Author Topic: Increasing Mars' atmosphere with magnetic field @ Sun-Mars L1  (Read 7540 times)

Offline 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

Offline 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 »

Online 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/m²-s... let's say 100 years, so 1,36e-5 kg/m²... so to produce 5e17kg would require 3,67e22m²... aka 3,67e16km²... 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

Online Robotbeat

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




Online 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.
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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/m²-s... let's say 100 years, so 1,36e-5 kg/m²... so to produce 5e17kg would require 3,67e22m²... aka 3,67e16km²... 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/m²-s... let's say 100 years, so 1,36e-5 kg/m²... so to produce 5e17kg would require 3,67e22m²... aka 3,67e16km²... 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/m²-s... let's say 100 years, so 1,36e-5 kg/m²... so to produce 5e17kg would require 3,67e22m²... aka 3,67e16km²... 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

Online 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?

Offline Rei

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It's all a moot point.  If you're in the magnetotail, then there's a magnetic field gradient between you and particles coming from any angle, meaning incoming particles experience Lorentz force and change direction.

Offline Dalhousie

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

The Earth's magnetic field is run by continuous energy release in the Earth's core.  Why do you say this needs no power once it is set up?
"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

Online Robotbeat

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

The Earth's magnetic field is run by continuous energy release in the Earth's core.  Why do you say this needs no power once it is set up?
Because you'd use superconductors, and superconductors don't have any measurable DC resistance. Or at least, I would use superconductors. I think they may be considering a different mechanism involving inflating the artificial magnetosphere with plasma, and that'd take power and mass.
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Online Space Ghost 1962

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Actually the particle flux of the solar wind does represent a current flow. There's also field strength issues/needs too.

Also, the way in which the Mars atmosphere (as it is being explored by missions such as MAVEN) appears to "work", might require significant amplitude to "not leak".

And not everyone share's Jim Greens enthusiasm for this idea.

And there's other ideas for restarting the Mars dynamo that I've heard, which rely on a better understanding of the status of mantle/core.

My read of this is to deal with the idea that Mars is somehow a broken world little better than the Moon, and much further away. It isn't a broken world, it's very different than Earth, and likely it did once possess near Earth like conditions, including considerable water.

Our understanding of the Moon has also improved - there's water too. The real difference is that we are now beginning to "read Mars" like we've read Earth. Perhaps we'll "read the Moon". But from what we've read, we can now "do stuff" that makes sense. Mars is not so bad.

Offline Dalhousie

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

The Earth's magnetic field is run by continuous energy release in the Earth's core.  Why do you say this needs no power once it is set up?
Because you'd use superconductors, and superconductors don't have any measurable DC resistance. Or at least, I would use superconductors. I think they may be considering a different mechanism involving inflating the artificial magnetosphere with plasma, and that'd take power and mass.

Those super conductors would need cooling.  There would be energy losses through interaction with interplanetary fields.  No free lunch.  So, again, what power is required?
"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

Online Robotbeat

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

The Earth's magnetic field is run by continuous energy release in the Earth's core.  Why do you say this needs no power once it is set up?
Because you'd use superconductors, and superconductors don't have any measurable DC resistance. Or at least, I would use superconductors. I think they may be considering a different mechanism involving inflating the artificial magnetosphere with plasma, and that'd take power and mass.

Those super conductors would need cooling.  There would be energy losses through interaction with interplanetary fields.  No free lunch.  So, again, what power is required?
False! No active cooling needed if shielded from the Sun and using a high temperature superconductor. And no losses from interaction provided you're cold enough not to get near the critical current and field limits.!It is a "free lunch" provided by quantum mechanics. It's the same reason electrons can orbit indefinitely without falling in to the center of an atom.

NO power required except for the initial 10^19 Joules to get the field established, at least if you just use superconducting coils.
« Last Edit: 03/06/2017 12:55 AM by Robotbeat »
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Online Space Ghost 1962

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NO power required except for the initial 10^19 Joules to get the field established, at least if you just use superconducting coils.
Nope. Not just "bootstrapping the field".

Incident UV ionizes atmosphere, the ions are massive moving charges, they would damp the magnetic field through inductance load.

Also, the current sheath through the particle bow shock would also inductively load the field, creating a "jump shock" in the plasma around the center of the bow shock, and possibly shorting the displacement current of said inductance.

The jump shock condition would allow through charge transfer neutrals. You'd have to up the lost power from the bootstrap to maintain field intensity, and likely have to source a greater load to change the jump shock to a level where the shielding would work for the neutrals.

It's much more complex than you think. And I've not even discussed the atmospheric loss issues in the wake, nor factored in about a dozen other mechanisms.

Online Robotbeat

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No, it provides pressure on the magnetic field which compresses it, but does not reduce the current flowing in the magnets. You can't demagnetize an iron atom for the same reason.

Yeah, the inflated artificial magnetosphere idea has gas loss mechanisms, though. But the superconducting coils do not lose strength.
« Last Edit: 03/06/2017 02:06 AM by Robotbeat »
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Online Space Ghost 1962

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No, it provides pressure on the magnetic field which compresses it, but does not reduce the current flowing in the magnets. You can't demagnetize an iron atom for the same reason.

