - Rooster61 8y ago
    This was actually posted a bit further down in the post. It's a good one though :P
- dyukqu 8y ago
  Is it my English or is it 1.39 million kilometers across?
  - avar 8y ago
    You're right. This was from memory and I'd mixed up radius & diameter. Radius 695,700 km, diameter 1,391,400 km.
arayh 8y ago
This is probably an oversimplified explanation, but this xkcd what-if will probably help you quantify the problem of generating light on the level of Sun's output: https://what-if.xkcd.com/13/ https://what-if.xkcd.com/13/
simcop2387 8y ago
Part of the problem is that there's a multiple day reflection time that makes things complicated for sweeping across the sky. It's certainly solvable but with a minimum 26-140 hour (100-500 au round trip) lag between firing the laser and then receiving the signal, we'd need to observe each swept area of the sky for a week or two at a time to see the return blip. I'd also then guess that a radar setup instead of visible light might work better, but that's also got resolution issues and would need a gigantic and incredibly sensitive receiver to start picking things up. Space based systems for doing all this would probably work better since they wouldn't have to contend with the rotation of the earth itself, but it'd still be hard.
teraflop 8y ago
One possibly non-obvious problem with this idea is that laser light will also spread out over very long distances, just like the light from the sun does. Any light source is subject to diffraction. The narrower the beam aperture, the worse the diffraction. As a quick back-of-the-envelope calculation: assuming you have a gigantic 100-meter diffraction-limited aperture, the smallest possible beam divergence for green light (550nm wavelength) is roughly 2 nanoradians. In the limit of large distances from both the earth and the sun, the area covered by that beam would already be receiving about 1GW of total energy from the sun. To outshine this you would need a laser of comparable power or more; the most powerful continuous laser I could find a reference to is only about 1MW, and was designed to destroy satellites and cruise missiles. [1] I wouldn't rule out the possibility that you could detect a reflected laser beam, with sufficiently sensitive detectors, but it's hard to see what you would gain as opposed to just using sunlight. Furthermore, lasers are by nature monochromatic, so you wouldn't really be able to do spectroscopy. At best, you could collect a few different points on the spectrum using lasers of different wavelengths. [1]: https://en.wikipedia.org/wiki/MIRACL https://en.wikipedia.org/wiki/MIRACL
- Rooster61 8y ago
  The light wouldn't need to be continuous. You would only need a fraction of a second to catch the blip. So assuming that, and that the most powerful nanopulse laser on earth is ~500 Terawatts, I'd think you could have a laser with a discernible pulse at about the 100 Terawatt level (This is an ass-pull number, but so is this entire excercise so who cares anyways :D )
  - wongarsu 8y ago
    It doesn't need to be continuous, but we take pictures by looking at all the light over some short-ish time. Even at 60fps we are talking about up to 16ms of light, and with telescopes exposure times of minutes or hours are the norm. Unless the pulses are long or frequent you get more light from an equally long exposure with sun light
    - kondro 8y ago
hwillis 8y ago
> Why would it not be possible to fashion an extremely powerful laser and sweep it across the sky in an area where gravitational clues indicate a dark object may reside? Lasers are not the important part here- collimation is. The closer the illumination is to a perfect cylinder, the more visible it stays. Lasers often happen to be highly collimated, but it's not a defining feature. For instance laser diodes emit laser light in a cone 20+ degrees wide, and lenses are needed to produce a beam. Bottom line: lasers aren't a panacea- generating a collimated beam isn't trivial, and just because we can make immensely powerful lasers that doesn't mean they are collimated. > Unless the object was utterly black on its surface, I'd think we would be able to monitor wherever the beam traverses and look for tell-tale spectra coming back in our direction due to the laser being scattered by a solid object's surface. There isn't really any tell-tale spectrum available- obviously you need something below X-rays, because those will be absorbed rather than reflected. In practice you also need something above millimeter waves, because those are also absorbed but more importantly the collimation of an EM source is inversely proportional to its wavelength. So millimeter waves are ~2000x less collimated than visible light, unless you use a massive antenna. Just above millimeter waves are infrared waves, and just below X-rays are ultraviolet waves. Along that entire range, the sun is very active, so any light is going to be washed out against that background. You don't really get to rely on the spectra being unique. > the necessary power needed for a laser strong enough to travel to such a distant point (it might not be physically possible to create a laser large enough to exceed the light we would already see reflected back from the already formidable output of the Sun) The reflected light of the laser has to be brighter than the suns reflection over the entire area of the object you're trying to see. For an Earth-sized object at 1000 AU, that's 174 gigawatts. Not only do you have the impossible task of actually hitting the object, you need to have a hundred nuclear reactors powering a single source the entire time you do it.
