Subject: Space-tech Digest #86 Contents: Paul Dietz Re: gamma-ray bursters Paul Dietz Re: gamma-ray bursters Henry Spencer buckyball chains Bob Munck Rotating a hula-hoop into a retro-grade orbit Marc Ringuette Re: Rotating a hula-hoop into a retro-grade orbit Phil Fraering Re: Rotating a hula-hoop into a retro-grade orbit Dani Eder Re: Rotating a hula-hoop into a retro-grade orbit Marc Ringuette Re: Rotating a hula-hoop into a retro-grade orbit Bob Munck Re: Rotating a hula-hoop into a retro-grade orbit Dani Eder Re: Rotating a hula-hoop into a retro-grade orbit ------------------------------------------------------------ Date: Thu, 19 Sep 91 16:11:44 EDT From: dietz@cs.rochester.edu To: fermat!r@la.tis.com Subject: Re: gamma-ray bursters Cc: space-tech@cs.cmu.edu Dale Amon and I were discussing the possibility of detecting gamma emission from antimatter powered interstellar spacecraft. Rich Schroeppel asked me to post the discussion to space-tech. Here's a brief summary. Dale thought that beamed gamma emission from an antimatter rocket would make it appear very bright to anyone aligned with the rocket's axis. I disputed this, and argued that, in fact, not all that much energy would be emitted as gamma photons, and that which was would not be highly beamed. In a high-Isp antimatter rocket, protons and antiprotons are annihilated in a magnetic bottle (it is best to avoid using higher atomic number nuclei, I think, as neutrons liberated upon annihilation cannot be guided in a magnetic nozzle). The annihilation products are primarily pions (with some kaons). The negative and positive pions escape out the magnetic nozzle before they decay. The neutral pions decay almost immediately to photons, but these photons are isotropic in the spacecraft reference frame. The pions decay to muons, which travel about 1 km before decaying to electrons. Electrons and positrons annihilating in the reaction chamber would also produce isotropic radiation. Beamed gamma emission could arise if these particles further annihilate in the exhaust. However, the cross section for annihilation of electrons with positrons (for example) is on the order of r_0^2 (mc^2/E)^2 where mc^2 is the mass of the electron, r_0 is the classical radius of the electrons (about 2.8e-13 cm) and E the energy of the particles. The density of positrons and electrons in the exhaust is limited by the need for their kinetic pressure to be less than the magnetic pressure of the confining magnetic field. A rough calculation gives that roughly on the order of 10^-8 of the positrons annihilate in 1 km of exhaust (asuming the positrons have a residual random kinetic energy of about 1 MeV). Similar calculations obtain for muons, and I suspect for charged pions as well. One would also have to do the calculation for the reaction pi- + p --> n + pi0 (with the pi0 decaying) and for the corresponding mirror reaction, if the annihilation is not 100% efficient and/or if the mixture is proton-rich. What happens to positrons that don't immediately annihilate? They escape to interstellar space, where their trajectories are bent by interstellar magnetic fields. A 100 MeV positron in typical interstellar magnetic fields has an orbital radius on the order of millions of kilometers. It will complete many orbits, losing energy to synchrotron radiation and collisions, before finally annihilating, so by the time it does annihilate it will likely have its direction of motion randomized, at least perpendicular to the local magnetic field lines. This argues against beaming. Could radiation from the exhaust plume be detectable? I don't know; Dale cited a paper in JBIS that seemed pessimistic. Paul F. Dietz dietz@cs.rochester.edu ------------------------------ Date: Thu, 26 Sep 91 17:17:32 EDT From: dietz@cs.rochester.edu To: Ted_Anderson@transarc.com Subject: Re: gamma-ray bursters Cc: space-tech@cs.cmu.edu Ted Anderson asked if we know what the spectrum of bursters is... A lot of work has been done on this. The burster spectrum is mostly fairly hard gamma radiation, in the > 1 MeV range, with a nonthermal spectrum. Only a few percent of the energy is in soft (few KeV) xrays, a fact that severely constrains the models (since any hot, optically thick matter should radiate copious x rays). No photons have been detected around 1 GeV, which is, I think, a blow against antiproton annihilation models. Some models show lines that have been interpreted as cyclotron lines in a strong (10^12 gauss) magnetic field, and also lines that look like electron/positron annihilation lines