6 ms·

For those like me who need some kind of reference for the distances mentioned (548-900 AU), Voyager 1 is 156.5 AU from earth today.

by hamter 4y ago

quirkot 4y ago
For those looking for a live look at Voyager status: https://voyager.jpl.nasa.gov/mission/status/ https://voyager.jpl.nasa.gov/mission/status/
- alexpotato 4y ago
  The solar system explorer view they have on that page is fantastic! I recommend clicking on the "solar system" toggle in the bottom middle of the view. It gives you a real sense of the planets, probes, asteroids etc that are flying around our solar system. Also reminds me of looking at air traffic control maps and what that might look like once intra-solar system space travel becomes routine.
- deleted 4y ago
  [deleted]
scotty79 4y ago
I wonder how much faster we could get there with solar sail and laser boost from Earth
- ben_w 4y ago
  Without specifying the size of the laser, anything from "even slower" to "whole trip in just under 16 days". https://en.wikipedia.org/wiki/Breakthrough_Starshot https://en.wikipedia.org/wiki/Breakthrough_Starshot http://www.wolframalpha.com/input/?i=548%20AU%2F0.2c http://www.wolframalpha.com/input/?i=548%20AU%2F0.2c
  - lazide 4y ago
    Hah, the latter number requiring some significant percent of earths mass being converted to energy or something?
    - vilhelm_s 4y ago
      Not really, it's the velocity the "Breakthrough Starshot" probes would reach. They propose[1] that launching each probe would take 84 GWh, which is not super much (about 15 times more than a space shuttle launch), but of course the Starshot probes would be much lighter than this proposed telescope so it's not directly comparable. [1] https://youtu.be/KIDuXQHt8pk?t=1562 https://youtu.be/KIDuXQHt8pk?t=1562
    - zardo 4y ago
      Accelerating a meter class telescope to .2 c is well beyond our current capabilities, but it's nothing like that fast.
      - lazide 4y ago
        If we assume the meter class telescope + power supplies and whatever else masses ~ 1 metric ton, it would take 1,853,298,442,530,598,439 Joules - 1.8 quintillion - 1.853 * 10^18 - to accelerate it to .2c. Only ~ 442 Megatons of TNT. Keep in mind, that is assuming 100% efficiency, which would be impossible with a light sail or any other known technology. With all the various inefficiencies in power collection/generation, laser generation, momentum/power transfer, etc. we'd be talking probably somewhere around 5% end-to-end power transfer - if we were lucky. Which is still way better than the rocket equation (probably). So to get the required 1.853 * 10^18 J at 5% efficiency, we'd need say 20x more power at earth to accelerate it (if 5% efficiency). So 3.706 * 10^19 J. Which starts get more concerning, at 8,840 gigatons of TNT. Let's take the most efficient means we can imagine to produce energy, direct matter annihilation. Annihilating 1KG of mass (using 500g of Anti-matter, 500g of Matter) produces 8.986 * 10^16 J of energy. If we could somehow feed the resulting energy directly into the laser for accelerating the craft, and assume near 100% efficiency in doing so, before laser losses - we'd only need roughly 1000 KG of matter/anti-matter to do so. Not bad! But wait, our more likely end-to-end efficiency is at best 1%. Hmm. Which would require 100 times the input energy to spacecraft acceleration. 44 Gigatons of TNT or 1.853 * 10^20 J. 5KG of matter/antimatter. Which is definitely not a significant fraction of earths mass, but yikes. I wouldn't want to pay that energy bill! 51,480,512,292,500 kWh (the 'wall plug equivalent') at my current rates would be $26 trillion dollars!
  - kelnos 4y ago
    That's 16 days if the average speed is 20% lightspeed. In reality you need to speed up, and then when you're halfway there, slow down again. Which raises the other question: how would we slow this thing down so it just doesn't keep going past the 900AU mark? Regardless, I guess we would be fine picking a slower speed that would get it there in a few years, which might be significantly easier to achieve.
    - ben_w 4y ago
      Breakthrough Starshot aims to do all that acceleration in 10 minutes. It’s an ambitious project.
- pavel_lishin 4y ago
  One issue would be slowing down when you get there. You'd need to carry a deployable/detachable mirror with you to reflect the laser back at the craft, but that mirror itself would also get accelerated further out, which means having to correct for that, etc., etc.
  - sp332 4y ago
    For this mission, you don't need to slow down or stop. Just keep taking image data starting at 548 AU and keep going until you're at 900 AU.
    - Cerium 4y ago
      I thought you have to move laterally to get the pixels? "The data are acquired pixel-by-pixel while moving an imaging spacecraft within the image."
      - samus 4y ago
        You can already do that while slowing down. Might require some image correction though to account for that in the pictures.
  - tokamak-teapot 4y ago
    With a solar sail you can just start tacking, right?
    - tgarv 4y ago
      I think a solar sail mostly works at a broad reach or a run, so it would be more of a jibe than a tack ;)
    - KineticLensman 4y ago
      Probably not at 540 AU. Not many photons out there for the solar sail to grip.
    - klyrs 4y ago
- scottmsul 4y ago
  The fastest way there would probably be with a nuclear propulsion engine, as researched in Project Orion. https://en.wikipedia.org/wiki/Project_Orion_(nuclear_propulsion) https://en.wikipedia.org/wiki/Project_Orion_(nuclear_propuls...
  - ncmncm 4y ago
    Princeton FRC
- MattPalmer1086 4y ago
  There's a nice article on centauri dreams just published looking at solar and nuclear options for such a mission. https://www.centauri-dreams.org/2022/07/26/getting-there-quickly-the-nuclear-option/ https://www.centauri-dreams.org/2022/07/26/getting-there-qui...
dublin 4y ago
Yeah, the solar lensing point is WAY the hell out there: since 1 AU is 8 light-minutes, so the 548 AU minimum works out to 73 light-hours away. When LIGHT takes half a week to make a one-way trip, you're in the deep space boondocks, folks. I haven't read the paper yet, but this thing would have to have a fair amount of nuclear power, and comms would be a challenge as well. As the abstract mentions, though, while the project has a high degree of difficulty, there appear to be no complete technology showstoppers to actually doing this, so it's at least as doable (and considerably cheaper than) a von Braun-style centrifugal space station in Earth orbit. It'll be interesting to see if the idea gets any traction...
danw1979 4y ago
and 900 AU is 0.014 light years. so you can get 10km resolution on a ~7100:1 object to focal-point distance (what’s the right term for this?). seems pretty good.