Yeah, the inflated artificial magnetosphere idea has gas loss mechanisms, though. But the superconducting coils do not lose strength.

There is no static field to compress. Only remnant crustal fields in a few, rare locations.

And yes there is inductive loss. You can tell this and dust and a few other things from the Langmuir probes on existing SC.

There's also a peculiar SEP bounce at about 100km which could be helpful if better understood. Lots of stuff.

But its a way overreach that a bootstrap field is enough.

Online Robotbeat

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I'm talking about enormous superconducting coils.
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I'm talking about enormous superconducting coils.
And I'm talking about a planet. Lots bigger. More complex system. Planetary plasma's are not simple.

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Just wrap the superconductor around the entire planet.

Trying to build a magnetosphere by restarting the core is like trying to fly by flapping your arms and strapping wings on. Not only will you not possibly be able to put enough energy into the system to make it work, but there's a much easier method that ultimately is higher performing anyway (fixed wings and an engine).
« Last Edit: 03/06/2017 03:11 AM by Robotbeat »
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Just wrap the superconductor around the entire planet.

Trying to build a magnetosphere by restarting the core is like trying to fly by flapping your arms and strapping wings on. Not only will you not possibly be able to put enough energy into the system to make it work, but there's a much easier method that ultimately is higher performing anyway (fixed wings and an engine).

One method I've heard involves arranging an asteroid impact with a nickel iron asteroid of sufficient size. I've even seen the simulation, which depends on the asymmetry of the Tharsis "bulge".

What limits a lot of this is the lack of understanding of the pre Noachian Mars.

There are magnetic spots, that in some projections appear as "stripes" that are unexplained. Some are trying to argue that they might represent a form of plate tectonics. Others think it might be impacts of asteroids that shut down the dynamo. Through the physics/geodynamics of invertability (unproven at this scale), one could argue that a "cancelling" impact could restart same.

Now you already have crustal fields, which could be enhanced with superconducting fields. Also, you could diffract SEPs and UV radiation passively with a L1 located item. To the degree that Mars might become more habitable. There are a number of approaches.

We just don't know enough to play with Mars for a suitable ROI.

Offline as58

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Jim Green's talk (~15 minutes) is available at https://livestream.com/viewnow/vision2050/videos/150701155 starting at about 1:36.

I think the tweet quoted in the original post (about the time scale) is based on some kind of misunderstanding.

« Last Edit: 03/06/2017 09:39 PM by as58 »

Offline meekGee

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

I don't know if that's true.

It depends whether work is being done while deflecting the solar wind.  If there is, then maintaining the magnetic field will require power.
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Online Robotbeat

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

I don't know if that's true.

It depends whether work is being done while deflecting the solar wind.  If there is, then maintaining the magnetic field will require power.
No work is being done, just like a regular magnet. Unlike a regular magnet, you will not lose magnetism unless you get to a critical current or critical field value, which you won't in a properly engineered system.
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Offline meekGee

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

I don't know if that's true.

It depends whether work is being done while deflecting the solar wind.  If there is, then maintaining the magnetic field will require power.
No work is being done, just like a regular magnet. Unlike a regular magnet, you will not lose magnetism unless you get to a critical current or critical field value, which you won't in a properly engineered system.
So in a simple scenario, a particle going through a magnetic field experiences a force perpendicular to its velocity vector, so the dot product is zero, and no work is done.

But if the particle experiences a resistance to that change I think you're now having to expend work.

Also - consider the change in the momentum vector.  If you deflect particles, even without changing their speed, you need a reaction force.  How much impulse do you need to provide to keep the field generator in place?  We are talking about a sizeable magnetic solar sail here...  good thing it's an inefficient one...

Then there are practical inefficiencies.

I still think it's a huge idea...

Just not sure it won't require a power source...
« Last Edit: 03/07/2017 02:51 PM by meekGee »
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Online Space Ghost 1962

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A charged particle encountering an axial magnetic field will be deflected around a gyradius. The incident flux of such massive particle flows will act as if a column buckling around the axis. The cross sectional area/volume of the field lines, along with the moment tensor, gives you the total work that the "magnet" does on the "current".

Also, please realize that the particle flow "current" has itself a magnetic field, like the created field has a displacement current - they interact.

For no work to be done, no interaction / deflection must be done, thus no flux. So yes, you could have that situation, but in that case it wouldn't do anything, so why would you do it?

The effect of a ring current on the boundary of the geomagnetic field in a steady state solar wind
Quote from: Spreiter and Alksne
The present  paper  reports  the results of an  extension  of  the  theoretical study  reported  in  references 1, 2, and 3 in which  approximate results are determined for  the  traces,  in  the  geomagnetic  equatorial  plane  and  in  the  geomagnetic meridian  plane  containing  the  sun-earth  line,  of  the  cavity  carved  out  of a steady  neutral  ionized  solar  corpuscular stream by  interaction  with a magnetic dipole  representing  the  geomagnetic  field. The novel  feature  of  this  extension is the  inclusion of the  effect  of  an  equatorial  ring  current  having  properties similar
to  those of the model  proposed  by  Smith,  Coleman,  Judge,  and  Sonett in reference 4 to represent  the magnetometer  data  from  Pioneer V and  Explorer VI. These properties are that there exists, during  quiet times, a westward  flowing current of about 5x10^6 amperes distributed  over a large volume having  the  form of a toroidal  ring 3 earth  radii  in  cross-sectional  radius  with its center  line situated  in  the  geomagnetic  equatorial  plane at a distance of  approximately 8 to 10 earth  radii.
During not quiet times, up to 8 orders of magnitude larger BTW.

