PB666

No, because its half life is 150 days and the trip to Jupitor or Saturn takes a 1 to 2 and a half years. Both Curium and its decay products would have decayed. If you want to generate ice melting heat, use a dirty low level fission bomb. That is too say at the top of the bomb you have the detonator, structural protected underneath is fissile uranium that has sub critical mass. this is then driven into the ice wear its heat melts it down but not enough to destroyed. Screw the microbes, if they are worth their weight they should have evolved radiation resistance!!! Another way to do this is to create the Galactic Empire and devise a planet destroying death star, then blow Europe to bits and collect water containing fragments from space. [snip] On a more serious note . . . . . . . The way to do this is to have two materials that undergo fission when put together but only when put together if the mass of either is kept subcritical. Then you place the isotopes in a stable plastic material. Once the PL has landed the two 'tapestries' are rolled together to form a chimeric role where the rate of the reaction is governed by the rate of the role. This then generates heat which melts the ice to water. The problem with this design is to control the direction of the melt and because of heat transfer with depth the water on top (if it has not evaportated) will refreeze on top of the craft, which means its now logically stranded and what value would it have. To fix that problem it would have to eject an antenna on the surface and a wire that rolls out as the ship travels down. Again the vessel is under the ice, the ice has mass, the water has mass, eventually the crafts instruments need to be able to withstand the pressure, which adds to the spacecrafts weight. There is a third way, carry 20 or 30 bunker busting bombs and blast a hole in the ice from orbit. Next send down shape charges and level out the cavity. This needs to be done from craft that have pretty much zeroed horizontal velocity before impact . . .equals weight in drag equipment. Once the water level is reached allow a few hours for the ice to refreeze. Then send down your probe and land on the frozen ice and drill though the much thinner ice. The problem with this design is that you need to have a space craft directly overhead in geosynchronous orbit to communicate so that you need to limit latitudes to those close to the equator. Another way is to send a 1 MW fission reactor into ESO and focus the heat output of the reactor i down at the surface and try to tune the frequency of heat to the wobble frequency of water, this would be akin to boiling a cup of coffee in the microwave, At ground level you have water pumps that pump the ice out of the hole. As you approach the natural ceiling of the buried water layer you land your probe and sample the water. So the problem with this method is focusing the heat and reflecting it at the desired frequency avoiding cloud formation over the hole would require alot of pumping. Lets say the hole is 10 meters wide. Your out put power is 50 % efficient, so you get 500 KW (very very optimistic). 10 meters wide is about 80 cubic meters of water per meter of depth, this converts to 80,000 liters = kg. So lets say we start at -150'C there is a specific heat of 2108 J * Kg-1 * K-1. To get ice to its melting point requires 316200 J * Kg-1 * K-1 then to convert the ice to water and raise its temperature to a stable 4'C requires 334000 J * Kg-1 * K-1and 16000 J * Kg-1 * K-1a for at total of 660,000 J * Kg-1 * K-1. For 80,000 kg would require 53.6 MJ per meter in depth. Thus latent heat drill would convert ~ 0.01 mm per second of ice to water. The pumps would need to pump 0.8 liters per second (not to bad, a good bilge pump could do it) we need a power supply on the surface to pump. In a day the system could drill 0.8 meters. If the ice was 500 meters thick it could drill it out in 2 years. However most bilge pumps cannot pump more than a few meters in height. We know that energy = m*g*h since the pumps move 0.8kg of water per second we know that m = 0.8. g Europas surface gravity is 1.3 m/s and of course height varies. The energy required to pump a liter of europa water to the surface = 1 J * L-1 * depth-1/ sec or about 1 watt of power to remove that last meter of water is 500 watts. On Europa the surface solar power is about 1/25th that of earth and here on earth it would require about 1.5 meters of solar panel to achieve that. So you would need ~40 square meters of solar panels. In addition to pump the water 500 meters you would need approximately a 8 cm tube with a rigid wall capable of holding the pressure of 500 kg of water that the pump needs to pump. I estimate this to weight about 0.2 kg per meter so around 100 kg. The problems with this method is controlling the diameter of the hole from ESO.

When did SpaceX start hiring pygmies . . . . . .

