Author Topic: Increasing Mars' atmosphere with magnetic field @ Sun-Mars L1  (Read 8596 times)

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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Offline 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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Offline 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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Offline 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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Offline Space Ghost 1962

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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.

Online 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).
« Last Edit: 03/06/2017 03:11 AM by Robotbeat »
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Offline Space Ghost 1962

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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 »

Online 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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Online 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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Offline 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.

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?
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Online 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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