Now, for a reasonable discussion we'd have to introduce space borne plasmas and MHD equations. Some are even investigating how to harvest the solar wind as a source of power. Suffice to say, if there was no "work", there would be no point, as well as no geomagnetic storms, aurora, and other disruptions.

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For no work to be done, no interaction / deflection must be done, thus no flux. So yes, you could have that situation, but in that case it wouldn't do anything, so why would you do it?
..
This is completely wrong.
If I have an electric motor with permanent magnets, the magnets are obviously interacting. But they are not themselves doing work. The permanent magnets do not appreciably lose their magnetism, they last for the life of the vehicle.

If I place an object on the hard floor and the object is sitting there, is the floor doing work? No. If I bounce a ball off the hard floor, is the floor doing work? Again, no. The floor does expend energy in order to deflect and interact with the ball. Same for a magnet.

I can't believe we're having this discussion. Can someone else help me explain this really basic physical concept?
Chris  Whoever loves correction loves knowledge, but he who hates reproof is stupid.

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Offline meekGee

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For no work to be done, no interaction / deflection must be done, thus no flux. So yes, you could have that situation, but in that case it wouldn't do anything, so why would you do it?
..
This is completely wrong.
If I have an electric motor with permanent magnets, the magnets are obviously interacting. But they are not themselves doing work. The permanent magnets do not appreciably lose their magnetism, they last for the life of the vehicle.

If I place an object on the hard floor and the object is sitting there, is the floor doing work? No. If I bounce a ball off the hard floor, is the floor doing work? Again, no. The floor does expend energy in order to deflect and interact with the ball. Same for a magnet.

I can't believe we're having this discussion. Can someone else help me explain this really basic physical concept?

RB - your analogy is not quite analogous.

The floor has a "normal force" mechanism. L1 is not stable.  So it can't "support" the field generator, and so has to continuously thrust against the delta-impulse it imparts on the solar wind.  (That's the "planet-sized but inefficient solar sail" I was describing.

It's the difference between putting load on a floor and putting it on a helicopter.  The helicopter clearly uses power just to perform zero work on the payload.

As to the electro-magnetics, I'm out of my depth.  I am with you as far as "single charge in a magnetic field doesn't experience a change in energy and requires no work".  But I am not sure the situation here is completely described by this.  It may be (for example) that unless you perform work, the solar wind generates a magnetic field that nulls your original field, for example, and so the simple-magnet analogy doesn't hold either.

Or that electro-magnets behave a bit like the helicopter - while in theory a simple magnet would do (analogous to the floor), in reality you have to use an active magnet (analogous to the helicopter) and have to invest electric power which just like the downwash of the helicopter, gets wasted.

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Offline Danderman

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My favorite idea is a cycler between Venus and Mars that scoops up excess Co2 and methane from Venus and transports it over to Mars, back and forth, shampoo, rinse, repeat, until both planets are optimized.

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My favorite idea is a cycler between Venus and Mars that scoops up excess Co2 and methane from Venus and transports it over to Mars, back and forth, shampoo, rinse, repeat, until both planets are optimized.
How would you scoop up a large mass of atmospheric gases without it seriously aero-braking your cycler at Venus?

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For no work to be done, no interaction / deflection must be done, thus no flux. So yes, you could have that situation, but in that case it wouldn't do anything, so why would you do it?
..
This is completely wrong.
If I have an electric motor with permanent magnets, the magnets are obviously interacting. But they are not themselves doing work. The permanent magnets do not appreciably lose their magnetism, they last for the life of the vehicle.

If I place an object on the hard floor and the object is sitting there, is the floor doing work? No. If I bounce a ball off the hard floor, is the floor doing work? Again, no. The floor does expend energy in order to deflect and interact with the ball. Same for a magnet.

I can't believe we're having this discussion. Can someone else help me explain this really basic physical concept?

RB - your analogy is not quite analogous.

The floor has a "normal force" mechanism. L1 is not stable.  So it can't "support" the field generator, and so has to continuously thrust against the delta-impulse it imparts on the solar wind.  (That's the "planet-sized but inefficient solar sail" I was describing.
You can just offset slightly from the Lagrange point and use gravity (of the Sun in this case) to react against the force of the solar wind. Just like a solar sail except acting on the wind and not just the light.
Chris  Whoever loves correction loves knowledge, but he who hates reproof is stupid.