The core it going to be sacrificed right. IMHO they should take a look at Russian rockets. Q is not going to increase (or at least it doesn't have to unless you increase the cross-sectional area of the PL), just T x Q for the bottom of the second stage. That is not a problem. So the only real problem is the skin of the core. To handle this problem and keep F9 core just create bolt attachment point on the skin. Then when you need to add pairs of booster, bolt an aluminum bar to the bolt holes and the bar itself has the attachment points, then on the F9 head have a riser that rides the bars once they are attached that adds the additional structure. If you look at Russian rockets they do not pay alot of attention to form drag in the mid-section, what for you are just going to dump it. Some of their rockets don't even use fairing. There is a lesson there in that if you got all kind of crowdy junk around the core, detailing aerodynamics has no reward. You can see the boundary layer on the nose cone, once it starts getting close to Mach the boundary begins to separate and pretty much all the dynamic resistance is transferred to the nosecone. As to the other point, yes fire up the engine to get rocket off the pad then throttle it down to 5%, you can do the same thing as asparagus just by carefully timing throttle down along the core, dispose of two boosters, then the second pair and then rethrottle the core. The core cannot be salvaged anyway so no need to worry about having landing legs or spoilers. The deal is that if they want to land the core, I guess then they need to due more in aerobraking department before adding thrust to land. The whole prospect is bad business IMHO, if you get 1000 more dV out of the launch stages, your ability to land-target the core becomes more difficult across the board. Landing predictions are harder, there is more heat on reentry, less control of craft on renentry, more weight on the core because of structure (although the structural outmounts can be bolted and theoretically unbolted to land). This isn't silly blue origin stuff, they are trying to sell PLs (heavy's in LEO). For all intents in purpose the LEO is the contract point, any other service provided is fluff. So they need to decide the most efficient way to get a rocket in the cheapest, immediately stable LEO. The heavier the payload the better. If you can quadruple the PL at the expense of a core, you take it.

There is only one RTG left I think?

The elliptical mark ups as we use them today were not developed in Keplers time, so the equations used today are sort of projections of the elliptical math that Kepler knew of or expanded up[on. The required elements of a conic section are a a = semi-major axis, e= eccentricity from which one can draw an ellipse. Major axis = the longest diameter of the orbit. The specific problem here is a reference point. . . how do you know. Eccentricity: Angular width of it face at full moon from its closest and furthest points, we would need to know the radius of the earth. The part astonomers work with such as inclination (i) require the choice of a reference plane, then you need to know the inclination line in that plane and create a cross-reference measure of distance from that line. From that you need to measure the argument of Periapsis (w). Once you get this far fabrication of an orbital model would be pretty strait forward. The constant sweep rate is pretty good intuition. For Newtonian orbital mechanics you have the modern vectors r and v. If the barycenter of the system is xyz = 0,0,0 then magnitude r is dot product (x,y,z) of its cartesian coordinates and v is the direction of motion perpendicular direction of r, r x v = h (a vector) which is invariant (dot products and cross products are from the 1890s). The sweep rate is roughly 1/2 h. In this system if the ellipse is in X,Y then h projects into Z (its own dimension). Thus the two systems are congruent, in the Newtonian system its more refined however because specific (mass-reduced) angular momentum has a vector. This means any addition of energy to the orbit therefore would change that vector h2/u is however the semi-latus rectum (l) of the orbit with reference to the systems barycenter, this particular component is not derivable from Kepler, but is derivable from Newtonian celestial mechanics. l is important since b is a trivial aspect of the orbital ellipse, and it is possible to predict the orbits using r = l / (1 + e Cosine O). Since both the system center and its satellites would obey this it is possible to map orbits and predict future positions. In the Newtonian 2-body problem since h can also be made a function of angle r2dO/dt (where V = wr and w = dO/dt) and r = (h2/u) / (1 + e Cosine O): specifically the periapsis is the angular reference point and h2/u = L/m, there is the immediate consequence that since h is constant and can be related to r that the ellipse cannot evolve in any direction without the change in energy. To see this, for the inclination of an orbit to change h (which is also a direction vector) would have to change, this is a constant. The periapsis then could change in the inclination plane. There are two ways to do this, the periapsis could change its magnitude, but that changes v without changing r (at the change point) this means magnitude of h changes, you would have to a another change at 180' to change v again and r to maintain h' this however would change eccentricity (making it circular). You could then go to a third point and burn out to the original rmax, then go to rmax and draw in the rmin to the original pe. h would not have remained but would have been restored but no r = (h2/u) / (1 + e Cosine O) would be invalid because O = 0 is no longer the periapsis. To be clear apsidal precession occurs in orbits, its one of the factors that contributes to the ice ages. And it occurs without any energy being put into the system, but the system in n-body, which means that energy is being transferred in the system from one orbit to the next and back again. Basically every year Ope, new = Ope, old + .000056 radians. This precession occurs as a consequence of the effects of the major planets in the solar System, namely Jupiter and Saturn. So this goes to a frequent disclaimer about h (or L) that h is local and not infinite. If we forget about relativistic differences, the two major problems are, for satellites traveling in LEO around the Earth, that the earth is not a point mass nor is it of uniform density, this alters u/r2, that external, other non-centric celestials, perturb orbits. Finally for LEO orbits as orbital speed increases so does the effect of drag, both increase in proximity to planetary surfaces, so that loses of h can be permanent.

