PB666

Woodchucks cant chuck wood.

Although the universe is less than 13.9 billion years in age, its believed to be at least 42 billion light years in diameter. The Methuselah star has an estimated age older than the Universe.

Well NK is right no at about 1956 space technology, moving about 1/2 the speed of the US space program. Its the equivalent of bankrupt . . . . . . They could send people to Mars against their will.

Again if we use that logic the colony is necessarily dependent on Earth. Without cedar trees, rubber trees or graphite mines there are no pencils. It would be cheaper to import these items assembled than to attempt to grow them. After the first manned mission and at least for decades thereafter the only manned missions will be land, gather, some manipulation and leave. Robots will have to build the infrastructure. Where on earth are circumstances comparable to Mars, Antarctica has lots of land, but in the interior of the continent only a handful of people stay the entire Antarctic winter. Obviously people could build greenhouses and whatnots for extended human survival in Antarctica, but they don't. I can pick an infinite number of cheaper places on earth to buy land and develop. The country of western Sahara has a long beach you can use solar panels to desalinate water that can be used to grow plants and animals where it is 100 times easier to do than Antarctica, and yet people still aren't doing it there, either. Either in Sahara or in Antarctica or on the coast of Namibia, someone needs to make the initial expensive investment to turn minimal outposts into places suitable for colonization. Mars is the same but the transportation costs are 100s of times more expensive, and if you decide to build X at t=0 then finding a rocket, building X, getting it into LEO and then a 9 month mission followed by landing on Mars, disassembly of the shipping fairing, reassembly, etc. Maybe 5 years for building an 18th century outhouse.

" Rotary Rocket failed due to lack of funding, but some[who?] have suggested that the design itself was inherently flawed. The Rotary Rocket did fly three test flights and a composite propellant tank survived a full test program, however these tests revealed problems. For instance, the ATV demonstrated that landing the Rotary Rocket was tricky, even dangerous. Test pilots have a rating system, the Cooper-Harper rating scale, for vehicles between 1 and 10 that relates to difficulty to pilot. The Roton ATV scored a 10 — the vehicle simulator was found to be almost unflyable by anyone except the Rotary test pilots, and even then there were short periods where the vehicle was out of control.[citation needed] Other aspects of the flight plan remained unproven and it is unknown whether Roton could have developed sufficient performance to reach orbit with a single stage, and return – although on paper this might have been possible. These doubts led some of the aerospace community to dismiss the Rotary Rocket concept as a pipe dream.[citation needed] Whether the concept would have worked successfully remains open to speculation.[citation needed] " -Wikipedia. Not to mention the fact that it has never been tested on Mars or in an atmosphere of similar density.

I see, so your tilting at windmills. I have to tell you that it will not provide the retrograde thrust you need to land.

No it wouldn't, the cave system is likely cracked and has buried tubes going to the surface. YOU would put air in and it would either leak out or blow-up at somepoint with the stress fractures finally give. Use the Anglo-Roman philosphophy of divide an conquer. Don't bite of more than you can chew, certainly not more than the resulting xxxx pile would cover your head (as in caveins and blowing of precious little colonist into the vacuum of space)

I have thought about this for about a millsecond, I would not go to Mars if someone paid me lots of money, not if they made me lord king. I would rather have a grass hut in a mosqutio infested jungle in Central America with a mangy dog, a lice infested wife and precocious drug dealing kids .........than spend 14 months in a tin can watching my already bad vision go to pot, taking more blood pressure medication, eating unfresh food, cleaning up poop scatter, breathing stale air, wearing sweat saturated clothes, then landing on a dusty planet my only real function is to clean solar panels after every dust storm. You can add the lack of mental exercise giving way to idle-fears of resupply failure, my general failure of growing plants with LED, etc, etc, etc. Im sure the uninformed can come up with rather robust reasons for traveling to Mars, truth lies at the intersection of valid perspectives, I just wonder how many of them are using valid perspectives in their decision making process.

