4 ms·

Excellent page discussing the physics of colonizing mercury: https://einstein-schrodinger.com/mercury_colony.html https://einstein-schrodinger.com/mercury_colon

Excellent page discussing the physics of colonizing mercury: https://einstein-schrodinger.com/mercury_colony.html https://einstein-schrodinger.com/mercury_colony.html "Colonization of Mercury appears to be a very real and practical possibility, whereas colonization of Mars or the other planets, moons or asteroids is really more in the realm of fantasy." A big advantage of Mercury over mars is higher light intensity: "One very important advantage is the high solar light intensity, which is stronger than on Earth by a factor of 10.6 at perihelion and 4.6 at aphelion"

jrussino 6y ago
Cool! Personally, I like the idea of building a Dyson swarm out of it (sorry Mercury!) https://www.youtube.com/watch?v=pP44EPBMb8A https://www.youtube.com/watch?v=pP44EPBMb8A
lmilcin 6y ago
Big disadvantage of Mercury is its depth in Sun's gravitational well. This is worse than planetary escape velocity, because you have to pay it on both ways (to get to Mercury as well as get back). Another problem is radiation which this close to Sun and with no atmosphere is pretty huge issue. Light intensity is irrelevant. Energy in the form of light is pretty much free within solar system unless it is somehow obstructed. Free in that you don't need to take it with you and it is completely renewable. If you have less light you just gather it from a larger area. Mars has actually pretty good light intensity because it has thin atmosphere (it blocking much less electromagnetic spectrum). For a colony much more important are parameters: how difficult and long to get there, how much need to be taken with you (ie. do you need to have water/air with you), how much protection you will need to live there (temperature, differences, atmospheric pressure, radiation, gravity, etc.)
- codeflo 6y ago
  Can you explain why you have to pay for the Sun's gravitational well on the way to Mercury? From the famous XKCD illustration[1] I get the impression that it might be hard to "stop at" Mercury since the pull of the Sun is relatively strong. Is there no clever trajectory that comes close enough to Mercury to get caught in its orbit? [1] https://xkcd.com/681/ https://xkcd.com/681/
  - forand 6y ago
    You want to stop when you get to Mercury. In the XKCD illustration you have fallen quite a bit from Earth to Mercury and now have that as kinetic energy. You can dissipate that by hitting Mercury but that would likely have significant negative side effects.
    - lmilcin 6y ago
      Loosing the energy on Mercury is only part of the problem. You also have to get there. Contrary to popular belief, going closer to the Sun is as much difficult as going further away from it. Orbits require energy to change them regardless which direction you want to do it. You can go to tables and see going to Mercury costs almost the same as going to Jupiter: https://en.wikipedia.org/wiki/Delta-v_budget https://en.wikipedia.org/wiki/Delta-v_budget There are handy maps of Sol system if you are interested: https://forum.kerbalspaceprogram.com/index.php?/topic/131682-is-there-a-delta-v-chart-for-real-solar-system-mod/ https://forum.kerbalspaceprogram.com/index.php?/topic/131682... You can of course minimize the energy even further using gravity assists, but this would require multi-year journey. There is another trick when you want to go very close to the Sun, which is to go further away first, and then use much less energy to actually stop your orbital speed and fall however deep into Sun well you want. Unfortunately, Mercury is just far enough from the Sun to make it unprofitable and it also takes an awful amount of time to execute.
  - AllegedAlec 6y ago
    A better image (imho): https://i.imgur.com/SqdzxzF.png https://i.imgur.com/SqdzxzF.png
  - lmilcin 6y ago
    I think Randall's image is a bit misleading. I mean it is correct and fun, but for an uninitiated person it might cause impression that going down the well costs zero energy. That of course is true, if you are stationary. Then you just start falling like apple from apple tree. But almost no objects in space are stationary, typically everything orbits something and frequently multiple things at the same time. Think about a rocket that is on an orbit around Earth. It is not free to fall to Earth, it must expend delta V to slow down enough so that its perigee is within Earth atmosphere (and typically well within atmosphere or it will just bounce off it). At the same time, the rocket orbits the Sun, so even if it leaves Earth it still is not free to "fall" into Mercury. It must expend delta V to slow down so that its perihelion lowers at least to Mercury orbit. So the graph on xkcd is correct, but with regards to bowling balls you might want to shoot up or let fall down from a position that is stationary in some celestial body's frame of reference.
satori99 6y ago
Kim Stanley Robinson wrote a book, 2312, which featured an equatorial city on Mercury mounted on massive concrete rails. The expansion of the concrete rails on the day side permanently keeps the city in the planets relatively pleasant pre-dawn zone. The city only needed to move at walking pace to stay ahead of the sunrise. IIRC one of the major mercurian industries was maintaining massive solar reflectors that were used to boost light going to specific asteroid greenhouses.
- oefnak 6y ago
  What if the city would have engine trouble, though...
  - satori99 6y ago
    It had no engines. The sun-side tracks thermal expansion provided the energy to keep pushing the city into the night-side.