I say we send dal, why waste a good robotic space craft, a appropriate time DAL's head goes out the sampling hatch he puts out is official BSoA bug catcher net, catches what he can . . .we carefully store the net and send it back to Earth in the return ship. Did I forget something on this mission, . . . .oh well.

You probably would not be able to recycle the core F9. it would end up as space junk. (Its not impossible, but probably difficult) Nice rocket, you need more struts, you have not boosted enough your rocket until you have at least 200 or 300 struts.

A surface strike from below to the inside of the shell is always a strait line along the vector of travel. Here is the reason. https://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation The net gravity acting inside of a shell is zero (r < Rshell). Consequently, there is a direction vector from the center. The crafts position is different from the center. In this case the inside radius is r'. Current position is x, y, z from center and velocity is dx/dt, dy/dt, dz/dt. Consequently r'= ((x0+tidx/dt)2 + (y0+tidy/dt)2 + (z0+tidz/dt)2)1/2 where ti is the time of impact. Please do your math homework yourself.

Yes but how are they going to extract water with only 10 kW of power after it has been distributed for personal use?

Where do you get this stuff? A few centimeters down from what, a mile thick layer of ice?

Thanks for the link. Newton's application works for slow moving bodies around distant stars, it tends to break down as things speed up. The page needs an awful lot of work to be complete. First GM = u. Assuming that one is a infinitesimally small satellite around a massive central (and far enough away such that relativity does not need to be taken into account) and get rid of m, temporarily. r = 1 / (u/L) (1 + e Cos O)) versus r = l / (1 + e Cos O) where l is the semi-latus rectum or r = a * (1-e2) / (1 + e Cos O) where l = a * (1-e2) or l = b2/a for keplerian mechanics a = is the arithmetic mean of Pe and Apo as measured from the center of massive bodies, b is the geometric mean. Both represent the axis of an ellipse. Therefore the semi-latus rectum goes undefined as L/u. To truly bridge the gap between Kepler and Newton (although neither formulated their math to the above stated description we assume that their understanding was equivalent) you need to convert one understanding to the other. Here it is, the X axis is the axis that runs from Pe though CB to Apo and also runs along O = 0(2π)- π. It is the axis where Y = 0. The Y axis runs through the systems center of mass therefore it runs through π/2 and 3π/2 each radial to the radius of the semi-latus rectum; it is the axis were X = 0. When Cos (O = π/2) then e Cos O = 0 thus r = 1/(u/L) = L/u (where L is the angular momentum of the orbit, mass reduced in our example). The problem between Kepler and Newton is a discrepancy in the central body that is related to the planets mass, for example Jupiter would pull on the sun gives the sun orbital moment, so that the semi-latus rectum is not at right angles to Jupiter's periapsis through the sun but through the system center. Thus if you are looking at binary stars you cannot reduce the mass. Therefore there are three (x2) other orbital landmarks. There are a, the radii in which V is equal (u/r)1/2 [Note that for a circular orbit all radii = a]. In terms of the angular distance from the periapsis a is variable. The limit of Oa as e-->0 is Ol this means that as eccentricity increases a is almost always closer the minor focus than the central body. This has to be the case, because for a to have a velocity of (u/r)1/2the and also be eccentric means that the absolute value of the radial velocity vector is large and thus the tangential component of velocity must be small and similar to the velocity at the apoapsis. There is the l radii, where (reduced angular momentum)/u. The third points are r = b, which is more or less trivial since b is a measure from the elliptical center to along the minor axis. I should point out also the link is incomplete in one regard for the stated purpose. There is no reason that upon leaving O = 0 after completion of an orbital cycle O since the conversion from polar coordinate system makes the assumption that the process is a cycle (IOW its a circular definition), which of course it is, but the Newtonian systems assumes it is a static cycle, in order to do that you need to know whether that stasis has an energy dependence or not; it does. No total energy (information) change is what causes the cycle to be 'a standing wave'. Here is the reason using the Newtonian system. Lets first set some boundaries, 2- body orbits the point-center mass move in a plane x, y. There are two O for which X = rO Cosine O and Y = rO Sine O accept for x where y = 0 or y where x = 0. While it is true that Apo does not have to be π radians from the Pe, if one follows the function it does. Likewise if one folllows Pe it is generally very close to Oapo + π. As one rids emperical systems of perturbations the deviations decrease to zero. Although there are several definitions of Pe we want to use the definition that the distance of the planets center at orbital minimum from the systems center, the X = Pe] Given the Pe is the minimum approach the dX/dt is zeroed out, momentarily, it is also one of two unique radial lengths around the major foci in an ellipse, the other is the apoapsis. Consequently the Pe is traditionally assigned X = rO cos (O =0). In zeroing all radial velocity (dX/dT = 0) at Pe, the potential energy has also reached its minimum. Mechanical energy is key to understanding orbital motion. For 2 point masses in orbit about each other Mechanical Energy is always conserved. At any position in the orbit the application of a moment of kinetic