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Online Robotbeat

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My favorite idea is a cycler between Venus and Mars that scoops up excess Co2 and methane from Venus and transports it over to Mars, back and forth, shampoo, rinse, repeat, until both planets are optimized.
How would you scoop up a large mass of atmospheric gases without it seriously aero-braking your cycler at Venus?
Youd have to input energy here.
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Offline meekGee

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For no work to be done, no interaction / deflection must be done, thus no flux. So yes, you could have that situation, but in that case it wouldn't do anything, so why would you do it?
..
This is completely wrong.
If I have an electric motor with permanent magnets, the magnets are obviously interacting. But they are not themselves doing work. The permanent magnets do not appreciably lose their magnetism, they last for the life of the vehicle.

If I place an object on the hard floor and the object is sitting there, is the floor doing work? No. If I bounce a ball off the hard floor, is the floor doing work? Again, no. The floor does expend energy in order to deflect and interact with the ball. Same for a magnet.

I can't believe we're having this discussion. Can someone else help me explain this really basic physical concept?

RB - your analogy is not quite analogous.

The floor has a "normal force" mechanism. L1 is not stable.  So it can't "support" the field generator, and so has to continuously thrust against the delta-impulse it imparts on the solar wind.  (That's the "planet-sized but inefficient solar sail" I was describing.

You can just offset slightly from the Lagrange point and use gravity (of the Sun in this case) to react against the force of the solar wind. Just like a solar sail except acting on the wind and not just the light.

Did you figure out how "slightly" this would be?

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Offline Rei

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My favorite idea is a cycler between Venus and Mars that scoops up excess Co2 and methane from Venus and transports it over to Mars, back and forth, shampoo, rinse, repeat, until both planets are optimized.
How would you scoop up a large mass of atmospheric gases without it seriously aero-braking your cycler at Venus?

Well, you could probably make up for the delta-V with gravity assists.  But I certainly wouldn't want to be the one tasked with finding a viable trajectory  ;)

Either way, that would be a painfully slow process.  And I'm not sure how much of Venus's atmosphere you could remove before you'd hit its other, opposite problem.  Venus's surface is a large percent (I want to say something like 7%?) FeO. If you start terraforming it and producing oxygen, it's just going to rust away your oxygen.  It needs at least part of the oxygen that's bound up in its carbon dioxide.  Before you can reach stability, you basically need to recreate what happened on Earth with the formation of the banded iron deposits..
« Last Edit: 03/08/2017 11:02 AM by Rei »

Online Robotbeat

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For no work to be done, no interaction / deflection must be done, thus no flux. So yes, you could have that situation, but in that case it wouldn't do anything, so why would you do it?
..
This is completely wrong.
If I have an electric motor with permanent magnets, the magnets are obviously interacting. But they are not themselves doing work. The permanent magnets do not appreciably lose their magnetism, they last for the life of the vehicle.

If I place an object on the hard floor and the object is sitting there, is the floor doing work? No. If I bounce a ball off the hard floor, is the floor doing work? Again, no. The floor does expend energy in order to deflect and interact with the ball. Same for a magnet.

I can't believe we're having this discussion. Can someone else help me explain this really basic physical concept?

RB - your analogy is not quite analogous.

The floor has a "normal force" mechanism. L1 is not stable.  So it can't "support" the field generator, and so has to continuously thrust against the delta-impulse it imparts on the solar wind.  (That's the "planet-sized but inefficient solar sail" I was describing.

You can just offset slightly from the Lagrange point and use gravity (of the Sun in this case) to react against the force of the solar wind. Just like a solar sail except acting on the wind and not just the light.

Did you figure out how "slightly" this would be?
It depends on how massive the magnetosphere device is.
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Offline Hop_David

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For no work to be done, no interaction / deflection must be done, thus no flux. So yes, you could have that situation, but in that case it wouldn't do anything, so why would you do it?
..
This is completely wrong.
If I have an electric motor with permanent magnets, the magnets are obviously interacting. But they are not themselves doing work. The permanent magnets do not appreciably lose their magnetism, they last for the life of the vehicle.

If I place an object on the hard floor and the object is sitting there, is the floor doing work? No. If I bounce a ball off the hard floor, is the floor doing work? Again, no. The floor does expend energy in order to deflect and interact with the ball. Same for a magnet.

I can't believe we're having this discussion. Can someone else help me explain this really basic physical concept?

RB - your analogy is not quite analogous.

The floor has a "normal force" mechanism. L1 is not stable.  So it can't "support" the field generator, and so has to continuously thrust against the delta-impulse it imparts on the solar wind.  (That's the "planet-sized but inefficient solar sail" I was describing.
You can just offset slightly from the Lagrange point and use gravity (of the Sun in this case) to react against the force of the solar wind. Just like a solar sail except acting on the wind and not just the light.

Normal L1 is an impasse in a 3 man tug of war. Sun's gravity on one side, Mars gravity and centrifugal force on the other side.

If pressure from the solar wind were constant, you could set up the magnet a little sunward of Sun Mars L1. Then you'd be striving for a balance in a 4 man tug of war. Sun on one side, Mars gravity, centrifugal force, and pressure from solar wind on the other side.

But solar wind isn't constant. The solar sail man in this tug of war would be napping sometimes, other times it'd seem like he was on crack.

Station keeping for this L1 solar sail would be a nightmare.