For those like me who need some kind of reference for the distances mentioned (548-900 AU), Voyager 1 is 156.5 AU from earth today.

by hamter 4y ago

For those like me who need some kind of reference for the distances mentioned (548-900 AU), Voyager 1 is 156.5 AU from earth today.

quirkot 4y ago

For those looking for a live look at Voyager status: https://voyager.jpl.nasa.gov/mission/status/ https://voyager.jpl.nasa.gov/mission/status/

alexpotato 4y ago
The solar system explorer view they have on that page is fantastic! I recommend clicking on the "solar system" toggle in the bottom middle of the view. It gives you a real sense of the planets, probes, asteroids etc that are flying around our solar system. Also reminds me of looking at air traffic control maps and what that might look like once intra-solar system space travel becomes routine.
deleted 4y ago
[deleted]

scotty79 4y ago

I wonder how much faster we could get there with solar sail and laser boost from Earth

ben_w 4y ago
Without specifying the size of the laser, anything from "even slower" to "whole trip in just under 16 days". https://en.wikipedia.org/wiki/Breakthrough_Starshot https://en.wikipedia.org/wiki/Breakthrough_Starshot http://www.wolframalpha.com/input/?i=548%20AU%2F0.2c http://www.wolframalpha.com/input/?i=548%20AU%2F0.2c
- lazide 4y ago
  Hah, the latter number requiring some significant percent of earths mass being converted to energy or something?
  - vilhelm_s 4y ago
    Not really, it's the velocity the "Breakthrough Starshot" probes would reach. They propose[1] that launching each probe would take 84 GWh, which is not super much (about 15 times more than a space shuttle launch), but of course the Starshot probes would be much lighter than this proposed telescope so it's not directly comparable. [1] https://youtu.be/KIDuXQHt8pk?t=1562 https://youtu.be/KIDuXQHt8pk?t=1562
  - zardo 4y ago
    Accelerating a meter class telescope to .2 c is well beyond our current capabilities, but it's nothing like that fast.
    - lazide 4y ago
      If we assume the meter class telescope + power supplies and whatever else masses ~ 1 metric ton, it would take 1,853,298,442,530,598,439 Joules - 1.8 quintillion - 1.853 * 10^18 - to accelerate it to .2c. Only ~ 442 Megatons of TNT. Keep in mind, that is assuming 100% efficiency, which would be impossible with a light sail or any other known technology. With all the various inefficiencies in power collection/generation, laser generation, momentum/power transfer, etc. we'd be talking probably somewhere around 5% end-to-end power transfer - if we were lucky. Which is still way better than the rocket equation (probably). So to get the required 1.853 * 10^18 J at 5% efficiency, we'd need say 20x more power at earth to accelerate it (if 5% efficiency). So 3.706 * 10^19 J. Which starts get more concerning, at 8,840 gigatons of TNT. Let's take the most efficient means we can imagine to produce energy, direct matter annihilation. Annihilating 1KG of mass (using 500g of Anti-matter, 500g of Matter) produces 8.986 * 10^16 J of energy. If we could somehow feed the resulting energy directly into the laser for accelerating the craft, and assume near 100% efficiency in doing so, before laser losses - we'd only need roughly 1000 KG of matter/anti-matter to do so. Not bad! But wait, our more likely end-to-end efficiency is at best 1%. Hmm. Which would require 100 times the input energy to spacecraft acceleration. 44 Gigatons of TNT or 1.853 * 10^20 J. 5KG of matter/antimatter. Which is definitely not a significant fraction of earths mass, but yikes. I wouldn't want to pay that energy bill! 51,480,512,292,500 kWh (the 'wall plug equivalent') at my current rates would be $26 trillion dollars!