energy changes the total mechanical energy but not r=f(O) or potential energy. We can see this by applying a moment of force on an object in either Y direction and resulting in a moment of radius no different from r = f(O) [Done for simplicities sake] so we could change energy or information in the system without changing the function .. . . .errrrr, no the function was right but for the wrong reason. If we track the function from that pulse of energy, the x, y coordinates with respect to T, O, y = f(x) or x = f(y) have all changed but one, the position from which we pulse, all except one point. So in looking back at the equation f0(O0) = r = f1(O1). In the pulsing of prograde/retrograde thrust at periapsis precisely we retained O0= O1 at only a single position, O = 0 [If O = 0 such that Pe = rO cos O ]. The function did not change, its no longer valid at all points for expressing r. However if we had pulsed kinetic energy in any other direction other then dVx = 0 we quickly find that r = f1(O1) where O <> [Pe referenced] 0 *. IOW the periapsis has moved. The reason is very simple if you introduce velocity in the X at pY = 0 then thermodynamic energy is no longer at the minimum and consequently it cannot be the Pe. If X, Y is referenced to Pe then the system changes or the reference between O and x, y has to change (creating a 3rd reference system). There is no singular means of kinetically pulsing the system that does not either change one or more informational aspects of the system using any static reference system *. Thus if function change requires energy input, the reverse must also be true the function will not change without some sort of informational input. *Note: For a Newtonian system (a benefit) you can choose a coordinate reference that is not, say, going to be effected by change, say the position of three far off galaxies that appear at right angles to each others position and whose positions evolve slowly. From that coordinate system you would need two O to define every position in the orbit, one in X-Y and other in X-Z (or Y-Z or all three). This is actually the most thorough way of testing the system, the problem is that it is the most complicated particularly so if you don't place the systems barycenter at X=0, Y =0, Z=0. Here are the flaws in the Newtonian system. Information is always leaving and entering the system, the system is never in stasis. Also, the Newtonian system only assumes that matter can contribute to gravity and that energy can change the state. However energy and matter both contribute to gravity. Energy is always entering and leaving the system. For example, the moon's orbit is affected by the water on the surface of the earth (The earth is not a point mass, who knew). The sun is loosing mass via d(hv)/dtc2 . If the sun were to stop converting mass into hv then the water on Earth would freeze, and the moon's orbit would change more slowly (but still change due to perturbations in elevations on Earth). This is an example of the complexity of the system. Another example, a comet behave by Newtonian orbital mechanics until is reaches a certain radius, but strictly Newton does not reduce mass, and yet the comets mass is reduced by solar winds and hv. As a consequence the comets orbit is subject to change or alternatively disappear. In this 2-body problem the larger body is converting mass into both hv and energetic ejecta. The comet in response is converting its mass into the circumsolar atmosphere (compositely know as the wind). A third example, a satellite orbiting the sun is taking pictures of deep space, in doing so most of its reaction wheels and can no longer maintain attitude, by using the remaining reaction wheels and solar panels it braces itself again photon pressure therefore able to take pictures at specific positions in space as the satellite rotates around the sun, by bracing itself against photon pressure it is changing dr/dt. A forth example, a super nova is an object whose fusion pressure is no longer sufficient to keep heavier atoms from accumulating in its deepest core, the core collapses, in that immediate moment the core converts tremendous amounts of a matter to energy. In the exact moment before the star explodes a pulse of space-time energy leaves the core of the star and as energy is converted to light in the subsequent explosion another larger pulse of space-time density radiates from the star. Any object orbiting that star would observe a change of its position relative to the systems center of mass. Finally as the mass ejecta fly by this superman planet the gravity that once held the planet in orbit diffuses into the deepness of space, if it was in a circular orbit it would be free to orbit as an ellipse or a hyperbola and wander the galaxy as an orphan planet. In each of these scenarios the information of the system, but the kinetic energy in the objects has not changed in complete response. IN the first the moving tides are providing information from the sun transmitted to the earth revealing differences in the viscosity of matter on the surface that rigid matter would not provide. In the second the objects mass is dividing itself and while some of the mass remains in predicted motion, much of the mass is taking a different motion, it is doing so as a consequence of non-gravimetric information the central body is transmitting. In the third example information is provided from humans to adjust the random position of the spacecraft such that different forces balance each other. In the forth example simple newtonian function is not static, the warping of space-time changes several times before the central tenant of the function, that the system center acts as a point mass, becomes invalid and after a time, becomes valid again but with a different mass.