- regularfry 8y ago
  Not to say this is in any way practical, but lasers used for nuclear fusion research are up in the petawatt range. They've just got extremely short pulses. But if all you need to do is detect that pulse coming back, like a radar return, it starts to sound within the range of conceivable engineering, not a total physical impossibility.
mikeash 8y ago
It’s worth noting that this sort of thing is done for closer objects: https://en.wikipedia.org/wiki/Radar_astronomy https://en.wikipedia.org/wiki/Radar_astronomy The problem is that the power requirements go up sharply with distance. The amount that hits the object decreases with the square of the distance, as does the amount that’s returned. The end result is that your power requirements are proportional to the fourth power of the distance. Twice as far means you need sixteen times more power. Now extrapolate that to something hundreds of times farther away than the closer planets....
perkee 8y ago
For a sense of the scale of this problem, the retro reflectors that we left on the moon return 1 photon out of the 10^17 we shoot at them and that's not even that reliable. This thing would be a mind bendingly further distance away and would not have any retro reflectors on it (oh man it'd be WILD if it did though!!). https://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment#Principle https://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment...
- Rooster61 8y ago
  This is the experiment that put the idea in my head in the first place. And also the reason I prefaced the question with the fact that I KNOW it's going to need to be a ridiculously powerful laser. Making something detectable that's hitting something with an unknown albedo and angle of whatever the photons are impacting makes the requirements start to bump up against the laws of physics, much less what we are technically capable of.
stmfreak 8y ago
Assuming you can build a laser bright enough for your detector, you are forgetting about the time & distance problem. We don’t know how far away this Planet X sits at any given point in time. You would have to pulse your laser at a spot or set of spots long enough to irradiate the possible region of space during the time it takes for Planet X to traverse said space. Then wait hours or days on the off chance the pulse comes back. If you look away too soon to examine the next spot, you miss the more distant object. If your pulse is too short, the object may pass in front or behind the pulse and be invisible. The Sun on the other hand is irradiating all space at all times. Aside from eclipses, we know Planet X is currently reflecting sunlight. We just have to look at it, compare to background, look again, compare to background, and notice the movement.
- Rooster61 8y ago
  > you are forgetting about the time & distance problem Not forgotten, just glossed over. You'd need to have a receiver that is pointing at a given (admittedly massive) part of the sky over a given range of possible times that a return photon would take to traverse the gap back to earth and strike the receiver. > We just have to look at it You're right, but the problem is that something orbiting at that distance follows an EXTREMELY slow orbit, so it's difficult to use the method you are describing. It's quite effective for looking for asteroids/space junk, but at that distance it becomes pretty tough to spot.
yongjik 8y ago
Imagine you're sitting relaxed in a dark, forgotten corner of asteroid X, five billion kilometers from the sun. Imagine looking up and seeing the feeble sunlight. Some of it gets reflected back and starts a long journey toward some telescope on Earth. Now imagine seeing "extremely powerful laser" from earth. Some of it gets reflected ... (ditto). For your method to work, the latter must be more powerful than the former. In other words, if you point your laser at Mars, a Martian should suddenly find the Earth shining brighter than the sun. I don't believe mankind possesses such power yet (thank god).
dotancohen 8y ago
Even assuming a super-strong light beam (brighter than the sun), and ignoring beam divergence, the time that the light returns to Earth would be dependent on the distance to the object. So if we don't know how far away it is, we won't know when to look back at the spot that the laser was aimed at.
TheOtherHobbes 8y ago
Empirical law: “Whenever there’s an engineering problem in space, at least one person on HN will suggest it can be solved with a big enough laser.”
- Rooster61 8y ago
  I aim to please :)