red shifted by an amount consistent with the gravitational redshift of a neutron star. The integrated flux of the strongest burst, GB790305, was 10^-4 ergs/cm^2 (the time structure of this pulse was consistent with a rotating or precessing neutron star; the period is about 8 seconds). A lethal dose to unshielded astronauts would be about 4 x 10^6 ergs/cm^2, so anyone 200,000 closer to the burster than we were had better have good shielding. This means that a burster at 1 light year (say) the energy released would be equivalent to the annihilation of 11 million kilograms of matter, and would be lethal to unshielded astronauts within about .3 AU. At 100 km from this burst source, the gamma intensity would be 10^15 J/m^2. This would be a bit much for any starship to withstand! If the burster is further away, the intensity at 100 km would be still larger. If the burster was at 5 billion light years (say), the lethal radius for unshielded astronauts would be around 25,000 light years. I hope one doesn't go off in our galaxy soon. For more details, see a recent issue of Physics Reports that contained a review article on various models for gamma ray bursters (mostly now obsolete, I'm afraid). Gamma ray bursters are located by triangulation, using differential time delay between widely space detectors. In addition to BATSE on (C)GRO, there have been gamma ray detectors on the Venera spacecraft, as well as a detector on Ulysees (BATSE has better sensitivity and and time resolution (.1 ms, I'm told)). We should be able to determine if burst sources are close to the solar system using triangulation. If the burst source is closer to us than about d^2 / (2 c t), where d is the interdetector spacing, t is the time resolution of the detectors and the pulse, and c is the speed of light, the detectors should be able to distinguish the source from one at "infinity" (if the detectors are properly arranged in space). GB790305 had a rise time of .2 ms, so if d = 1 AU one could determine if it were within roughly 10 light years of the solar system; current data, I am told, may be able to test if the sources are in the Oort cloud (~ 1 light year). If we could put high resolution detectors on spacecraft out at 1000 AU, we could directly determine distances out to millions of light years, as well as determine the direction to the sources with very high precision. Paul F. Dietz dietz@cs.rochester.edu ------------------------------ From: henry@zoo.toronto.edu Date: Mon, 7 Oct 91 18:01:06 EDT Cc: space-tech@cs.cmu.edu Subject: buckyball chains To: munck@STARS.Reston.Unisys.COM >I've been wondering about Buckyballs, much in the news recently. >Suppose you could line a bunch of them up in a row, and then >somehow break some bonds along their mutual diameter (at points of >tangency) so that they rebonded to each other, forming a tube. >Wouldn't it have a very high tensile strength? ... Yes. It would also be very stiff, so we're talking about a beam material, not a cable material. That would be my guess, anyway. However, "if wishes were horses, beggars would ride". Actually *making* such a material is very non-trivial. Lining things up in a row and then breaking bonds so the molecules rebond to each other is very non-trivial. Polymer-manufacturing chemists spend their lives figuring out how to make much more ordinary molecules do this in an orderly way. A milder form of the same thing -- with the side branches full of hydrogen atoms rather than other carbons -- is already marketed, tradename "Spectra" I think, as an alternative to Kevlar. Henry Spencer at U of Toronto Zoology henry@zoo.toronto.edu utzoo!henry ------------------------------ To: space-tech@cs.cmu.edu Reply-To: munck@STARS.Reston.Unisys.COM Return-Receipt-To: munck@Stars.Reston.Unisys.COM Subject: Rotating a hula-hoop into a retro-grade orbit Date: Fri, 04 Oct 91 14:19:05 -0400 From: Bob Munck I've been playing around with the hula-hoop idea (a ring all the way around the earth, spinning faster than orbital velocity, and thus able to carry the weight of "spokes" down to the surface and payloads coming up them). One of my thoughts involves several pairs of rings with those in a pair going in opposite directions. In other words, one spinning east to west and the other west to east. My problem is in getting the first two up without having a great deal of equipment going opposite ways in the same general orbit. What would be involved in rotating a W-E hula-hoop 180 degrees around an axis through the Equator up and over the Poles until it is going E-W? Remember