The video was see-rap, its was absolutely a sales pitch for fluff and nothing more. They referenced putting solar arrays in GTO to power earth, just as they were going to mine the moon and send all wonderful minerals they extracted back to Earth, and a number of other fluffy things. @DAL59 If you see NASA planning to launch a fully versatile space factory you might take some of the accounts seriously. All they are talking about are gateways, basically way-points between different potential energy points in space. If there is no factory there is zero ability to build in space, which means there is zero ability to build panels that could significantly impact Earths currently electrical output.

To explain, difficult. We need to do a little history, Issac Newton basically described a mysterious force that acted at a distance, and although he did not particularly like the idea he did not have a better explanation and the concept of gravity has stuck since then until Einstein. The concept of general relativity, and so gravity as we are familiar with is a faux force, there is no force acting at a distance. Another faux force we are all familiar with is centripedal acceleration which is not acceleration at all, but comes from the fact that objects are in a non-intertial reference frame. Our first example exemplifies the first. You have a world that has zero rotation (like venus) and you are standing on the surface. In a moment all the mass becomes a quantum singularity at the center. As you are standing you begin to fall and you are absorbed into the singularity. While you are falling you were in a inertial reference frame, since there was no forces acting on you. Therefore the force you were feeling on the ground was the ground and the electrostatic interactions holding the ground up, all the way to the center of planet, and cooperate as a force acting against your inertia. The reason you feel is because you had potential energy but no kinetic energy. The exact metric is the specific potential energy (SPE it means the energy per unit of mass). Lets say you were standing on a planet that had a radius of 10,000km and standard gravimetric parameter of 1E16 (=Mass * universal gravity constant= mu, written like u but is greek m). The equation actually comes from the integral of MGH which at the microscale determines the moment of energy exchanged at each radius. SPE=uo/r-u1/r Between any two radius we can know the specific potential energy gained from a movement to the other. For example if you move from (2-body problem) deep space to 10,000 km SPE is equal 1,000,000 joules per kilogram. So lets say the you fall from 10,000km to the 'new' singularity, which happens to be 0.01M you would be approaching c.Your apoapsis on this orbit would be the point from which you fell and if that point was infinitesimally small your random motions would have created a periapsis at that point and retruned you to your apoapsis, a perfect line from your starting position to the center of the planet and back. The reason is that the space-time next to the point mass is highly warped by the nearby energies as you approach you need to increase your velocity (inertia) but in fact the only way for inertia to increase in a free fall is our faux friend gravity. You have to think of it like this, the energy in space creates space-time (this is a combination of mass and other forms of energy). If you lack kinetic energy then it draws you in, but if you have enough kinetic energy there is an energy level by which you can leave. As you fall you convert potential energy to kinetic energy allowing you to escape. The kinetic energy you gain is Specific Kinetic energy (SKE). Specific potential and kinetic energy compose the specific mechanical energy of an orbit (as in orbital mechanics). dSKE=-(uo/r-u1/r) Total SKE = starting V^2/2 + dSKE resulting in a velocity of (2*tSKE)^0.5 If we however pushed your body 1 meter tangential to the radius you would not hit the surface of the point mass, but instead the second faux force would kick in, centripedal acceleration. So there are a couple of logics here. In our reference frame (think stars and galaxies far away) we can define a state which is not in rotation. If you fell from a dead stop circled the point mass and returned back to a different point from which you fell relative to the point mass, then it means the point mass has imparted information . . . energy would not have been conserved. That explains why you return to the same position and altitude but why does the periapsis need to be 180' opposed. The reason is that as you fall through each radius you impart potential energy and gain kinetic energy and as you rise on the other side of the point mass in an exact reversal of the process you gain potential energy at the expense of kinetic energy. Consequently both sides look the same, in practice however this is seldomly true but approximate. This is because in our universe, there are no true 2-body problems. The other issue is that relativity applies, so what appears to be Mercury's starting position to us, is not the starting position when viewed from mercury. The second logic is an empirical one. Before Newton had applied gravity to the planetary motions Kepler had already noted the elliptical motion of the planets. He also noted that 'massive' bodies were typically close to a foci. IF the central mass (say the sun) is at one of the foci it means that every planets orbit has two symmetrical parts, one that approaches the sun and the other part that moves away. The equation is this r = l / 1 + e cos (angular position). l is the semilatus rectum, e is the eccentricity. As you know cos 0 = 1 , 0 is the periapsis, the reference position for the ellipse. As you move in either direction from zero, so for example π and -π radians place a satellite exactly at the same distance from the point mass that it orbits (roughly). In an ellipse there are only two unique radius, those that intersect the semi-major axis, all other radii mirror the other half of the ellipse. How do we connect Kepler and Einstein. [See general relativity]. As you move about an orbit there is an inter-conversion of energy, you fall into a warping of space time but at some point a (the semi-major axis) you have enough energy to remain at a static potential energy level. If by chance SKE has a tangential velocity vector where speed = Vtan and Vtan = SQRT(u/r) then you have a circular orbit. If however you have at a some radial velocity then the satellite will 'cycle' from pE to Apo. There are two ways we could have orbits. if you took a cylinder and sliced it along a diagonal you would have a elliptical shaped object with two equivilant foci. In such a state we assume the mass is at the elliptical center and the satellite would approach the central body twice per orbit. You should note in the first example that change in PE from deep space (Infinity) to 10,000M is much, much less than the change of PE from 10,000 M to 0.1 M. This 'gravity' acts as a radiative force F=k/r^2. This implies that the warping of space-time is not linear with respect to radius. The decrease in the warping of space slows as one measures further and further away. Consequently the type of ellipse that is more suited is a conic section where the center of the cone is the major foci (central body) and the minor foci is at some distance from the central body. So because the rate of warping relative to distance is not uniform we need a conic section to explain orbital motion, and with a conic section the major body allows only one pass at close distance per orbit. I should point out the Kepler's laws of planetary motion are a conservative guess. In fact with 100s of years of study since there are lots of n-body corrections. There are a great many places in our solar system where planetoids and other things could reside in stable orbits, but the problem is that massive bodies have a tendency to tweek the orbits of smaller objects, potentially catastrophically, but often throwing them out of our system. The la grange points are examples of places where we can find 'trojan' asteriods, and for jupiter and saturn there are a fair number. But earth and venus have few of these objects. So even n-body solutions are not perfect explanation.