Nope, you trying to equivocate mid 17th century technology on a 2000 km journey to mid/late 21st century technology on a 550,000,000 km journey. The pilgrims were not the first europeans in temperate North Americas, those came 300 years earlier. The vinland colonies were a complete loss. The Jamestown colony was a complete loss. If you want a better 17th century comparable lets consider the mortality risk of pilgrims creating a nudist colony at the south pole during the middle of the southern winter on a snow pack that has a 20,000 meter elevation, with an arrival method by catapult from elephant island and parachute as the only braking device. No I think we will gain negative growth much sooner, something big will happen cause we have alot of idiots in powerful positions now. You can make beads on Mars and sell them to the Natives! Viva Columbus!

For just about every journeys there are questions that need to be made. Just about all of us have taken off from Kerbin with 10k dV in the Moho's orbital direction and finding out, when we go close to Moho that we did not have enough fuel left to circularize (or some other nice oversight like pointing the solar panel in the direction of the sun for 6 months). Can we sit down with a spreadsheet and make decisions that factor the choices that are presented. I decided to basically write these posts because of the eccentricity thread in order to illustrate what the real value of eccentricity is when all is said and done. To make the answer short, energy is much more important, but eccentricity gives us on the fly information. For example if you are circularizing from an eccentric orbit close to Pe or Apo (whichever is the burn point) and your delta-e/t is too low compared to time to pe/apo, this informs you that you need to increase thrust or should have carried a more powerful engine. Another situation is during reentry from distal targets, the delta-e/t tells you how rapidly or effective your entry-theta was. If your ship is overheating and your e is not likely to approach zero at some point during reentry then you probably should have used a bigger shield(or kept some retro fuel) and choosen a steeper entry angle (lower no-ATM Pe). The eccentricity argument has an effective range of 0.0005 to 1.0000 below or above which are meaningless in the game. In this case an eccentricty of 0.5 = 0.4995 to 0.5005. This differs from other stats such as dV which are accurate over a 100,000 fold range in the game, altitudes are accurate to >10 decimal places. IOW, the values used to derive e are much more precise than e itself. Does it make a difference, yes and no, theoretically if you had a TWR = infinity, an exactly angle to prograde for a perfect burn (with perfect thruster control) there is no wasted dV and you end up intersecting the minimum orbit of the target planet. In reality the dV calculated at best puts the craft in a range were RCS thrusters in the departure orbit can be used to get within about 5000 meters of the perfect arrival orbit (on a good day). However knowing energy makes some logical sense of what is going on, for example by the Oberth effect works, why burn from low orbit, why use kicks on lowTWR craft in low orbits (versus spiralling away from the celestial). When we are using e for on-the-fly decision making accuracy is not really an issue, however in the formulation of travel strategies we do want to use as accurate as possible starting information. So what about everything else? The procedure is this. Step one. For a target planet orbital ap _and_ pe (meaning two parallel analyses), assign a departure and arrival altitudes relative to kerbol, transform to radius, derive a. Step two. Assign u/a (Escape energy) and u/2a (SKE at a) Step three. Assign SPE changes from kerbin-to-a and from a-to-target. Step four. Assign deltaSPE changes (changes in Kinetic energy) from a to kerbin or target. Step