Solar-pumped laser power transmission, a way to dramatically decrease launch costs?
in [Physics]
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From: ASCEND Internet Marketing Group on 15 Feb 2010 02:47 On Dec 17 2009, 7:12 pm, "Jonathan" <H...(a)Again.net> wrote: > I like this idea, Relatively small mirrors would power > the lasers, not huge solar cell arrays. The lasers would > transmit their beams to other satellites that convert it to, and > beam it down, as microwaves. No need for mile-size > collectors in orbit. > > Proceedings of the ASCE Earth&Space 2006 Conference > April 2006 > > Space Power Grid- Evolutionary Approach To Space Solar Power > > "At a higher level, a direct solar-pumped laser could be used to > convert solar energy on the LEO satellites, and transmit the laser > beams to other satellites where the demand for power is greater > (e.g., satellites over the dark side of earth). Recently, development > of such lasers has reached a stage where efficiency of up to 38% > has been shown. These satellites would receive incoming > laser energy using their high-efficiency narrow-band photovoltaic > cells, convert it to microwave, and beam it to Earth. This > architecture has two advantages: the beaming to Earth > could be done at optimal microwave frequencies for maximum > transmission through the atmosphere, without requiring excessive > transmitter size. The laser beams would propagate with very > high efficiency, and require only small collectors. Thus the mass > and overall cost per unit power of the system with this architecture > may be substantially lower than the lower-risk option > presented before." > > http://www.adl.gatech.edu/archives/adlp06040601.pdf > > And it should be noted, the SPS start up company, Space Energy Inc, > maybe one of the more legitimate commercial attempts at SPS, has > as one of it's technical advisors this guy, and his /current/ specialty > might be a clue of things to come..... > > Dr. Richard Dickinson > > Space Energy Inc technical advisors > > "Mr. Dickinson is one of the world's foremost experts on Wireless > Power Transmission (WPT). President of OFF EARTH-WPT, > Mr. Dickinson was Group Supervisor of the High-Power Transmitter > Group at Goldstone and was NASA's microwave power transmission > specialist on the Solar Power Satellite Reference System team.... > > .....he is currently involved in studying and designing the solar pumped > laser-power beaming phased array for interstellar missions."http://www.spaceenergy.com/s/TechnicalAdvisors.htm > > s Dude, I've been promoting this concept for years. It's freakin' solid state, lighweight. You can park one at a Lagrangian point and still get tons of energy on a cost/kw basis.
From: Pat Flannery on 15 Feb 2010 06:17 Scott M. Kozel wrote: >> Solar sails are ideally suited for this mission. I am considering >> fully reusable vehicles capable of putting 1,000 metric ton payloads >> on orbit for very little cost per ton. This sort of vehicle is >> required to build any sort of orbital or lunar or Martian >> infrastructure. Its well within our capacity to build it. > > The orbital velocity of a 24-hour circular orbit 2 million miles > radius from the center of the Sun -- (3.14 * 4,000,000) / 24 = 523,333 > miles per hour. > > That is the velocity you would have to attain. Can a solar sail even be tacked towards the Sun? Ships tack by using the force generated by the wind on their sails to head into the wind at around a 45 degree angle as their keel keeps them from being blown backwards by passing through the denser medium of water. But in space there is nothing like a keel or denser medium to pass it through. Pat
From: Pat Flannery on 15 Feb 2010 13:06 David Spain wrote: > Mr. Mook, > > How do you prefer to be addressed, Bill, Will, William? I hate formal last > name greetings. I've found "pathetic Earthling" to be a good all-around form of address. Ming
From: David Spain on 15 Feb 2010 14:23 Pat Flannery <flanner(a)daktel.com> writes: > David Spain wrote: >> Mr. Mook, >> >> How do you prefer to be addressed, Bill, Will, William? I hate formal last >> name greetings. > > I've found "pathetic Earthling" to be a good all-around form of address. > > Ming You are merciless... ;-) Dave
From: William Mook on 15 Feb 2010 15:06
On Feb 14, 9:56 pm, "Scott M. Kozel" <koze...(a)comcast.net> wrote: > William Mook <mokmedi...(a)gmail.com> wrote: > > > "Scott M. Kozel" <koze...(a)comcast.net> wrote: > > > "Androcles" <Headmas...(a)Hogwarts.physics_u> wrote: > > > > "William Mook" <mokmedi...(a)gmail.com> wrote in message > > > > > That is absolutely correct. It will be on solar orbit. The orbital > > > > period will be 24 hours at 3 million km radius. > > > > ============================================== > > > > You may find it a tad warm 2 million miles from the sun. It will gather > > > > rather more energy than you wanted, and having a 24-hour year it won't > > > > be in sight of Earth for more than 14 hours each Earth day as it > > > > disappears behind the Sun for 10 hours of that period. > > > > It would take an enormous amount of energy to put a satellite that far > > > down into the Sun's gravity well. > > > > It has been difficult enough to get a satellite into the same orbit as > > > Mercury, which is 35 million miles from the Sun. > > > Solar sails are ideally suited for this mission. I am considering > > fully reusable vehicles capable of putting 1,000 metric ton payloads > > on orbit for very little cost per ton. This sort of vehicle is > > required to build any sort of orbital or lunar or Martian > > infrastructure. Its well within our capacity to build it. > > The orbital velocity of a 24-hour circular orbit 2 million miles > radius from the center of the Sun -- (3.14 * 4,000,000) / 24 = 523,333 > miles per hour. > > That is the velocity you would have to attain. BASIC ASTROGATION The vis-viva equation gives you the velocity of an object in an orbit; V = SQRT( mu * ( 2/r - 1/a )) If r = a then the equation becomes simplified V = SQRT ( mu * (1/a)) a=semi-major axis r=body position V=velocity mu=gravitational constant. The Earth takes 1 year (31.557 mega-seconds) to orbit the sun at 1 AU (150 million km) which translates to an average speed of 29.865 km/ sec. Squaring this and multiplying by a (150 million km) obtains a mu of; mu = 1.33791e+11 km3/sec2 for the sun. DIRECT APPROACH Now, an orbit that has a perihelion of 3.5 million km and an apohelion of 1 AU (150 million km) has a semi-major axis (a) of (3.5 + 150) /2 = 76.75 million km. So, AT EARTH - the orbital velocity of an object orbiting the sun following this orbit - would have V = SQRT( mu * (2/150e+6 - 1/76.75e+6)) = 6.3776692166 km/sec ~ 6.378 km/sec Now we just saw the the Earth moves at an average speed of 29.865 km/ sec. So, subtracting these two numbers obtains; 29.865 - 6.378 = 23.488 km/sec. Well below the figure you cite. This will bring the perihelion to 3.5 million km which we're talking about. When the distance is 3.5 million km the speed rises to V = SQRT(mu*(2/3.5e+6 - 1/76.76e+6)) = 273.328681 ~ 273.329 km/sec That's because when an object falls in a gravity field, it speeds up!! Now, to STAY in that orbit, it must be SLOWED DOWN to V = SQRT(mu/3.5e+6) = 195.51441 ~ 195.514 km/sec Subtracting these two numbers gives us the DELTA VEE required at perihelion to maintain the orbit 273.329 - 195.514 = 77.815 km/sec Which when added to the first figure obtains the TOTAL DELTA VEE FOR THE MISSION; 77.815 + 23.488 = 101.303 km/sec Which is about half the speed and one quarter the energy. INDIRECT APPROACH Jupiter is 5.2 AU from the sun. 780 million km. So, to fly from Earth to Jupiter requires that you enter an orbit with (150e+6 + 780e+6)/2 = 465e+6 semi-major axis. So, at Earth you must have a speed of V = SQRT(mu*(2/150e+6 - 1/465e+6)) = 38.680 km/sec Which must be added to the orbital speed of Earth which is 29.865 km/ sec. So the DELTA VEE is dV = 38.680 - 29.865 = 8.815 km/sec Which is less than 4% of the speed and 1/6th % of the energy of your analysis. Why go to Jupiter? For gravity assist! You can approach Jupiter in such a way as to ZERO OUT ALL MOTION RELATIVE TO THE SUN IN THE SATELLITE! Now gravity on the solar surface is 27.94 gees. Light intensity on the solar surface is 64.1 MW. So a reflective film feels 445.9 pascals of pressure. That's 45.49 kgf per square meter. Dividing by 27.94 gees obtains an areal mass density of 1.63 kgm per square meter. Now, gravity force falls off with distance as the inverse square. Light intensity falls off with distance as the inverse square. Light intensity is proportional to light pressure. http://books.google.com/books?id=8n8OAAAAIAAJ&pg=RA5-PA327&dq=torsion+balance+radiation#v=onepage&q=torsion%20balance%20radiation&f=false So, anything that has less areal density than this can be lifted away from the sun at any distance. By changing the orientation of a thin film structure - one can then dive into the sun, and by re-orienting to intercept light - it may be slowed. Now, at Earth the solar pressure is 9.2 micro-pascals (9.2 micro- newtons per meter squared) and at 3.5 million km it is 16.9 pascals - with 100% reflection. With 25% reflection is 4.2 pascals. 25% of the ineffective photons are reflected to keep the system cool - while 75% of the effective photons are absorbed, and drives a highly efficient free electron laser system - and then the rest of the waste heat is radiated away by radiators operating at 1,600K. At 3.5 million km attraction due to the sun is a little less than 1 gee. So, with 4.2 pascals pressure, and 430 grams per square meter at 25% reflectivity, we have the ability to hover. With 100% reflectivity at this distance, we have the ability to execute a 4 gee maneuver! At Earth we can only execute a 1/10,000th gee maneuver! So, this gives us what we need to do. 1) we put the payload into LEO. 2) we deploy the payload to fly to Jupiter by adding 12 km/sec (adding hyperbolic excess to escape velocity) - at 1/10,000th gee this takes 4.65 months of thrusting and another 2.72 years of transit time. 3) at Jupiter we do a gravity boost to drop the speed down to nearly nothing, allowing it to fall into a 3.5 million perihelion from Jupiter. This takes 2.28 years 4) at 3.5 million km we must slow down about 200 km/sec to stay at that altitude. So, we deploy a solar sail near perihelion - and execute a 4 gee maneuver using light pressure. This takes 1.4 hours - which occurs around 20 degrees before reaching perihelion. If we drop straight in at zero speed - we must slow down about 280 km/sec and continue to hover above the Sun. 5) we then eject the reflective layer, and begin operations - using the reflected ineffective photons to provide continuous propulsive adjustments. |