- kelnos 4y ago
  That's 16 days if the average speed is 20% lightspeed. In reality you need to speed up, and then when you're halfway there, slow down again. Which raises the other question: how would we slow this thing down so it just doesn't keep going past the 900AU mark? Regardless, I guess we would be fine picking a slower speed that would get it there in a few years, which might be significantly easier to achieve.
  - ben_w 4y ago
    Breakthrough Starshot aims to do all that acceleration in 10 minutes. It’s an ambitious project.
pavel_lishin 4y ago
One issue would be slowing down when you get there. You'd need to carry a deployable/detachable mirror with you to reflect the laser back at the craft, but that mirror itself would also get accelerated further out, which means having to correct for that, etc., etc.
- sp332 4y ago
  For this mission, you don't need to slow down or stop. Just keep taking image data starting at 548 AU and keep going until you're at 900 AU.
  - Cerium 4y ago
    I thought you have to move laterally to get the pixels? "The data are acquired pixel-by-pixel while moving an imaging spacecraft within the image."
    - samus 4y ago
      You can already do that while slowing down. Might require some image correction though to account for that in the pictures.
- tokamak-teapot 4y ago
  With a solar sail you can just start tacking, right?
  - tgarv 4y ago
    I think a solar sail mostly works at a broad reach or a run, so it would be more of a jibe than a tack ;)
  - KineticLensman 4y ago
    Probably not at 540 AU. Not many photons out there for the solar sail to grip.
  - klyrs 4y ago
scottmsul 4y ago
The fastest way there would probably be with a nuclear propulsion engine, as researched in Project Orion. https://en.wikipedia.org/wiki/Project_Orion_(nuclear_propulsion) https://en.wikipedia.org/wiki/Project_Orion_(nuclear_propuls...
- ncmncm 4y ago
  Princeton FRC
MattPalmer1086 4y ago
There's a nice article on centauri dreams just published looking at solar and nuclear options for such a mission. https://www.centauri-dreams.org/2022/07/26/getting-there-quickly-the-nuclear-option/ https://www.centauri-dreams.org/2022/07/26/getting-there-qui...

dublin 4y ago

Yeah, the solar lensing point is WAY the hell out there: since 1 AU is 8 light-minutes, so the 548 AU minimum works out to 73 light-hours away. When LIGHT takes half a week to make a one-way trip, you're in the deep space boondocks, folks. I haven't read the paper yet, but this thing would have to have a fair amount of nuclear power, and comms would be a challenge as well. As the abstract mentions, though, while the project has a high degree of difficulty, there appear to be no complete technology showstoppers to actually doing this, so it's at least as doable (and considerably cheaper than) a von Braun-style centrifugal space station in Earth orbit. It'll be interesting to see if the idea gets any traction...

danw1979 4y ago

and 900 AU is 0.014 light years. so you can get 10km resolution on a ~7100:1 object to focal-point distance (what’s the right term for this?). seems pretty good.