Offline meekGee

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For no work to be done, no interaction / deflection must be done, thus no flux. So yes, you could have that situation, but in that case it wouldn't do anything, so why would you do it?
..
This is completely wrong.
If I have an electric motor with permanent magnets, the magnets are obviously interacting. But they are not themselves doing work. The permanent magnets do not appreciably lose their magnetism, they last for the life of the vehicle.

If I place an object on the hard floor and the object is sitting there, is the floor doing work? No. If I bounce a ball off the hard floor, is the floor doing work? Again, no. The floor does expend energy in order to deflect and interact with the ball. Same for a magnet.

I can't believe we're having this discussion. Can someone else help me explain this really basic physical concept?

RB - your analogy is not quite analogous.

The floor has a "normal force" mechanism. L1 is not stable.  So it can't "support" the field generator, and so has to continuously thrust against the delta-impulse it imparts on the solar wind.  (That's the "planet-sized but inefficient solar sail" I was describing.
You can just offset slightly from the Lagrange point and use gravity (of the Sun in this case) to react against the force of the solar wind. Just like a solar sail except acting on the wind and not just the light.

Normal L1 is an impasse in a 3 man tug of war. Sun's gravity on one side, Mars gravity and centrifugal force on the other side.

If pressure from the solar wind were constant, you could set up the magnet a little sunward of Sun Mars L1. Then you'd be striving for a balance in a 4 man tug of war. Sun on one side, Mars gravity, centrifugal force, and pressure from solar wind on the other side.

But solar wind isn't constant. The solar sail man in this tug of war would be napping sometimes, other times it'd seem like he was on crack.

Station keeping for this L1 solar sail would be a nightmare.
.. and the direction of thrust would vary, with the direction of the solar wind, and with the shape of the sail, which would drag behind the craft in a hard to predict direction.

Still though, this doesn't mean you can't counter it with an acceptable amount of thrust.

Numbers would be useful at this point, but electromagnetic dynamics is not my field (!)
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Offline Hop_David

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Normal L1 is an impasse in a 3 man tug of war. Sun's gravity on one side, Mars gravity and centrifugal force on the other side.

If pressure from the solar wind were constant, you could set up the magnet a little sunward of Sun Mars L1. Then you'd be striving for a balance in a 4 man tug of war. Sun on one side, Mars gravity, centrifugal force, and pressure from solar wind on the other side.

But solar wind isn't constant. The solar sail man in this tug of war would be napping sometimes, other times it'd seem like he was on crack.

Station keeping for this L1 solar sail would be a nightmare.
.. and the direction of thrust would vary, with the direction of the solar wind, and with the shape of the sail, which would drag behind the craft in a hard to predict direction.

Still though, this doesn't mean you can't counter it with an acceptable amount of thrust.

Numbers would be useful at this point, but electromagnetic dynamics is not my field (!)

Say we did terraform Mars, it'd take millions of years for atmospheric erosion via solar wind to make an appreciable difference. It's a very slow process.

I wouldn't be surprised if propellent spent in stationkeeping would exceed mass of atmosphere lost via solar solar wind.

Offline meekGee

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Wiki says:

Quote
The wind exerts a pressure at 1 AU typically in the range of 1–6 nPa (1–6×10−9 N/m2), although it can readily vary outside that range.

The dynamic pressure is a function of wind speed and density. The formula is

P = 1.6726×10−6 * n * V2

where pressure P is in nPa (nanopascals), n is the density in particles/cm3 and V is the speed in km/s of the solar wind.

For the sake of discussion, let's work with P(Earth)=3 nPa.

I'd expect n to go by inverse r-squared, and v (based on gravitational energy) by inverse root r.  So just as a starting point, let's take P(Mars)=1 nPa, nice and round.

Therefore we get a thrust of 1 mN for each square kilometer of effective sail area.

"effective" being the key word here.  Most of the wind flux continues with very little disturbance, but still, we're talking about shielding an entire planet (~7E6 km2) and if the wind catches the long magnetotail at an angle, you have a humongous cylindrical cross-sectional area.

Let's work with a starting point assumption of Aeff = 1E6 km2, so the required force is 1000N.  (This is an absolute WAG.  I'd say it's conservative for "head-on" wind, but might be liberal for "sideways" wind.

If the thruster ve is 3000 m/s, the mass burn rate is (F = m-dot * ve) = 1000/3000 = 0.3 kg/sec, or 25 tons per day.  Yuck.  But not completely crazy, so there's hope.

----------------

Can it be compensated for with orbital positioning?  Let's see what we can work with.

A big component of this acceleration will be radial, away from the sun.  So positioning the craft closer to the sun could help, as commented above.

L1, the article says, is 320 Mars radii away from Mars, so Mars' gravity there is 1/100,000 of Mars surface gravity, or 0.00004 m/s2.  By varying the position of the spacecraft (towards the sun(, we can recoup a fraction of that. Let's say 10%.  So 0.000004 m/s2.

We want to generate 1000N, so the mass of the spacecraft has to be m=F/a=25,000 tons.  (Did I get that right?)

k.  That's more than yuck.

Thrust it is then.

----

Since we're talking about terraforming an entire planet, 25 tons/day is not crazy.

