Dr. Phil Metzger @DrPhiltill 17m17 minutes agoCool! Jim Green: put a magnetic shield at the Sun-Mars L1 point to shield Mars' atmosphere. Gigantic atmospheric test on Mars. #V2050
Dr. Phil Metzger @DrPhiltill 10m10 minutes agoWoah. Really? Modeling say this magnetic field can raise Mars' atmosphere to 1/2 of Earth's pressure in just years, not centuries. #V2050
Sheyna Gifford @humansareawesme 13m13 minutes ago"Mars might look like this in 700 million years - OR SOONER." Jim Green models a planetary shield to heat up #Mars #v2050 @AstrobiologyMaghttps://t.co/6TuVxFvt1o
How much power is needed to generate such a field?
What's the means of generating and distributing such a field? Large magnetoplasma?Or how about a large swarm of laser-formation flying satellites with superconductive coils?
... because the solar wind comes in at a large angle (45 degrees?) at Mars.
Let's see, if they're talking about using solar wind to make atmospheres... at 1AU it's 6e8 particles per square centimeter per second... so 6e12 per square meter per second... Mars is 1,52, so with (overly) simplistic quadratic scaling assumed that'd be 2,6e12 per square meter per second... Earth's atmosphere is ~5e18kg... let's say you want 5e17 for Mars... solar wind is about 1.1g/mol... and we have 4,3e-12 mol/mē-s... let's say 100 years, so 1,36e-5 kg/mē... so to produce 5e17kg would require 3,67e22mē... aka 3,67e16kmē... equivalent to a disc 108 million km in radius... aka 0,72 AU in radius.... aka a pretty dang large segment of the solar wind Somehow I doubt that's what's being referred to here. They must just be talking about how much atmosphere would accumulate from Mars itself when you stop stripping. Could it really be that fast? Even if so, Mars has had its nitrogen stripped long ago, so I'm not sure where one could get a replacement.
Quote from: Rei on 03/02/2017 03:11 pmLet's see, if they're talking about using solar wind to make atmospheres... at 1AU it's 6e8 particles per square centimeter per second... so 6e12 per square meter per second... Mars is 1,52, so with (overly) simplistic quadratic scaling assumed that'd be 2,6e12 per square meter per second... Earth's atmosphere is ~5e18kg... let's say you want 5e17 for Mars... solar wind is about 1.1g/mol... and we have 4,3e-12 mol/mē-s... let's say 100 years, so 1,36e-5 kg/mē... so to produce 5e17kg would require 3,67e22mē... aka 3,67e16kmē... equivalent to a disc 108 million km in radius... aka 0,72 AU in radius.... aka a pretty dang large segment of the solar wind Somehow I doubt that's what's being referred to here. They must just be talking about how much atmosphere would accumulate from Mars itself when you stop stripping. Could it really be that fast? Even if so, Mars has had its nitrogen stripped long ago, so I'm not sure where one could get a replacement.From this article: https://www.nasa.gov/press-release/nasa-mission-reveals-speed-of-solar-wind-stripping-martian-atmosphere, Mars loses about 100 g of atmosphere per second. If you could tune that down to zero somehow, and assuming all the atmospheric sources on Mars are in equilibrium with loss, you would just gain 0.1 * 86400 * 365 = 3 million kg per year. The total mass of the Atmosphere or Mars is about 2.5 x 10^10 million kg. So I don't understand where that suggestion ("500 mbar in 5 years") is coming from. Also, how does it square with the suggestion that higher pressures than 50 mbar are unstable? A bit more context would be nice.EDIT: Article on phys.org: https://phys.org/news/2017-03-nasa-magnetic-shield-mars-atmosphere.html
Quote from: Bynaus on 03/04/2017 11:44 amQuote from: Rei on 03/02/2017 03:11 pmLet's see, if they're talking about using solar wind to make atmospheres... at 1AU it's 6e8 particles per square centimeter per second... so 6e12 per square meter per second... Mars is 1,52, so with (overly) simplistic quadratic scaling assumed that'd be 2,6e12 per square meter per second... Earth's atmosphere is ~5e18kg... let's say you want 5e17 for Mars... solar wind is about 1.1g/mol... and we have 4,3e-12 mol/mē-s... let's say 100 years, so 1,36e-5 kg/mē... so to produce 5e17kg would require 3,67e22mē... aka 3,67e16kmē... equivalent to a disc 108 million km in radius... aka 0,72 AU in radius.... aka a pretty dang large segment of the solar wind Somehow I doubt that's what's being referred to here. They must just be talking about how much atmosphere would accumulate from Mars itself when you stop stripping. Could it really be that fast? Even if so, Mars has had its nitrogen stripped long ago, so I'm not sure where one could get a replacement.From this article: https://www.nasa.gov/press-release/nasa-mission-reveals-speed-of-solar-wind-stripping-martian-atmosphere, Mars loses about 100 g of atmosphere per second. If you could tune that down to zero somehow, and assuming all the atmospheric sources on Mars are in equilibrium with loss, you would just gain 0.1 * 86400 * 365 = 3 million kg per year. The total mass of the Atmosphere or Mars is about 2.5 x 10^10 million kg. So I don't understand where that suggestion ("500 mbar in 5 years") is coming from. Also, how does it square with the suggestion that higher pressures than 50 mbar are unstable? A bit more context would be nice.EDIT: Article on phys.org: https://phys.org/news/2017-03-nasa-magnetic-shield-mars-atmosphere.htmlWell, that conference abstract/paper provides a bit more context. It doesn't say much about how long all that takes (except that it will be studied), so that "years, not centuries" seems to be some sort of (optimistic) extrapolation/guess.
Right, I missed that. Link: http://www.hou.usra.edu/meetings/V2050/pdf/8250.pdfQuite vague, especially the part on the "new equlibrium". But it is quite remarkable that a temperature change of only 4 K would be enough to start melting the polar caps and resulting in a higher atmospheric pressure as a consequence.
Quote from: Robotbeat on 03/04/2017 02:13 am... because the solar wind comes in at a large angle (45 degrees?) at Mars.Do you have a source for this?
Quote from: as58 on 03/04/2017 10:54 amQuote from: Robotbeat on 03/04/2017 02:13 am... because the solar wind comes in at a large angle (45 degrees?) at Mars.Do you have a source for this?I was wrong. At worst, it comes in at about 10 degrees at Earth. Also, the Earth's magnetotail is wide, such that Earth-Sun-L2 is usually inside the magnetotail.The interplanetary magnetic field (embedded in the solar wind), however, is at 45 degrees, and that is the rough direction tha solar energetic particles come at.