avar 8y ago

Relative to planet #9 we're right next to a giant nuclear fusion powered light 700,000 km across. If it isn't lighting up the surface of planet #9 well enough for us to see it already, then we're not going to be able to construct some laser that makes a difference. That's before you get to the problem that space is ridiculously big, and sweeping it across the entirety of the surface of the imaginary sphere that might contain planet #9 would probably take millions of years (citation needed).

Rooster61 8y ago
> That's before you get to the problem that space is ridiculously big, and sweeping it across the entirety of the surface of that imaginary sphere would probably take millions of years. The scale isn't lost on me. I'm aware of how much ground would need to be covered when you get out that far, and I figured that might be a pretty damning factor. That said, you wouldn't need to sweep it across the entire surface, only enough of the surface to rule out any interfering debris between us and the object. Wouldn't take much. We would need VERY accurate models of the orbit based on effects on other bodies, but it's hard to totally rule it out in my head.
- snissn 8y ago
  I recently listened to an episode of Sean Carroll's podcast with the planet nine astronomer interviewed. Basically they're working on finding it, but their math can only limit the location of the planet to a specific orbit, and they need to check all possible angles of that orbit and it is a lot of the sky to look through. Edit - podcast link https://www.preposterousuniverse.com/podcast/2018/08/27/episode-11-mike-brown-on-killing-pluto-and-replacing-it-with-planet-9/ https://www.preposterousuniverse.com/podcast/2018/08/27/epis...
  - Rooster61 8y ago
    Damn, you just put a hole in my already utterly impractical scheme. CURSES!!! (twirls mustache) :) In all seriousness, that would definitely kill any possibility. Sweeping an entire orbit at the resolution needed to hit something smaller than Uranus/Neptune would take eons.
    - stcredzero 8y ago
      Space is so hugely, mind-bogglingly big, that even paring it down to small fractions using clever orbital analyses still yields hugely mind-bogglingly big regions of space. I mean, you may think it's a long way down the road to the chemist's, but that's just peanuts to space. I think someone should re-imagine all of Star Wars as hard Sci-fi set in the vicinity of a system like Alpha Centauri. That's still a vast setting.
      - masonic 8y ago
        "Space is big. You just won't believe how vastly, hugely, mind- bogglingly big it is. I mean, you may think it's a long way down the road to the chemist's, but that's just peanuts to space." Douglas Adams, The Hitchhiker's Guide to the Galaxy
        stcredzero 8y ago
        Even as old as it is, I didn't think it necessary to explain my reference.
LeifCarrotson 8y ago
Regarding those two points: A very good laser has a beam divergence of tenths of milliradians, or 0.01 degrees. At the radius of planet 9, this is a very large beam, and we'd have a hard time putting out enough power to compete with the sun. Furthermore, the sky has an area of 40,000 square degrees. Our little laser beam projects a circle with an area of 0.0000785 square degrees. You'd need to aim at 500 million spots to see the whole sky.
- avar 8y ago
  > You'd need to aim at 500 million spots to see the whole sky. That's smaller than I expected. To convert that into human scale sqrt(500 million) = 22360, in millimeters that's around 22 meters (72 ft). So if we had a square wall 22 meters by 22 meters we could paint a grid of 1 mm by 1 mm squares on it and end up with around 500 million squares. Then we'd need to stand there in front of the wall with a laser pointer and shine the laser at each individual square and check the light bouncing back at us.
  - hughes 8y ago
    Don't forget - due to the speed of light it takes 5-10 hours to get feedback on if the square illuminated, but maybe more because you don't know exactly how long you'll need to wait. And there's uncertainty in whether your measurement is accurate. And there's a background of other illuminated squares that can interfere with your readings. And you're spinning while this is going on. And the target square is continually moving. You might have just missed it.
- burtonator2011 8y ago
  this has an easy solution. 1. Build a Dyson sphere around the sun. 2. Capture all the mass in the solar system to build a giant freaking laser. 3. Use the energy to from the Dyson sphere to power the laser. 4. Melt your entire civilization.
  - nerfhammer 8y ago
    If it's a super-earth it probably has an iron core. We should be able to use magnets to crash it into the sun. The impact event will be hard to miss. We'll know where it is for sure then. This is a lot less work than building a Dyson sphere.
thaumaturgy 8y ago
For those interested in the notion of using lasers to look at things in space, this XKCD what-if might be helpful: https://what-if.xkcd.com/13/ https://what-if.xkcd.com/13/
- Rooster61 8y ago
  This was actually posted a bit further down in the post. It's a good one though :P
dyukqu 8y ago
Is it my English or is it 1.39 million kilometers across?
- avar 8y ago
  You're right. This was from memory and I'd mixed up radius & diameter. Radius 695,700 km, diameter 1,391,400 km.