David Hop above points out that this is more than today's mass loss rate (which is 0.1 kg/sec) but I'd argue that this mass loss rate would increase the minute the atmosphere starts thickening, so this will become a net positive.

Also, my Aeff=1E6 km2 is really a wild guess.  I think I'm on the conservative side.  If I'm off by even 1 OOM, this goes from "yuck" to "ho hum".

Much will depend on the shape of the magnetosheath.  If it can be shaped to be more "aerodynamic", Aeff will decrease.

Here's an idea: use multiple generators.  A small one near the sun, followed by a larger one downstream, etc.  This will make the magnetosheath more streamlined, with less "head" area, and hopefully less drag.
« Last Edit: 03/08/2017 06:35 PM by meekGee »
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Online Space Ghost 1962

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Back to the "work" argument. The root of the confusion is the presumption of constants not time varying.

E.g. that the tensor isn't present. It is.

Offline meekGee

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Back to the "work" argument. The root of the confusion is the presumption of constants not time varying.

E.g. that the tensor isn't present. It is.

You lost me...

I see to places where work is needed.

- Maintaining thrust to oppose the impulse change in the solar wind (since the orbital tricks seems much much too weak)
- Maintaining the magnetic field (which I'm not sure if it really requires work) - and is an energy, not momentum, consideration.

Which of these are you addressing?
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Online Robotbeat

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For no work to be done, no interaction / deflection must be done, thus no flux. So yes, you could have that situation, but in that case it wouldn't do anything, so why would you do it?
..
This is completely wrong.
If I have an electric motor with permanent magnets, the magnets are obviously interacting. But they are not themselves doing work. The permanent magnets do not appreciably lose their magnetism, they last for the life of the vehicle.

If I place an object on the hard floor and the object is sitting there, is the floor doing work? No. If I bounce a ball off the hard floor, is the floor doing work? Again, no. The floor does expend energy in order to deflect and interact with the ball. Same for a magnet.

I can't believe we're having this discussion. Can someone else help me explain this really basic physical concept?

RB - your analogy is not quite analogous.

The floor has a "normal force" mechanism. L1 is not stable.  So it can't "support" the field generator, and so has to continuously thrust against the delta-impulse it imparts on the solar wind.  (That's the "planet-sized but inefficient solar sail" I was describing.
You can just offset slightly from the Lagrange point and use gravity (of the Sun in this case) to react against the force of the solar wind. Just like a solar sail except acting on the wind and not just the light.

Normal L1 is an impasse in a 3 man tug of war. Sun's gravity on one side, Mars gravity and centrifugal force on the other side.

If pressure from the solar wind were constant, you could set up the magnet a little sunward of Sun Mars L1. Then you'd be striving for a balance in a 4 man tug of war. Sun on one side, Mars gravity, centrifugal force, and pressure from solar wind on the other side.

But solar wind isn't constant. The solar sail man in this tug of war would be napping sometimes, other times it'd seem like he was on crack.

Station keeping for this L1 solar sail would be a nightmare.
Now THIS is a very good point!

The solar wind typically has about 10 MN of force on a magnetosphere the size of the Earth's. That's about an F1 or two's worth of thrust.
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Online Space Ghost 1962

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Could a huge magnet turn the Red Planet green?

Quote from: Bruce Jakosky
At current loss rates driven by the sun and solar wind as measured by MAVEN, it would take about 2 billion years to remove the present atmosphere…. If we put up an artificial magnetosphere, it would be very intriguing in terms of physics.  But it would have minimal effect on the thickness of the atmosphere, the global temperature, or the behavior of the polar caps. At least not for billions of years.

Offline meekGee

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For no work to be done, no interaction / deflection must be done, thus no flux. So yes, you could have that situation, but in that case it wouldn't do anything, so why would you do it?
..
This is completely wrong.
If I have an electric motor with permanent magnets, the magnets are obviously interacting. But they are not themselves doing work. The permanent magnets do not appreciably lose their magnetism, they last for the life of the vehicle.

If I place an object on the hard floor and the object is sitting there, is the floor doing work? No. If I bounce a ball off the hard floor, is the floor doing work? Again, no. The floor does expend energy in order to deflect and interact with the ball. Same for a magnet.

I can't believe we're having this discussion. Can someone else help me explain this really basic physical concept?

RB - your analogy is not quite analogous.

The floor has a "normal force" mechanism. L1 is not stable.  So it can't "support" the field generator, and so has to continuously thrust against the delta-impulse it imparts on the solar wind.  (That's the "planet-sized but inefficient solar sail" I was describing.
You can just offset slightly from the Lagrange point and use gravity (of the Sun in this case) to react against the force of the solar wind. Just like a solar sail except acting on the wind and not just the light.

Normal L1 is an impasse in a 3 man tug of war. Sun's gravity on one side, Mars gravity and centrifugal force on the other side.

If pressure from the solar wind were constant, you could set up the magnet a little sunward of Sun Mars L1. Then you'd be striving for a balance in a 4 man tug of war. Sun on one side, Mars gravity, centrifugal force, and pressure from solar wind on the other side.

But solar wind isn't constant. The solar sail man in this tug of war would be napping sometimes, other times it'd seem like he was on crack.