five. Calculate SKE at kerbin or target. Step six. determine dV required to achieve kerbin or target orbits without entering kerbin or targets SOI. Step seven. determine the SKE at planets SOI entry or exit based on step six. Step eight. Add this to planets minimum orbit radius escape energy, this give energy to reach minimum orbit around the planet and free fall to planet. Step nine. Convert this to dV required to free fall from minimum stable orbit Step ten. Subtract the circular orbital velocity from freefall at minimum orbit dV requirement. Step eleven. Add the two dV (kerbin and target planet) together and get total dV. At 6 specific points in the 11 step process unique energy parameters were used to derive decision making information. The table below compares the Total dV (m/s) cost of intersecting orbits (values rounded for clarity) and also compares to inclination dV performed in circumkerbol orbit. Planet Target Intrcpt δV inclination at Apo at Pe dV at a Moho 4724 4001 723.1 1818 (Depart from Kerbin at the Kerbin-Moho inclination node closest to Moho's Apo, inclination nodes are priorities) Eve 2911 2913 002 400 (Eve's orbital inclination nodes are priority) Duna 1928 3009 1081 78 (Depart from Kerbin close to Duna's Pe) Dres 2819 4837 2081 466 (Depart from kerbin closest to Dres's Pe, inclination nodes should be also considered) Jool 5202 5686 484.0 79 (more analysis of satellites requires) Eeloo 3416 3449 32.7 386 (Eeloo's orbital inclination nodes are the priority). As we can see above the analysis is devoid of any consideration of the e parameter, although it is easily obtained from the information we have. How can we get those pesky inclination nodes. One way is to place a satellite in a Kerbinesce orbit at theta = 2/3 pi and 4/3 pi relative to kerbin (in the same orbit as Kerbin but at maximum distance. Then target a planet, the nodes will show up also relative to kerbin. Such satellites can have a dual function since one can also place a deep space array on the satellite. That allows communication to objects that current orbit is on the other side of Kerbol. [Another set of energy and dV calculations that involve the equation The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit. in which we create an orbit a which is 2/3 or 4/3 the period of kerbin by generating a periapsis or apoapsis (respectively) whose average with kerbin orbit gives us a]. Disclaimers. Above assumes two sets of reciprocal process, launch/ascent & descent/land and transfer commence and complete that have a boundary at the minimum stable orbit of both planets. 1. There is a simpler mechanic, single step "star trek" mechanic, in which your ship has so much power that you travel from departure x,y,z, vx, vy, vz to arrival x, y, z, vx, vy, vz before either of which have significantly changed position (arrival point and departure points are oriented to each other). Given that human life could never survive the dV/dt and the dV & TWR required do not exist this scenario can be disregarded and physically impossible and should not be considered. Deep space probes such as New Horizons need not depart into a circular orbit. That if they reach an angle to prograde of theta>090' while still in upper atmosphere while at the same time burning the dV required for a highly eccentric orbit does not require circularization and may use less dV. In such a burn the base assumption is that the best delta-SKE on hill sphere exit is obtained from the lowest altitude burn even if that burn starts at a suborbital trajectory with pe always below safe circular orbit altitude. To convert this to KSP the ascent angle is say inclination is at >15' and you are at alt=45 you simply can burn to Eeloo Apo intersect as you are crossing some theta~90 and have a large plume of overheated gas momentarily burning through the fairings before they are deployed. 