Author Topic: Increasing Mars' atmosphere with magnetic field @ Sun-Mars L1  (Read 8511 times)

Online FutureSpaceTourist

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From  Planetary Science Vision 2050 workshop:

Quote
Dr. Phil Metzger‏ @DrPhiltill 17m17 minutes ago

Cool! Jim Green: put a magnetic shield at the Sun-Mars L1 point to shield Mars' atmosphere. Gigantic atmospheric test on Mars. #V2050

https://twitter.com/DrPhiltill/status/836959848868810752

Quote
Dr. Phil Metzger‏ @DrPhiltill 10m10 minutes ago

Woah. Really? Modeling say this magnetic field can raise Mars' atmosphere to 1/2 of Earth's pressure in just years, not centuries. #V2050

https://twitter.com/DrPhiltill/status/836961787090202624

Online FutureSpaceTourist

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Quote
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 @AstrobiologyMag
https://t.co/6TuVxFvt1o

https://twitter.com/humansareawesme/status/836961705745858560

Edit: added slides attached to tweet
« Last Edit: 03/01/2017 02:54 PM by FutureSpaceTourist »

Offline Robotbeat

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Wouldn't work. Solar wind comes in at a significant angle from the Sun, not a straight line.
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Offline Rei

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Is this stuff that's being talked about peer-reviewed?  Is there a paper anywhere I can check out?  So they're looking at a hydrogen-with-a-bit-of-helium atmosphere for Mars (aka, solar wind)?  When you start adding oxygen, that would be  kind of... what's the word I'm looking for here... ah yes, "explosive"  ;)    (Really, though, in practice, I'd just expect any produced oxygen to be consumed by the hydrogen as fast as it's made).  Or do they mean the atmosphere comes from somewhere else, and they're just blocking stripping?

If they're talking solar hydrogen, I'd think that would be much more interesting for Venus.  Large amounts of added hydrogen would result in the Sabatier reaction, simultaneously consuming Venus's carbon dioxide and creating seas.  Bonus points if the tech can inject hydrogen in a manner that imparts relevant angular momentum (doubtful).  But not having a paper to look at, I have no clue what technology exactly is being proposed.

@Robobeat: Interesting comment - as I've never looked into solar wind anisotropy, I generally just assumed that within the heliosphere, when not modified by solid objects, it was relatively isotropic due to the high particle energy (aka not prone to being distorted by gravity).  If there is a relevant anisotropic component, surely it's far less than the anti-sunward component, is it not?  I should read up more on this.  Of course, if  you're creating a magnetic field then you're going to be creating a magnetosphere, with a magnetotail (presumably the goal is to have Mars within the magnetotail).   Any particles moving sideways would be affected by this, experiencing Lorentz force as they cross the field lines there just as they would any other magnetic field lines.

 Note that as per the slides the point of the shield is to prevent stripping.  But it's not clear whether it's also supposed to impart solar hydrogen and helium, or if not, where exactly the atmosphere is supposed to come from.
« Last Edit: 03/02/2017 03:38 PM by Rei »

Offline Rei

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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.
« Last Edit: 03/03/2017 10:00 AM by Rei »

Offline Rei

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Now that has however gotten me curious... what would be needed to capture solar wind?  Normally it's more of an erosive process than a depositional one.  searching for "solar wind" and "capture" on Google Scholar mainly leads me to papers talking about how "this model captures the details of the solar wind..." and things like that.  Has anyone seen any papers on the subject?  I can find a good number of papers talking about natural precipitation of solar wind ions, but nothing about enhanced precipitation.

The hardest part of Venus terraforming concepts is getting it moist again - aka, importing obscene amounts of hydrogen.  While clearly hundred year timescales might be overly ambitious, timescales 1-3 orders of magnitude longer might be plausible.  But it's just speculation unless someone has studied it.
« Last Edit: 03/02/2017 03:38 PM by Rei »

Offline Dalhousie

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How much power is needed to generate such a field?
"There is nobody who is a bigger fan of sending robots to Mars than me... But I believe firmly that the best, the most comprehensive, the most successful exploration will be done by humans" Steve Squyres

Offline Robotbeat

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How much power is needed to generate such a field?
None, once established.
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Online sanman

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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?




Offline Robotbeat

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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?
Or just a big, spinning superconducting coil structure.

Anyway, this whole idea wouldnt work because the solar wind comes in at a large angle (45 degrees?) at Mars.
Chris  Whoever loves correction loves knowledge, but he who hates reproof is stupid.

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Offline as58

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... because the solar wind comes in at a large angle (45 degrees?) at Mars.
Do you have a source for this?

Offline Bynaus

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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.

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
« Last Edit: 03/04/2017 11:49 AM by Bynaus »

Offline as58

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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.

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

Well, 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.

Offline Bynaus

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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.

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

Well, 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.pdf

Quite 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.

Offline as58

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Right, I missed that. Link: http://www.hou.usra.edu/meetings/V2050/pdf/8250.pdf

Quite 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.

It looks like the idea is to stop solar wind erosion so that outgassing from crust/interior will start increasing atmospheric pressure/temperature. Once temperature has increased 4 K, polar caps should start melting and atmospheric density should increase rapidly(?).

However, current outgassing rate seems so low that even if the solar wind erosion were stopped completely, it would take a long time (certainly more in the centuries than years) before there would be any appreciable change in the atmosphere. There would probably be a positive feedback so that even a small increase in temperature would increase outgassing, but I'd like to see some kind of quantitative estimate of this.  Otherwise is feels just silly to talk about achieving this 'in a lifetime'.

Offline Bynaus

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Of course, one could always help bring the warming about with some super-greenhouse gases. But this is true regardless of whether there is an artificial magnetic field or not. Long-term loss of an atmosphere thick enough to sustain humans is not a big issue, as it should be very slow, at least, compared to human time-scales.

Offline Dalhousie

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There seems to be an assumption that Mars is still outgassing so that if solar wind ablation is reduced total atmospheric mass increases.
« Last Edit: 03/05/2017 06:35 AM by Dalhousie »
"There is nobody who is a bigger fan of sending robots to Mars than me... But I believe firmly that the best, the most comprehensive, the most successful exploration will be done by humans" Steve Squyres

Offline Robotbeat

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... 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.
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Offline as58

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... 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.

Thanks, that matches what I thought I remembered from a space physics course many years ago. But my memories of the Parker spiral and all that are quite hazy by now...

Offline mikelepage

Surely Coronal Mass Ejections come directly from the sun though?

By preventing those from reaching Mars (how much energy would that take?), this would stop most of the atmosphere loss, no?

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