that the hoop is flexible, and therefore may need to be spinning a fair amount faster than orbital velocity to keep it relatively rigid. Another thought involves hula-hoops at intervals all the way up to geosynch. If we can't do a cable that can reach the entire way down, maybe we can do it in shorter segments. Just a little food for thought. Bob Munck ------------------------------ Date: Fri, 4 Oct 1991 16:45-EDT From: Marc.Ringuette@DAISY.LEARNING.CS.CMU.EDU To: space-tech@cs.cmu.edu Subject: Re: Rotating a hula-hoop into a retro-grade orbit [ If anybody wants to review the previous discussions on this, send me mail. ] Bob, Ick! Ugh! Changing orbital planes takes truly huge amounts of energy, and I bet it's even worse for a loop in orbit. I don't think the idea of turning a loop 180 degrees should even enter your brain. Begone, vile thought! Instead, how about this: 1. Launch the loop cable to LEO. Pay it out with zero tension. Start it spinning to generate a little tension. For the retrograde loop, either 2a. Pay out a smaller loop, right next to the first, then decelerate it to zero and back to orbital velocity in the other direction. Loop #1 holds it up during the time it's going less than orbital velocity. Repeat. or 2b. Just launch the second loop into a retrograde orbit and pay it out. Either one seems like a practical & not-too-expensive method for bootstrapping the first two small-capacity loops. Accelerating them is probably done most cheaply by using solar power and electric propulsion. Once there are two counterrotating loops and a tiny space station with a cable to the ground, use ground-based electric power and magnetic induction to simultaneously accelerate the two loops in opposite directions. Then winch up more loop material, use induction launching to speed it up, and off you go. Regarding multiple hula hoops up to geosynch...I don't think we ever want to winch ourselves all the way out. Just winch up the cable to LEO, then zing yourself into orbit by magnetic induction on the LEO hula hoop. Or use pinwheels. --- This is assuming we're using a kevlar hula hoop with bits of metal impregnated in it to use for magnetic induction. However, please do remember that the Launch Loop style string of iron rods or pellets, moving ballistically and deflected by magnets, is capable of transmitting much more force across orbital distances than a cable is. Last time when I worked it out, the Kevlar loop could only spin at maybe 120 km/s above orbital velocity. Some careful thinking may be in order here, about whether the tension actually helps us hold up any significant mass. There's another problem with our hopes to use magnetic induction launching on a solid loop. In Lofstrom's proposal, the iron rods are brought closer together when you induction-launch off them, and are re-accelerated on the next passes round the loop. With a solid loop, induction launches will create longitudinal waves that may get really scary. I still prefer the Loftstrom ballistic loop concept to the solid hula hoop. I think the simplicity of the solid loop is mostly illusory, since we would have to actively stabilize it anyhow. Marc ------------------------------ Date: Fri, 4 Oct 1991 19:41:43 -0500 From: Fraering Philip G To: Marc.Ringuette@DAISY.LEARNING.CS.CMU.EDU, space-tech@cs.cmu.edu Subject: Re: Rotating a hula-hoop into a retro-grade orbit Of course, there's another advantage of the Loftstrom-type loop: if built in Texas, there's the chance that the main regulatory agence you'll have to put up with is the Texas Rail Commision. Phil Fraering pgwres01@ucs.usl.edu ------------------------------ Date: Mon, 7 Oct 91 11:57:11 CDT From: eder@hsvaic.boeing.com (Dani Eder) To: space-tech@cs.cmu.edu Subject: Re: Rotating a hula-hoop into a retro-grade orbit Sender: mnr@DAISY.LEARNING.CS.CMU.EDU Uh, you gentlemen realize that a loop like that is gravitationally unstable? I prefer static structures for applications like an elevator to 200 km. For example, Amoco Carbon fiber/epoxy composite (T40/ ERL-1906 to be specific) has a compressive strength of 1930 MPa, and a spacific gravity of 1620 kg/m^3. With no taper or design margin, it can support a column 121.5 km high made of itself, under 1 gravity. Allowing for a factor of 2.5 for design margin and non-loadbearing structural overhead, and a 3% average reduction in gravity over the 200 km height, we require a taper ratio of 54 in cross section from bottom to top. It is slightly worse than this in reality, since you have to design for wind loads over the first 