2. KSP provides perfect examples, kerbin is in a zero inclination, zero eccentricity orbit about kerbol and as it so happens it common departure planet or arrival planet for most of the transfers. The argument of periapsis differs for all other planets so you are not going to be able to match a Apo/Pe angle of a planetary departure with 180 degree Pe/Apo of a planetary arrival except from Kerbin because kerbins orbit a=Pe=Apo. This has some relevance for the Eeloo, Jool transfer which is damn cheap relative to traveling back to kerbin and traveling to Jool (given some dV spent on plane matching). 3. The smallest-sweep area stable orbit about a planet may be unpreferential. Jool being the example. This infers there is complex decision making involved in getting to a system moon in which the planet is not the target. mechanical thermodynamics can be used to burn less than the amount needed to circularize at the planets rMin, that dV would be used to circularize at an apo that intersects the moons orbit. In the case of Jool, all three inner planets have a=pe=apo, so this is not too much of a problem. In these instances you want to compare intersecting the moons orbit directly (using a different planetary Rx,orbit) versus a hohmann transfer to 200k Jool-altitude and a partial circularization burn to intersect the target orbit and circularize. 4. If we make the assumption that inclination nodes are approximate to r = a (semi-major axis), that the dV required for inclination burn (see table) is low enough not to be a priority. In these cases we can, if we desire burn at a bearing above or below the departure planets equitorial plane on depart to send the inclination node to r = a and get rid of some inclination. In comparing the table below the difference between an Kerbin-Moho transfer Pe-target and Apo-target is delta-dV = 724 but the inclination change dV averages at 1814. Therefore its simply intelligent to set a priority on changing planes over departing theta = Moho's apo theta (fortunately Moho-apo is relatively close to the inclination node). The cost of changing inclination at kerbol is reduced by 100s of dV. The same logic is also true for Eve, and Eeloo. For the other planets a departure window closest to the target planets periapsis is a better choice than choosing a departure window closest to an inclination node. 5. Entry burns particularly on planets like Jool need transfers that seldomly overlap with their pe or Apo, consequently there is a triangulation between time to get good window for efficient inclination change, or close to Jool theta. In other instances like Moho, which is so small oberth effect is minimal, free burn times at pe near a kerbin inclination node is going to occur separately than the moho circularization burn. 6. Depending of kerbol relative altitude of the target the true burn altitude is different from the planets altitude. Our burn starts 670,000 meters closer but a maximum efficiency burns leaves kerbin's SOI at the moment of crafts circumkerbol Apo or Pe (depending on an interior or exterior target). We always want the exit trajectory to be parallel to kerbins path of travel even if the line is not identical with Kerbin, otherwise predicting intercept could be off and correcting dV would be required. This occurs both on kerbin exit and on target arrival. For example a the flat part of the escape curve to moho should generally be at a final angle to prograde ever so slightly more than 180 at kerbin SOI otherwise the Apo for the circumkerbol orbit will occur in the future. This means that the numbers for apo and pe differ slightly relative to the calculation. If the target was exactly one SOI in front or behind Kerbin, the difference would be zero, on an interstellar trajectory that the difference is nominal, from Kerbin 670,000 radius is 0.99995 that of the calculated. On such a trajectory 670000 = 85000000 sin theta, translates to an angle to prograde of is 180.45'.

