Dr. Phil Metzger @DrPhiltill 17m17 minutes agoCool! Jim Green: put a magnetic shield at the Sun-Mars L1 point to shield Mars' atmosphere. Gigantic atmospheric test on Mars. #V2050
Dr. Phil Metzger @DrPhiltill 10m10 minutes agoWoah. Really? Modeling say this magnetic field can raise Mars' atmosphere to 1/2 of Earth's pressure in just years, not centuries. #V2050
Sheyna Gifford @humansareawesme 13m13 minutes ago"Mars might look like this in 700 million years - OR SOONER." Jim Green models a planetary shield to heat up #Mars #v2050 @AstrobiologyMaghttps://t.co/6TuVxFvt1o
How much power is needed to generate such a field?
What's the means of generating and distributing such a field? Large magnetoplasma?Or how about a large swarm of laser-formation flying satellites with superconductive coils?
... because the solar wind comes in at a large angle (45 degrees?) at Mars.
Let's see, if they're talking about using solar wind to make atmospheres... at 1AU it's 6e8 particles per square centimeter per second... so 6e12 per square meter per second... Mars is 1,52, so with (overly) simplistic quadratic scaling assumed that'd be 2,6e12 per square meter per second... Earth's atmosphere is ~5e18kg... let's say you want 5e17 for Mars... solar wind is about 1.1g/mol... and we have 4,3e-12 mol/mē-s... let's say 100 years, so 1,36e-5 kg/mē... so to produce 5e17kg would require 3,67e22mē... aka 3,67e16kmē... equivalent to a disc 108 million km in radius... aka 0,72 AU in radius.... aka a pretty dang large segment of the solar wind Somehow I doubt that's what's being referred to here. They must just be talking about how much atmosphere would accumulate from Mars itself when you stop stripping. Could it really be that fast? Even if so, Mars has had its nitrogen stripped long ago, so I'm not sure where one could get a replacement.
Quote from: Rei on 03/02/2017 03:11 pmLet's see, if they're talking about using solar wind to make atmospheres... at 1AU it's 6e8 particles per square centimeter per second... so 6e12 per square meter per second... Mars is 1,52, so with (overly) simplistic quadratic scaling assumed that'd be 2,6e12 per square meter per second... Earth's atmosphere is ~5e18kg... let's say you want 5e17 for Mars... solar wind is about 1.1g/mol... and we have 4,3e-12 mol/mē-s... let's say 100 years, so 1,36e-5 kg/mē... so to produce 5e17kg would require 3,67e22mē... aka 3,67e16kmē... equivalent to a disc 108 million km in radius... aka 0,72 AU in radius.... aka a pretty dang large segment of the solar wind Somehow I doubt that's what's being referred to here. They must just be talking about how much atmosphere would accumulate from Mars itself when you stop stripping. Could it really be that fast? Even if so, Mars has had its nitrogen stripped long ago, so I'm not sure where one could get a replacement.From this article: https://www.nasa.gov/press-release/nasa-mission-reveals-speed-of-solar-wind-stripping-martian-atmosphere, Mars loses about 100 g of atmosphere per second. If you could tune that down to zero somehow, and assuming all the atmospheric sources on Mars are in equilibrium with loss, you would just gain 0.1 * 86400 * 365 = 3 million kg per year. The total mass of the Atmosphere or Mars is about 2.5 x 10^10 million kg. So I don't understand where that suggestion ("500 mbar in 5 years") is coming from. Also, how does it square with the suggestion that higher pressures than 50 mbar are unstable? A bit more context would be nice.EDIT: Article on phys.org: https://phys.org/news/2017-03-nasa-magnetic-shield-mars-atmosphere.html
Quote from: Bynaus on 03/04/2017 11:44 amQuote from: Rei on 03/02/2017 03:11 pmLet's see, if they're talking about using solar wind to make atmospheres... at 1AU it's 6e8 particles per square centimeter per second... so 6e12 per square meter per second... Mars is 1,52, so with (overly) simplistic quadratic scaling assumed that'd be 2,6e12 per square meter per second... Earth's atmosphere is ~5e18kg... let's say you want 5e17 for Mars... solar wind is about 1.1g/mol... and we have 4,3e-12 mol/mē-s... let's say 100 years, so 1,36e-5 kg/mē... so to produce 5e17kg would require 3,67e22mē... aka 3,67e16kmē... equivalent to a disc 108 million km in radius... aka 0,72 AU in radius.... aka a pretty dang large segment of the solar wind Somehow I doubt that's what's being referred to here. They must just be talking about how much atmosphere would accumulate from Mars itself when you stop stripping. Could it really be that fast? Even if so, Mars has had its nitrogen stripped long ago, so I'm not sure where one could get a replacement.From this article: https://www.nasa.gov/press-release/nasa-mission-reveals-speed-of-solar-wind-stripping-martian-atmosphere, Mars loses about 100 g of atmosphere per second. If you could tune that down to zero somehow, and assuming all the atmospheric sources on Mars are in equilibrium with loss, you would just gain 0.1 * 86400 * 365 = 3 million kg per year. The total mass of the Atmosphere or Mars is about 2.5 x 10^10 million kg. So I don't understand where that suggestion ("500 mbar in 5 years") is coming from. Also, how does it square with the suggestion that higher pressures than 50 mbar are unstable? A bit more context would be nice.EDIT: Article on phys.org: https://phys.org/news/2017-03-nasa-magnetic-shield-mars-atmosphere.htmlWell, that conference abstract/paper provides a bit more context. It doesn't say much about how long all that takes (except that it will be studied), so that "years, not centuries" seems to be some sort of (optimistic) extrapolation/guess.