arayh 8y ago

This is probably an oversimplified explanation, but this xkcd what-if will probably help you quantify the problem of generating light on the level of Sun's output: https://what-if.xkcd.com/13/ https://what-if.xkcd.com/13/

simcop2387 8y ago

Part of the problem is that there's a multiple day reflection time that makes things complicated for sweeping across the sky. It's certainly solvable but with a minimum 26-140 hour (100-500 au round trip) lag between firing the laser and then receiving the signal, we'd need to observe each swept area of the sky for a week or two at a time to see the return blip. I'd also then guess that a radar setup instead of visible light might work better, but that's also got resolution issues and would need a gigantic and incredibly sensitive receiver to start picking things up. Space based systems for doing all this would probably work better since they wouldn't have to contend with the rotation of the earth itself, but it'd still be hard.

teraflop 8y ago

One possibly non-obvious problem with this idea is that laser light will also spread out over very long distances, just like the light from the sun does. Any light source is subject to diffraction. The narrower the beam aperture, the worse the diffraction. As a quick back-of-the-envelope calculation: assuming you have a gigantic 100-meter diffraction-limited aperture, the smallest possible beam divergence for green light (550nm wavelength) is roughly 2 nanoradians. In the limit of large distances from both the earth and the sun, the area covered by that beam would already be receiving about 1GW of total energy from the sun. To outshine this you would need a laser of comparable power or more; the most powerful continuous laser I could find a reference to is only about 1MW, and was designed to destroy satellites and cruise missiles. [1] I wouldn't rule out the possibility that you could detect a reflected laser beam, with sufficiently sensitive detectors, but it's hard to see what you would gain as opposed to just using sunlight. Furthermore, lasers are by nature monochromatic, so you wouldn't really be able to do spectroscopy. At best, you could collect a few different points on the spectrum using lasers of different wavelengths. [1]: https://en.wikipedia.org/wiki/MIRACL https://en.wikipedia.org/wiki/MIRACL

Rooster61 8y ago
The light wouldn't need to be continuous. You would only need a fraction of a second to catch the blip. So assuming that, and that the most powerful nanopulse laser on earth is ~500 Terawatts, I'd think you could have a laser with a discernible pulse at about the 100 Terawatt level (This is an ass-pull number, but so is this entire excercise so who cares anyways :D )
- wongarsu 8y ago
  It doesn't need to be continuous, but we take pictures by looking at all the light over some short-ish time. Even at 60fps we are talking about up to 16ms of light, and with telescopes exposure times of minutes or hours are the norm. Unless the pulses are long or frequent you get more light from an equally long exposure with sun light
  - kondro 8y ago

hwillis 8y ago

> Why would it not be possible to fashion an extremely powerful laser and sweep it across the sky in an area where gravitational clues indicate a dark object may reside? Lasers are not the important part here- collimation is. The closer the illumination is to a perfect cylinder, the more visible it stays. Lasers often happen to be highly collimated, but it's not a defining feature. For instance laser diodes emit laser light in a cone 20+ degrees wide, and lenses are needed to produce a beam. Bottom line: lasers aren't a panacea- generating a collimated beam isn't trivial, and just because we can make immensely powerful lasers that doesn't mean they are collimated. > Unless the object was utterly black on its surface, I'd think we would be able to monitor wherever the beam traverses and look for tell-tale spectra coming back in our direction due to the laser being scattered by a solid object's surface. There isn't really any tell-tale spectrum available- obviously you need something below X-rays, because those will be absorbed rather than reflected. In practice you also need something above millimeter waves, because those are also absorbed but more importantly the collimation of an EM source is inversely proportional to its wavelength. So millimeter waves are ~2000x less collimated than visible light, unless you use a massive antenna. Just above millimeter waves are infrared waves, and just below X-rays are ultraviolet waves. Along that entire range, the sun is very active, so any light is going to be washed out against that background. You don't really get to rely on the spectra being unique. > the necessary power needed for a laser strong enough to travel to such a distant point (it might not be physically possible to create a laser large enough to exceed the light we would already see reflected back from the already formidable output of the Sun) The reflected light of the laser has to be brighter than the suns reflection over the entire area of the object you're trying to see. For an Earth-sized object at 1000 AU, that's 174 gigawatts. Not only do you have the impossible task of actually hitting the object, you need to have a hundred nuclear reactors powering a single source the entire time you do it.