Check those airspaces. https://www.faa.gov/uas/getting_started/ If its over 1/2 a pound you have to register. YOu cannot operate within 5 miles of an airport (In general most airports have a glideslope that starts 2000feet 6 nm from the runway end. Craft can enter at multiple positions depending on ATC. I should say this, its probably never a good idea to fly anything over 2000 feet that is not in contact with traffic controllers. It cannot be over 55 lbs in launch weight, and can never leave the line of sight (which means you might want to have a bright light). - Public Law 112-95, Section 336 – Special Rule for Model Aircraft ----- FAA Interpretation of the Special Rule for Model Aircraft

Well he didn't exactly window dress the issue. The phrases "building rocket" and "terrorist could do" might make some security people paranoid.

F3 is ranges from white to blue white. https://en.wikipedia.org/wiki/Stellar_classification#/media/File:HR-diag-no-text-2.svg If it were any bluer it age would not be approaching the age of the universe . I think we are missing the point. Lets go back and try this again, Life does not exist around unstable stars for any number of reasons, you can wipe all these stars from the discussion as every single last one of them are impertinent. The triangulation is how to create bluer stars that are stable and last at least a billion years in age. Its easy to get a 10 billion year plus red star. Its not easy at all to get a blue star that is long lived Class 0 1 in 3,000,000 Short lived and associated with star forming clouds Class B 1 in 800 Short lived and associated with star forming clouds Class A 1 in 160 are longer lived are hydrogen rich but also have notable Fe , Mg, Si Class F 1 in 33 stars These stars represent only 1/26th of the stars in our local neighborhood. If we talk about stable stars over a billion years in age the list stops essentially with F. Most blue stars die within a few million years. The way to make them live longer is to decrease their metallicity but also decrease the likelihood of core formation. It has to be remember that supernova do not occur because a star burns all its hydrogen, simply, but has enough elements in the carbon-iron range to begin core formation and core collapse. The heavier metals in its structure can aid in core accretion within a stars very volatile structure. Population 1 stars and early population 2 stars must wait until surrounding eject for the SN size threshold to decrease to average modern levels. It buys time for these stars. Since these stars have to wait until formed and blasted their direction its very difficult for them to form iron. The problem is that in the current stellar nurseries the density of hydrogen is simply too low to form long lived blue stars. T0 get an environment that allows that you have to go backward in time to a period when the density of hydrogen in the universe was 1000s of time higher. Population III stars " One theory, developed by computer models of star formation, is that with no heavy elements and a much warmer interstellar medium from the Big Bang, it was easy to form stars with much greater total mass than the ones visible today.[citation needed]" I translate this as meaning it was much more difficult to form smaller stars and certainly not planets. The heat of gas means that the critical mass required for accretion and density facilitates that accretion. "Typical masses for Pop III stars are expected to be about several hundred solar masses, which is much larger than that of current stars." " Stellar Population" - wikipedia. Theory belies the observation since the intense blue stars anticipated as population III and early population II stars is absorbed during reionization and thus most are difficult to observed and refutable. The computer models predict that early star formation required 100 to 130 solar masses of hydrogen, whereas these sized stars represent the extreme heavy end of the current spectrum. These stars are largely not visible, even when red-shifted toward the CMBR limit because at the time of their formation and for considerable periods thereafter there was too much gas in space.

Azi, you haven't a clue to what I am talking about. HD 140283, these are the lowest metal stars that were abundant in the first 5 billion years of the universe,it is nonetheless a second generation star, this star get the name Methusela because of its age. Earlier stars could not be formed accretion around dust, they only formed in superdense bodies of gas and were very difficult to form. But when they spontaneously formed they typical did so creating new galaxies, and their collapse probably initiated GBH formation and allowed similar blue stars (2nd generation) in their wake. Because these stars were so difficult to form the situation of binary formation was all the more difficult. Unlike Spiza this star is 14.46 ± 0.8 billion years in age. I think consideration of Spiza as a life bringing star is not worthy of any discussion. The star is no-longer in its main sequence is very unstable it has an eliptoid shape and as it expands it will begin stripping gas from its companion star with a catastrophic out come. Such a system could never sustain life. Rather than having complex life stopping situation HD 140283 could have planets that could sustain life, such as at the bottom of an ocean. There are two primary problems. First the first generation or stars (population III stars) unlikely had planets and certainly no rocky planets. So on this account life comparable to that on earth could not exist. The second problem is that the energies in these stars suffices to erode the atmospheres of smaller planets (as might have been the case with venus). Such a planet would have to be further out relative to luminosity than the earth. The early second generation of stars has largely the same problem, gassy planets and smaller rocky planets. For the most part the blue star phase of our universe is over, yes a few younger galaxies have active blue star formation. The the bulk of accreted mass in our universe lies in large galaxies with lots of red stars. There are very short lived and unstable blue stars that appear now, but its not like looking at a galaxy and seeing all blue stars. https://en.wikipedia.org/wiki/UDFy-38135539

If you want power just tap into the differential charged region in a solar storm. All you need is a wire 25,000,000 kilometers in length. lol. The for real here. The earths magnetic field is due to the fact that the surface of the earth is slowing down faster then its molten iron core. Given that the cause of the slowing down is the moon-tide interaction. Thus the dE is primarily in the tides. If you want dE to tap into go after the diurnal tidal motion.

I was talking about early deep blue stars.