20 km in altitude. There are many uses for such a tower. Even a conventional rocket would do better launching horizontally from the top of the tower. By not spending thrust vertically, and by avoiding air drag, the delta V of a standard rocket would go from 9144 m/s to 7384 m/s, a 19% velocity savings. What's better, a cryogenic SSTO stasge would go from 7/8 fuel, 10% structure, and 2.5% payload to 81% fuel, 10% payload, and 9% payload, so a given payload could be launched by a rocket 72% smaller. Dani Eder ------------------------------ Date: Mon, 7 Oct 1991 13:31-EDT From: Marc.Ringuette@DAISY.LEARNING.CS.CMU.EDU To: space-tech@cs.cmu.edu Subject: Re: Rotating a hula-hoop into a retro-grade orbit > Uh, you gentlemen realize that a loop like that is > gravitationally unstable? Compared to the hellish dynamic instabilities when you hang a station off the hula hoop, gravitational instabilities are nonexistent! :-) > I prefer static structures for applications like an elevator to > 200 km. For example, Amoco Carbon fiber/epoxy composite... Speaking of instabilities...200 km towers have instabilities galore. There seem to be two solutions to the leaning-tower-of-pisa problem. One is to be able to provide enough restoring force, using a static structure, to keep the tower from falling...and the other is to use some sort of active control to keep the tower perfectly balanced. I get the feeling you're planning to use static restoring forces. Let's try this: give the top of the tower a 1 km/h sideways velocity. The restoring force will presumably come through leverage from the bottom of the tower. If the base is 100m in width, and the needed restoring force is X, the forces exerted on the base will be increased by something like the ratio of height to base; 200 km / 100 m = 2000X. Isn't this a huge problem? --Marc ------------------------------ To: space-tech@cs.cmu.edu Subject: Re: Rotating a hula-hoop into a retro-grade orbit Reply-To: munck@STARS.Reston.Unisys.COM Return-Receipt-To: munck@Stars.Reston.Unisys.COM Date: Mon, 07 Oct 91 16:08:19 -0400 From: Bob Munck >> Uh, you gentlemen realize that a loop like that is >> gravitationally unstable? Yes, yes. I remember the MIT kids in the halls at some Con chanting "The Ringworld is unstable!" Of course, it didn't have spokes or a counter-rotating other half. >Speaking of instabilities...200 km towers have instabilities galore. > ... give the top of the tower a 1 km/h sideways velocity. >The restoring force will presumably come through leverage from the >bottom of the tower. If the base is 100m in width, and the needed >restoring force is X, the forces exerted on the base will be increased >by something like the ratio of height to base; 200 km / 100 m = 2000X. It's just the old Firesign First Law: "If you push something hard enough, it will fall over." I've been wondering about Buckyballs, much in the news recently. Suppose you could line a bunch of them up in a row, and then somehow break some bonds along their mutual diameter (at points of tangency) so that they rebonded to each other, forming a tube. Wouldn't it have a very high tensile strength? And suppose you could do this continuously at each end, making an arbitrarily-long molecule? And likely a very cheap one, given the raw materials. Could knit a _very_ strong cable. Cold fusion didn't (yet) pan out, and high-temp superconductors are going to be very slow coming. I'd like to see a real, genuine scientific breakthrough, and this could be one. What would we do with it first? Bob Munck ------------------------------ Date: Tue, 8 Oct 91 08:29:53 CDT From: eder@hsvaic.boeing.com (Dani Eder) To: space-tech@daisy Subject: Re: Rotating a hula-hoop into a retro-grade orbit Sender: mnr@DAISY.LEARNING.CS.CMU.EDU If a tower is 100 m across at the base and 200 km tall, then it has a slenderness ratio of 2000. A column this narrow is unstable against buckling under a compressive load. For a graphite/epoxy, a slenderness ratio of about 16 produces buckling at the same load as failure under crushing. So the base would have to span 12.5km for elementary static stability. When dynamic loads (winds, reaction against whatever you are doing at the top) are considered, it may have to be wider. This does NOT mean the base has to be filled area. Think of the way a high-voltage power line tower is built. The structure spans a large base, but only makes contact at a few points, and the structure itself is mostly empty space. Dani ------------------------------ End of Space-tech Digest #86 *******************