Right, I missed that. Link: http://www.hou.usra.edu/meetings/V2050/pdf/8250.pdfQuite vague, especially the part on the "new equlibrium". But it is quite remarkable that a temperature change of only 4 K would be enough to start melting the polar caps and resulting in a higher atmospheric pressure as a consequence.
Quote from: Robotbeat on 03/04/2017 02:13 am... because the solar wind comes in at a large angle (45 degrees?) at Mars.Do you have a source for this?
Quote from: as58 on 03/04/2017 10:54 amQuote from: Robotbeat on 03/04/2017 02:13 am... because the solar wind comes in at a large angle (45 degrees?) at Mars.Do you have a source for this?I was wrong. At worst, it comes in at about 10 degrees at Earth. Also, the Earth's magnetotail is wide, such that Earth-Sun-L2 is usually inside the magnetotail.The interplanetary magnetic field (embedded in the solar wind), however, is at 45 degrees, and that is the rough direction tha solar energetic particles come at.
Quote from: Dalhousie on 03/02/2017 11:19 pmHow much power is needed to generate such a field?None, once established.
Quote from: Robotbeat on 03/02/2017 11:51 pmQuote from: Dalhousie on 03/02/2017 11:19 pmHow 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?
Quote from: Dalhousie on 03/05/2017 10:15 pmQuote from: Robotbeat on 03/02/2017 11:51 pmQuote from: Dalhousie on 03/02/2017 11:19 pmHow 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.
Quote from: Robotbeat on 03/05/2017 10:22 pmQuote from: Dalhousie on 03/05/2017 10:15 pmQuote from: Robotbeat on 03/02/2017 11:51 pmQuote from: Dalhousie on 03/02/2017 11:19 pmHow 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?
NO power required except for the initial 10^19 Joules to get the field established, at least if you just use superconducting coils.
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.
I'm talking about enormous superconducting coils.
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).
Quote from: Robotbeat on 03/02/2017 11:51 pmQuote from: Dalhousie on 03/02/2017 11:19 pmHow 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.
Quote from: meekGee on 03/07/2017 07:04 amQuote from: Robotbeat on 03/02/2017 11:51 pmQuote from: Dalhousie on 03/02/2017 11:19 pmHow 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.
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.
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?..
Quote from: Space Ghost 1962 on 03/07/2017 08:37 pmFor 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?
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.
Quote from: Robotbeat on 03/08/2017 01:20 amQuote from: Space Ghost 1962 on 03/07/2017 08:37 pmFor 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.
Quote from: Danderman on 03/08/2017 03:25 amMy 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?
Quote from: meekGee on 03/08/2017 03:20 amQuote from: Robotbeat on 03/08/2017 01:20 amQuote from: Space Ghost 1962 on 03/07/2017 08:37 pmFor 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.
Quote from: Robotbeat on 03/08/2017 04:16 amQuote from: meekGee on 03/08/2017 03:20 amQuote from: Robotbeat on 03/08/2017 01:20 amQuote from: Space Ghost 1962 on 03/07/2017 08:37 pmFor 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?
Quote from: Robotbeat on 03/08/2017 04:16 amQuote from: meekGee on 03/08/2017 03:20 amQuote from: Robotbeat on 03/08/2017 01:20 amQuote from: Space Ghost 1962 on 03/07/2017 08:37 pmFor 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.
Quote from: Hop_David on 03/08/2017 02:58 pmNormal 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 (!)
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.
The wind exerts a pressure at 1 AU typically in the range of 16 nPa (16Ũ10−9 N/m2), although it can readily vary outside that range.The dynamic pressure is a function of wind speed and density. The formula isP = 1.6726Ũ10−6 * n * V2where 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.
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.
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.
Quote from: Hop_David on 03/08/2017 02:58 pmQuote from: Robotbeat on 03/08/2017 04:16 amQuote from: meekGee on 03/08/2017 03:20 amQuote from: Robotbeat on 03/08/2017 01:20 amQuote from: Space Ghost 1962 on 03/07/2017 08:37 pmFor 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.
Could a huge magnet turn the Red Planet green?Quote from: Bruce JakoskyAt 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.
Quote from: Space Ghost 1962 on 03/09/2017 01:02 amCould a huge magnet turn the Red Planet green?Quote from: Bruce JakoskyAt 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.
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.
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.
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?
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.
Quote from: Stormbringer on 03/28/2017 12:07 pmbear 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.
bear with me here
"Despite stronger solar wind and EUV-radiation levels under the early Sun, ion escape can not explain more than 0.006 bar of atmospheric pressure lost over the course of 3.9 billion years," says Robin Ramstad. "Even our upper estimate, 0.01 bar, is an insignificant amount in comparison to the atmosphere required to maintain a sufficiently strong greenhouse effect, about 1 bar or more according to climate models."The results presented in the thesis show that a stronger solar wind mainly accelerates particles already escaping the planet's gravity, but does not increase the ion escape rate. Contrary to previous assumptions, the induced magnetosphere is also shown to protect the bulk of the Martian ionosphere from solar wind energy transfer.Read more at: https://phys.org/news/2017-12-mars-atmosphere-solar.html#jCp