Agreed, the dreamy-eyed Mars and Venus colonization dreams are just that. But to address one issue. If we come up with a constant tk where tk is the time each day dealing with the effects of low gravity on the human body. Here on earth tk = 16 hours per day and since we are out of bed 16 hours a day walking around at 9.8 g then our bodies do not need to set aside time to deal specifically with gravity we can create a metric of 9.8m/s2 x 43200 sec/day or 4.9m/s2 (averaged over a day). Mars has a gravity of 0.376 is a person works 8 hours per day that gives the person 1.222. This means they only spend 9 hours in the gravity machine to get the same quota as earth. In zero gravity they would have to spend 12 hours a day and then 4 hours working at zero gravity. A second issue is that construction is one hell of alot easier with some gravity versus zero gravity. The earth itself (or mars) is a foundation for all buildings, it provides traction, friction, a hard place to provide force against. A third issue is that water on mars is trapped on the surface and you apply force to get it out of its trap, given the preexisting colony on mars extractin water from martian soils in and of itself is expensive, but doable.. On asteroids water is not as abundant and on comets its abundant but its incredible unstable (as the recent asteroid landing mission shows) and most comets apogees require more dV than landing on Mars. If you land on an asteroid you have to bring your own hydrogen with you. Oxygen can be extracted from the metal salts. And for those who think we can extract hydrogen from space, the level of proton in space at 150,000,000 km from the sun is a factor of 1 billion of what we need to be to extract. There is a way to overcome the problem, take a comet, wrap it in a non-breathing material (and shield the sun expose side) and force it into an asteroid with a low relative dV. Now you have volatiles and solids, you can do something. Keep the asteroid side always sunfacing (tidally locked) and the comet side anti-locked. To do this we need an energy source of a way we currently don't have, because comets reside in the part of the system with </14th the sunlight that earth has. Solar panels produce 135 watt for 1 meter panel, at best 135 W/ kg of panel. The energy density require to move a comet down to earths orbit needs to be at least 2 magnitudes higher. YOu could move a asteroid into a comets orbit (again dV from earth is an issue) and once again we have the issue of a higher wattage density power plant. While comets provide a readily usable ejection mass, some manipulation of asteroid will have to be done (e.g. purifying magnesium or manganese). The correct way to think about the problem is How do you provide humans with proper radiation shielding and proper gravimetric exposure while simultaneously providing the opportunity to build a colony. If a martian colony is built into the surface and if only they need the surface as an energy source, and they can work at least 6 hours a day at martian gravity and some other hours in a centrifuge then it does seem to be easier. But if we ask the question, humans are to explore X. Yeah X should be a roid whose orbit is most similar to earths.

Think about correction burns that are made, for instance, that is the question to ask, how much fuel is reserved for correction burns and how much is typically spent. BTW, now a days, space craft that land on the moon would directly target their landing site to the last meter. This is what separates pioneers from colonist or repeate missions. I would ask the folks at the JSC or GFC how much of a flights fuel is reserved for correction burns and are these made with RCS or fuel remnants. My assumption is that for cryogenic rockets the fuel is burnt and correction and station keeping is up to the RCS system.

The blue giants are not only short lived but they evolve rapidly and produce tremendous amounts of ionizing radiation. Rocky planets need dust, the hydrogen that almost entirely makes blue stars is not associated with space dust. So that the only planets that could form are gas giants. Those that are close would burn away in the heat of the star. My suspicion is that blue binaries would be separated by considerable distance due to the difficulty of forming blue giants.

Shhhhh, quite, I was keeping 370 in my back yard, at least until I can launch my new web cast 'where in the world is MH 370'. hmmm, too soon? '

150000 ton was an error, I use kg in the calculations, no problem. There is also thermal loss on the engine in the form or radiation, at 5800'C you are talking about fairly intense light producer (yellow green). The way to calculate is to take your Energry transfer (1298) and the apply the efficiency from that. The correct and accepted equations is Thrust (N) = 2 * power(watts) * efficiency(%) / Ve (m/s) The ISP is mediocre for external power driven space craft. Although ION drives are critically limited in thrust, their advantage is high ISP. They are a technology in wait of a decent power supply (like a fusion power plant or matter/antimatter reactor (lol)). The technology you list is an external power plant in search of a more efficient engine. Finally, what type of material can form a nozzel that remains stable at 5800'?