Station keeping for this L1 solar sail would be a nightmare.
Now THIS is a very good point!

The solar wind typically has about 10 MN of force on a magnetosphere the size of the Earth's. That's about an F1 or two's worth of thrust.

If you read above, I got 1 kN as a ballpark estimate, at Mars distance, on what I estimated was drag on the sail.

This will be the equivalent of, oh, 10-100 kN for Earth.  A couple of OOMs from your number.

I want to find out if I have a mistake there - do you have the source?
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Online Robotbeat

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Could a huge magnet turn the Red Planet green?

Quote from: Bruce Jakosky
At current loss rates driven by the sun and solar wind as measured by MAVEN, it would take about 2 billion years to remove the present atmosphere…. If we put up an artificial magnetosphere, it would be very intriguing in terms of physics.  But it would have minimal effect on the thickness of the atmosphere, the global temperature, or the behavior of the polar caps. At least not for billions of years.
I totally agree with this. The magnetosphere is the last thing we would do (well, before full oxygen) while terraforming Mars.
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Offline meekGee

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Could a huge magnet turn the Red Planet green?

Quote from: Bruce Jakosky
At current loss rates driven by the sun and solar wind as measured by MAVEN, it would take about 2 billion years to remove the present atmosphere…. If we put up an artificial magnetosphere, it would be very intriguing in terms of physics.  But it would have minimal effect on the thickness of the atmosphere, the global temperature, or the behavior of the polar caps. At least not for billions of years.
I totally agree with this. The magnetosphere is the last thing we would do (well, before full oxygen) while terraforming Mars.

Without numbers, we got nothing, just bouncing back and forth between "it won't work" and "yes it will!".

It really brings us back to the required thrust levels, and energy consumption.  It also depends on how the planet/atmosphere will react to the increase in pressure that will follow - will it be a runaway effect?

This can be the greatest concept to hit the airwaves in a very long time, or just an interesting theoretical discussion.

It's interesting enough that I'm sure the follow-up math will be done.
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Online Robotbeat

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Wait, who is not looking at numbers? The erosion rates of the atmosphere are very small and would remain small even while terraforming. So a runaway effect (if present) wouldn't be greatly affected by the presence or absence of the field. As long as we can start it, we should be able to trigger the runaway through easier means.

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Offline meekGee

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Wait, who is not looking at numbers? The erosion rates of the atmosphere are very small and would remain small even while terraforming. So a runaway effect (if present) wouldn't be greatly affected by the presence or absence of the field. As long as we can start it, we should be able to trigger the runaway through easier means.

I was referring to the discussion upthread about energy and impulse, sitting "near" L1, etc.

As for the erosion rate, you clearly can't undo in 100 years what has taken millions (or billions) to occur - unless you have some positive feedback mechanism which they were alluding to.

Looking at the numbers, however:
- Mars loses 0.1 kg/sec (they say), which is 3.1 E6 kg/yr.
- Mars atmospheric mass is 2.5E16 kg.   (https://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html)
- Straight time constant for a mere doubling of the atmosphere is therefore E10 years.  10 Billion.

So clearly it's not linear, since Mars isn't that old even.  Or the erosion rate is wrong. Or magnetic fields have nothing to do with atmospheric loss.

Hence the need for modeling, and some of the work they hinted at.

EDIT:
This abstract arrives at a similar conclusion:
http://science.sciencemag.org/content/315/5811/501

EDIT2:
Looked some more, and there's some head scratching going on. If atmospheric loss has to do with lack of magnetic field, then there must have been much higher erosion rates in the past.  .
« Last Edit: 03/09/2017 05:43 AM by meekGee »
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Offline JasonAW3

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A few basic questions here;

      1. What strength of magnetic field would be needed to LARGELY eliminate atmospheric erosion by solar winda and coronal mass ejections?

      2. Assuming that it would be significantly less than that that is required for Earth, (radiation reduction rules would likely be in effect.  Radiation levels are reduced by the square of the distance from the source of radiation) how much power would be required to produce such a field effect?

      3. Utilizing high temperature superconducting ribbons, wrapped around the planet to produce such a field, could fields of solar panels produce enough power to produce such a field?  (Don't balk at this; producing such lengths of superconducting ribbon is possible, although likely at a lower rate than we do for conventional wire, but it is possible).

      4. Are the materials needed to produce such lengths of superconducting wire available on Mars, or would extraction from asteroids be needed?

      Obviously the above questions are fanciful speculation, but I suspect that such a system would be less expensive, more practical and less prone to failure than an L1 based magnetic shield.

      Not that I am faulting the initial reasoning, but an enveloping Magnetic Field would likely be FAR more effective that one protecting a planet from only one side.  Besides, you'd still have some spillage from around the edges of such a shield that would still affect the planet, unless you made the shield many times the diameter of Mars itself.
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Offline meekGee

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A few basic questions here;

      1. What strength of magnetic field would be needed to LARGELY eliminate atmospheric erosion by solar winda and coronal mass ejections?

      2. Assuming that it would be significantly less than that that is required for Earth, (radiation reduction rules would likely be in effect.  Radiation levels are reduced by the square of the distance from the source of radiation) how much power would be required to produce such a field effect?