regularfry 8y ago
Not to say this is in any way practical, but lasers used for nuclear fusion research are up in the petawatt range. They've just got extremely short pulses. But if all you need to do is detect that pulse coming back, like a radar return, it starts to sound within the range of conceivable engineering, not a total physical impossibility.

mikeash 8y ago

It’s worth noting that this sort of thing is done for closer objects: https://en.wikipedia.org/wiki/Radar_astronomy https://en.wikipedia.org/wiki/Radar_astronomy The problem is that the power requirements go up sharply with distance. The amount that hits the object decreases with the square of the distance, as does the amount that’s returned. The end result is that your power requirements are proportional to the fourth power of the distance. Twice as far means you need sixteen times more power. Now extrapolate that to something hundreds of times farther away than the closer planets....

perkee 8y ago

For a sense of the scale of this problem, the retro reflectors that we left on the moon return 1 photon out of the 10^17 we shoot at them and that's not even that reliable. This thing would be a mind bendingly further distance away and would not have any retro reflectors on it (oh man it'd be WILD if it did though!!). https://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment#Principle https://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment...

Rooster61 8y ago
This is the experiment that put the idea in my head in the first place. And also the reason I prefaced the question with the fact that I KNOW it's going to need to be a ridiculously powerful laser. Making something detectable that's hitting something with an unknown albedo and angle of whatever the photons are impacting makes the requirements start to bump up against the laws of physics, much less what we are technically capable of.

stmfreak 8y ago

Assuming you can build a laser bright enough for your detector, you are forgetting about the time & distance problem. We don’t know how far away this Planet X sits at any given point in time. You would have to pulse your laser at a spot or set of spots long enough to irradiate the possible region of space during the time it takes for Planet X to traverse said space. Then wait hours or days on the off chance the pulse comes back. If you look away too soon to examine the next spot, you miss the more distant object. If your pulse is too short, the object may pass in front or behind the pulse and be invisible. The Sun on the other hand is irradiating all space at all times. Aside from eclipses, we know Planet X is currently reflecting sunlight. We just have to look at it, compare to background, look again, compare to background, and notice the movement.

Rooster61 8y ago
> you are forgetting about the time & distance problem Not forgotten, just glossed over. You'd need to have a receiver that is pointing at a given (admittedly massive) part of the sky over a given range of possible times that a return photon would take to traverse the gap back to earth and strike the receiver. > We just have to look at it You're right, but the problem is that something orbiting at that distance follows an EXTREMELY slow orbit, so it's difficult to use the method you are describing. It's quite effective for looking for asteroids/space junk, but at that distance it becomes pretty tough to spot.

yongjik 8y ago

Imagine you're sitting relaxed in a dark, forgotten corner of asteroid X, five billion kilometers from the sun. Imagine looking up and seeing the feeble sunlight. Some of it gets reflected back and starts a long journey toward some telescope on Earth. Now imagine seeing "extremely powerful laser" from earth. Some of it gets reflected ... (ditto). For your method to work, the latter must be more powerful than the former. In other words, if you point your laser at Mars, a Martian should suddenly find the Earth shining brighter than the sun. I don't believe mankind possesses such power yet (thank god).

dotancohen 8y ago

Even assuming a super-strong light beam (brighter than the sun), and ignoring beam divergence, the time that the light returns to Earth would be dependent on the distance to the object. So if we don't know how far away it is, we won't know when to look back at the spot that the laser was aimed at.

TheOtherHobbes 8y ago

Empirical law: “Whenever there’s an engineering problem in space, at least one person on HN will suggest it can be solved with a big enough laser.”

Rooster61 8y ago
I aim to please :)