jrussino 6y ago

Cool! Personally, I like the idea of building a Dyson swarm out of it (sorry Mercury!) https://www.youtube.com/watch?v=pP44EPBMb8A https://www.youtube.com/watch?v=pP44EPBMb8A

lmilcin 6y ago

Big disadvantage of Mercury is its depth in Sun's gravitational well. This is worse than planetary escape velocity, because you have to pay it on both ways (to get to Mercury as well as get back). Another problem is radiation which this close to Sun and with no atmosphere is pretty huge issue. Light intensity is irrelevant. Energy in the form of light is pretty much free within solar system unless it is somehow obstructed. Free in that you don't need to take it with you and it is completely renewable. If you have less light you just gather it from a larger area. Mars has actually pretty good light intensity because it has thin atmosphere (it blocking much less electromagnetic spectrum). For a colony much more important are parameters: how difficult and long to get there, how much need to be taken with you (ie. do you need to have water/air with you), how much protection you will need to live there (temperature, differences, atmospheric pressure, radiation, gravity, etc.)

codeflo 6y ago
Can you explain why you have to pay for the Sun's gravitational well on the way to Mercury? From the famous XKCD illustration[1] I get the impression that it might be hard to "stop at" Mercury since the pull of the Sun is relatively strong. Is there no clever trajectory that comes close enough to Mercury to get caught in its orbit? [1] https://xkcd.com/681/ https://xkcd.com/681/
- forand 6y ago
  You want to stop when you get to Mercury. In the XKCD illustration you have fallen quite a bit from Earth to Mercury and now have that as kinetic energy. You can dissipate that by hitting Mercury but that would likely have significant negative side effects.
  - lmilcin 6y ago
    Loosing the energy on Mercury is only part of the problem. You also have to get there. Contrary to popular belief, going closer to the Sun is as much difficult as going further away from it. Orbits require energy to change them regardless which direction you want to do it. You can go to tables and see going to Mercury costs almost the same as going to Jupiter: https://en.wikipedia.org/wiki/Delta-v_budget https://en.wikipedia.org/wiki/Delta-v_budget There are handy maps of Sol system if you are interested: https://forum.kerbalspaceprogram.com/index.php?/topic/131682-is-there-a-delta-v-chart-for-real-solar-system-mod/ https://forum.kerbalspaceprogram.com/index.php?/topic/131682... You can of course minimize the energy even further using gravity assists, but this would require multi-year journey. There is another trick when you want to go very close to the Sun, which is to go further away first, and then use much less energy to actually stop your orbital speed and fall however deep into Sun well you want. Unfortunately, Mercury is just far enough from the Sun to make it unprofitable and it also takes an awful amount of time to execute.
- AllegedAlec 6y ago
  A better image (imho): https://i.imgur.com/SqdzxzF.png https://i.imgur.com/SqdzxzF.png
- lmilcin 6y ago
  I think Randall's image is a bit misleading. I mean it is correct and fun, but for an uninitiated person it might cause impression that going down the well costs zero energy. That of course is true, if you are stationary. Then you just start falling like apple from apple tree. But almost no objects in space are stationary, typically everything orbits something and frequently multiple things at the same time. Think about a rocket that is on an orbit around Earth. It is not free to fall to Earth, it must expend delta V to slow down enough so that its perigee is within Earth atmosphere (and typically well within atmosphere or it will just bounce off it). At the same time, the rocket orbits the Sun, so even if it leaves Earth it still is not free to "fall" into Mercury. It must expend delta V to slow down so that its perihelion lowers at least to Mercury orbit. So the graph on xkcd is correct, but with regards to bowling balls you might want to shoot up or let fall down from a position that is stationary in some celestial body's frame of reference.

satori99 6y ago

Kim Stanley Robinson wrote a book, 2312, which featured an equatorial city on Mercury mounted on massive concrete rails. The expansion of the concrete rails on the day side permanently keeps the city in the planets relatively pleasant pre-dawn zone. The city only needed to move at walking pace to stay ahead of the sunrise. IIRC one of the major mercurian industries was maintaining massive solar reflectors that were used to boost light going to specific asteroid greenhouses.

oefnak 6y ago
What if the city would have engine trouble, though...
- satori99 6y ago
  It had no engines. The sun-side tracks thermal expansion provided the energy to keep pushing the city into the night-side.