Risk of death is increased due to the effects of lack of gravity, exposure to radiation, the risk of being 'spaced' by micrometeor collisions or system failures, the risk of starvation, the risk of acquiring a disease that is treatable on earth but not in space, the risk of going technically blind and the consequences thereof . . . . . . . . . Space aint Oslo.

The amount of energy derived by the sun at any given distance from the sun is 2.5E25/r2 w/m2 He would not spiral past mars. The rate of acceleration is low, 6000/1200 seconds is 5N/kg For comparison at Launch a typical rocket is producing 14 N/kg and it goes all the way to 20 N/kg at max Q.

You do not want a greenhouse on the surface of the moon for the reasons you say. 1. The moon has a light cycle, 15 days dark, 15 days light. 2. Plants do not light direct exposure to sunlight. 3. The moons surface temperature goes from deep space cold to several hundred degrees over the monthly cycle. The whole reason for being underground is to modulate the extremes of space on the surface. Ground = insulation from heat and cold swings. cosmic and high energy hv radiation. A place to pressurize in which the chamber itself provides resistance to biogenic atmospheric pressures. IN addition the best modern greenhouses use LED (about 4 red for 1 blue) for growth. Some apply tiny amounts of UV to control bugs. Plants don't like green light, they slow growth in green light. Instead you need a 15 day power storage system such as a H2 02 fuel cell, or lithium ion batteries.

OK so here is the deal, lets take this by the numbers. Watts are good but ISP is mediocre, infrastructure is massive. 12km/sec is exhaust velocity, seemingly good except its much less than an ION drive and not much better than a NTR. Ion drives run from about 3500 to a practical 10,000. Lets say half the rocket is fuel. from this we have dV=12000*ln(Sm/fm) in this case 12000 ln2 = 8317 m/s. Second problem, like all solar driven ION drives this drives performance drops with distance from sun. Third problem, you see the poles, the also have to be accelerated, but those poles are not secured, as you apply thrust they move and are less efficient. Forth, 20 minute burn time from earth, yeah . . . .right. For a 150000 ton system and 12000 m/s ev. Lets see launch to mars in 20 minutes needing 7000 dv. That translates into 5.8333N/kg 5.83333N * 125000 average load needs a thrust around 700,000 Newtons. So in essence solar power generates maximum 1350 Watts per meter squared. Next the Power transferance rate is = 2 * watts *efficiency/EV so to get 700,000 N you need 700,000 x 12000 / (2 * efficiency = 1.6) = 5250000 Kilowatts requiring 3,888,888 meters of solar collector. Since most of the world plays soccer you might know that a soccer (futbol) pitch is minimally 6400 square meters. So that the area covered by the panels would be equal to 607,000 soccer pitches. To create these semispherical collectors you need a rigid mess and a reflective material. lets say the plastic is a thin 5 mil, about 0.00013 meter thickness. Assuming the plastic has a density of 1.2 (adding some density for metalic reflectant) the mass per meter would be about 0.153 kg per sq. meter. So that doing 595,000 kg (not 50,000 as the authors suggest). 5 mil is a good choice because of the forces involved in traveling in LEO, where there is significant heat from drag and at least some drag forces. The authors have chosen 5 um mylar sheets. The problem is that this sheeting is about 1/5th of a mil which is several fold thinner than the cheapest painters plaster (something like 3 mil). We havent even gotten to the rigid infrastructure to hold the plastic. On top of this 600,000 soccer pitches, thats a hell of alot of hv pressure to deal with, combine that with LEO to MEO drag and you have a wonderful application of murphy's law, Everything that can go wrong did go wrong. I could imagine this craft positioning itself getting hit by the sun slowing down orbit decaying, drag sets in, sheets melt the space craft spinning uncontrollably as it hits earths atmosphere burning up 100,000 kgs of payload streaking through the sky.