But that is not the sats they were talking about, they were talking GSO satellites, since the almost always are close to the equator and because they are 33 k km from the earth and general at 23' Eath sun axis relative they will seldomly block the sun. If anything beaming microwave energy from large solar panels in space will heat the surface of the Earth, stop propagating BS, please. You get your science from two-bit popular science sites.

The electronics were made with 1960-1970 level of durability. Made in USA and overengineered. By criticality you mean pulse subcriticality, allow the material to approach the prompt critical limit for a few seconds every hour just enough to create isotopes for heat generation. Again if you use a layered thermocouple design you would never get more than 30% of the heat energy from the reactor, the rest is waste. So as the units power output drops so that 70% of output < Wattage of radiant energy release at the desired radiator temperature you would pulse the reactor briefly and then let it idle. You never want a dry reactor to approach criticality, most of the metals have very low heat retention capability and will rapidly heat to disintegration temperature. If you want to know more about Uranium styled reactors you need to read up on the Russian designs, they are the leading experts on these types of reactors in space. Plutonium is very dangerous to work with, period, it can be rendered safe, but the biggest problem is that someone needs to work with really dangerous materials to get it into the safe state. I used to be a safety officer and we worked with more dangerous materials, suffice it to say whatever you can't imagine that might go wrong has the unfortunate problem of being the thing that happens. Here is the basic problem. We tend to think of cooling as something that has to be done to keep reactor cool. Cooling of a reactor is equivalent in function to turbocharging a rocket engine. The reason for this is that power is generated along the temperature potentials. If we removed steam from the equation then the best method of heat transfer and work generation is lost. So that cooling moves power from the reactor and to the surfaces that can generate power. If we remove all cooling you drop the output capacity of Uranium based reactors by magnitudes. The only thing that really limits the ability to produce more power in a fission electric steam generation system is that as you increase cooling capacity and thus power production your reactor coil inches closer and closer to prompt criticality and your margin of safety begins to disappear. As per chernobyl the single moment that caused everything to 'uncontrolled' was the building up of resident steam on the uranium pellets. IOW parts of the reactor could not get rid of steam fast enough. In water steam conversion steam always builds to a point and is released and so heat flow needs to stay below the point were steam becomes static.