      3. Utilizing high temperature superconducting ribbons, wrapped around the planet to produce such a field, could fields of solar panels produce enough power to produce such a field?  (Don't balk at this; producing such lengths of superconducting ribbon is possible, although likely at a lower rate than we do for conventional wire, but it is possible).

      4. Are the materials needed to produce such lengths of superconducting wire available on Mars, or would extraction from asteroids be needed?

      Obviously the above questions are fanciful speculation, but I suspect that such a system would be less expensive, more practical and less prone to failure than an L1 based magnetic shield.

      Not that I am faulting the initial reasoning, but an enveloping Magnetic Field would likely be FAR more effective that one protecting a planet from only one side.  Besides, you'd still have some spillage from around the edges of such a shield that would still affect the planet, unless you made the shield many times the diameter of Mars itself.

They worked it out to be only 1-2 Tesla.

The reason it is so small (IMO) is the fact that you're creating such a long magneto bubble, instead of a round one.  You only need to deflect those particles by a tiny amount at those distances, and they'll miss the planet.

I think if you try to do it right around the planet, you'll need to create a much higher magnetic field.

I am very uncertain about the BoE calculations above.  Magnetic fields behave in all sorts of unexpected ways.  Maybe this trick is very much feasible.  The thing is - the current atmospheric mass loss rates seem to be out of whack - something I'm sure hasn't escaped the proposal's authors' attention.

If there's meat to it, it's interesting enough that there will be a follow up
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Offline JasonAW3

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Are we talking too high a loss or too low?

      If too.high, it could be some atmospheric "fixing" mechanism at work.  If too low, could be someting causing a release of "fixed" atmospheric components from the soil.  Either way, something weird is going on.

      I also wonder i the "water releases" could be qn odd combinatin of atmospheric chemicals either  in solution or as a weird low temperature liquid that may be being broken down by the high UV at surface level.
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wait... it is my understanding that if you somehow returned Mars' atmosphere such that it had the Earth's distribution of gases and density and then did nothing to retain it in place it would still take a period of time longer than human kind has existed for it to erode/devolve again to it's present state. That's a heck of a long time. Plenty of time for civilizations to rise and fall and even for species to rise and fall. So i don't get it...Why exactly do we need an artificial magnetic field except for the sake of OCD completionists?
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Offline JasonAW3

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wait... it is my understanding that if you somehow returned Mars' atmosphere such that it had the Earth's distribution of gases and density and then did nothing to retain it in place it would still take a period of time longer than human kind has existed for it to erode/devolve again to it's present state. That's a heck of a long time. Plenty of time for civilizations to rise and fall and even for species to rise and fall. So i don't get it...Why exactly do we need an artificial magnetic field except for the sake of OCD completionists?

Ok, this is pretty simple, at least n the face of it.

      Restoring the pressure to hear Earth levels will be difficult enough and take at least a century or more, even without the constant erosion of the atmosphere as it I now.  With the addition of the solar winds eroding what little atmosphere there is now, in addition to what is extracted from the soil and rocks, the task would likely be extended to millennia, if not made impossible all together.

      Even assuming that one could speed the process by dropping every stray comet and wet asteroid that you could, (and that's a LOT of Delta v needed for that) it really wouldn't speed the process much without limiting the atmospheric erosion from the solar winds.
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bear with me here as i have dropped a few digits of memory precision on this; but I have read that the time for erosion from Earth-like levels of atmosphere for Mars from all sources of erosion and other factors like chemical sequestration and things like that is either 300 million years or else 300 thousand years from the time Mars' dynamism ended. Either way it's more time than humanity or human civilization or recorded history has existed.

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bear with me here as i have dropped a few digits of memory precision on this; but I have read that the time for erosion from Earth-like levels of atmosphere for Mars from all sources of erosion and other factors like chemical sequestration and things like that is either 300 million years or else 300 thousand years from the time Mars' dynamism ended. Either way it's more time than humanity or human civilization or recorded history has existed.

Once you have an Earth-like atmosphere it will take a long time for it to erode, but first you have to create that atmosphere. Reducing erosion will make it easier to build up the atmosphere.

Offline ppnl

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bear with me here as i have dropped a few digits of memory precision on this; but I have read that the time for erosion from Earth-like levels of atmosphere for Mars from all sources of erosion and other factors like chemical sequestration and things like that is either 300 million years or else 300 thousand years from the time Mars' dynamism ended. Either way it's more time than humanity or human civilization or recorded history has existed.

Once you have an Earth-like atmosphere it will take a long time for it to erode, but first you have to create that atmosphere. Reducing erosion will make it easier to build up the atmosphere.

Mars is losing atmosphere at  a rate of about 0.25 pounds per second. If you can't add atmosphere many orders of magnitude faster than this then you may as well not bother. If you can then the loss rate is a very distant secondary consideration.

I have seen some estimates that the Earth is losing atmospheric mass ten times as fast as Mars.

I really don't think they have all the science settled yet.

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bear with me here

GrrrrrRRrrrr! Raaaaahhrrr!!!   *mauls a camper*

(sorry, I couldn't help myself  ;)  )

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