https://www.nasa.gov/exploration/systems/sls/multimedia/flight-hardware-for-sls-on-its-way-to-cape BTW, hiding way back behind that rocket is an RL10b-2 with its 110KN of thrust. Thats an awfully big rocket for an engine rated at 220 seconds of burn time. 220 * 110,000 = 24M n*sec, Im guessing the payload is 50 tons of mass (a~2 m/s^2). ~600dV a*sec of thrust, where is that going, there version of the RL10b-2 must have a much longer burn time. In notice other versions that have 4 RL10b-2 thrusters, which seems more intelligent to me since the thruster only weight 277 kg.

Solar energy in orbit adds to earths energy budget. Solar energy uses energy that has already penetrated the atmosphere, it would have gone to heating a roof, instead it is diverted to an A/C system or the like that then goes into the air quickly where its heat is more rapidly lost to convection or radiation. Microwave energy is beamed through the atmosphere, heating the air as it passes, then heating elements on the ground and also heating the ground. Anything in its way that blocks sunlight, like clouds will be heated and will tend also to disappear. The best energy choices are wind, geothermal and hydroelectric power since they generate power on the backside of energy flow on earth. Nuclear is also a good choice but only in countries that have tight regulations. Conservation is the best choice. As for some of the critiques. Wind power is often available when the price per kilowatt*hour is zero, therefore it is prudent to convert wind power into hydrogen during these periods of power excess. The same is also true for solar power; however solar power rarely saturates is local market.

The Space Shuttle and Soyuz have lost statistically rate-equivalent number of individuals. There are no other human transport system that are notable. Launches into space are inherently dangerous, much more dangerous that operating a civilian nuclear power plant.

Peak energy demand occurs roughly at the same time as peak solar energy for much of the world. The number of deaths associated with nuclear power is well below deaths associated with other industries. In the continental US you can count the number of people who have died in commercial nuclear accidents involving radiation as primary cause of death . . . .well none. The number of people who have been killed in space based accidents. Lets see 2 crews of 2 shuttles. Apollo 4, 2 russians died from asphixiation . . . . .Doing business in space is dangerous much, much more dangerous than doing nuclear. The report is BS, period, hypey B.S. and nothing more. Edit: I have to retract part of that, in the early 20th century some factory girls were killed because they painted watch 'hands' using liquified radium at a time when the risk of radiation exposure was not clear and there was no occupational health safety administration. Since the end of WWII there are a few incidences, most of them military, SP-1 was the result of physical and not radiation exposure.

The patching they are referring to are semi-continuously fired SEP devices to keep orbits from, well, slamming into a hard object, like the moon. 1. Adding power to the earth does not take away global warming already created. It only adds to it. By the time this things gets going the carbon usage will already be on the fall. 2. We already have potentially all the non-carbon producing electric power we need, Wind, Solar and Nuclear. The only countries that can afford "maser" power are the countries that already have enough money to create their own carbon-free power. The countries that don't wont be able to afford space maser power. Finally, the life of solar panels in space is not infinite, the larger panels are subject to damage from sun-spot activity (the larger the panel the greater the cross-panel voltage potential). 3. The plan was only reasonable to a fool or a politician. 4. SpaceX will not have 100 people on the moon in five years. SpaceX has yet to place a single person in space.

The first bug was actually a bug. I think that the correct statement is that its error checking subroutines were not up to the task.

I was just watching this I was a spectacular pile of BS. 1. Earth powered by cis-lunar solar power stations. Uh, global warming plus GTO solar power stations = global crazy warming. 2. Resourcing space missions from lunar mining buses. Because getting your stuff of them moon is radically better than getting them off of earth. Ok I not a perfect rocket science, but dV to land on moon is about 3000, the dV to lift off of moon and orbit, also about 3000. Thats 6000 dV. dV required to get stuff off of earth, around 10,300. a. So at best the moon is like 60% the cost of launching from Earth. b. Elon already has a system of ships that can be recycled. c. The worst case scenario for water here on earth is you dip a garden hose about 400 feet away into a briney canal and use reverse osmosis. The best case scenario on the Moon is that you have to grind rocks and heat them to generate steam which you can consense. d. The worst case scenario of oxygen on Earth is you buy a concentrator from your local medical supply shop and concentrate oxygen from the air. The best case scenario on mars it to create oxygen from electrolysis or heating rocks to very high temperatures. e. THe worst case scenario of hydrogen on earth is that you have to perform electrolysis, but methane also can be used to generate hydrogen much cheaper. Hydrogen can also be produce by biological reduction processes. There is no biological reduction on mars. So lunar bases cryogenics are very expensive, much so than earth. 3. Resource gathering from the Moon. Not better than the earth The moon is believed to be composed of the ancient surface of pre-theran earth IOW almost all of the moon's composition is resembles the surface of the earth except without all the separative processes here on earth that make copper, iron, silver, uranium mining work. Somewhere on the moon is a better place than on the surface of the Earth and they are going to map it with satellites. Lets look at the earth. a. Hydrogen (about 10% of the worlds ocean), helium is rare - but even harder to extract from space, helium is the expected by product of fusion energy. lithium (14,000,000 t reserve), berylium (400,000 t reserve), boron (major component of sea water, common component of detergents), carbon (coal), nitrogen (70% of air), oxygen (29% of air, 80% of sea water), florine (mostly toxic to humans), sodium (sea water), magenisum (a better source is earth passing asteroids, lower dV requirement), aluminum (29,000,000 t reserves), silicon (beach sand), phosphorous (extensively mined), sulfer (sulfate is a major component of sea water), chlorine (sea salt), potassium (pot ash) calcium (lime stone), . . . . . .None of these elements are cheaper on the moon. The only thing that raises the brow a bit is titanium oxide about 12% of moon rocks. At that percent it might be a tenth as competitive as the price on earth. b. they better use a very low frequency, the heat of accretion probably put the low level of precious heavy metals at the center of the moon, what kind of drills do they have?

I imagined a rescue scenario for Martian expedition. Here is what it was You burn retrograde from earth (Atp something like 310 to 290) and descend to a submercurial orbit (27 days), then you burn prograde at an appropriate time to intersect with Mars (<136 days), assuming that Mars is say behind earth in its orbit, the craft then wraps around the sun and burns out to Mars intercepting mars with a excess radial motion nearly at right angle to Smars, orbital. Atmosphere entry velocity is ~20 km/sec. One always has to remember that if we can improve ION drive power systems, there is no ISP limit for ION drives, c is entirely possible for hydrogen (as it is done in a super-collidor). Therefore the fuel requirement can be zero (i.e meaning it could be the fuel waste of a nuclear reactor). Orbits that drop in the direction of the sun can utilize increase insolance and burn in the direction of Mars with very little mission lead time. For example suppose you had a situation were Mars was 120' degrees behind earth, there are 'diving orbits' that intersect Mars. Here is where ION drives ironically have some ability. Given adequate lead time an ION drive could, in priniciple, burn to nearly match the orbital velocity vectors of the target as it approaches. The problem with exiting earth is not a problem intersecting planets, its just a problem if one wants to insert into an orbit quickly. So that once you get to a planet the rescue ship would need to have some other propulsion. Presumably this would be a lander that would use the atmosphere to break its orbit, while the ION drive very slowly circularizes its orbit. The problem is a rescue ship needs also to leave Mars (maybe there is fuel on Mars?) and reach the same orbit as the rescue ship, whereby it could return to Earth.

July 28th 2035 buy your tickets now. lol. 128 days to mars, then I guess Elon will have a chance to wait on Mars for a ride back. Mars Earth Departure May-07-2033 4.13 km/s C3 = 20.7 km2/s2, DLA = -19°, 120-day transfer Mars Arrival Sep-04-2033 2.78 km/s* Mars Departure Sep-24-2033 3.92 km/s* 210-day transfer Earth reentry Apr-22-2034 14.72 km/s reentry 350-day total mission 10.83 km/s (add an additional 0.5 km/sec for other stuff like rendevous with landers) Solar range: 0.73 - 1.39 AU Earth range: 0 - 0.95 AU If we place the lander in Martian Orbit . . .apriori.... Trajectory Itinerary Earth Departure Jul-12-2035 3.77 km/s C3 = 12.2 km2/s2 DLA = -20° 176-day transfer Mars Arrival Jan-04-2036 716 m/s* Total = 4.48 km/s (allow 0.5 km/s for docking its lander with the Transfer habitat) If this ship has a lander then the lander needs 1000 to 2000 dV for landing on Mars and 5000+ dV for reattaining orbit.

The problem is that many Martian landings make the assumption that you are going to be able to aerobrake in the martian atmosphere. However if your transfer has too great of a Vrad relative to the ||Sun-Mars then what happens is you will have to spend extra dV to carry a comparable heat shield.