PB666

They all take off their clothes and tell everyone back home they skinny dipped with venusian women.

No species is moving itself faster from global domination to extinction than ours. Of course we still do not know what cause snowball earth. One has to remember that Lystrosaurus represented a higher percentage of vertebrate fossils following the Permian-Triassic extinction event, within 10 million years neither they or their descendants have been observed. In finance, past performance is no indication of future success.

We've been through that, the problem is that the amount of fuel required to launch the craft into a L1 orbit and block just a few percent of the suns light exceeds the benefit. To be able to do this would mean the wholesale redirection of asteroids into an L1 orbit (not only risky but hard to do) and ISRU conversion of aluminum, nickle and silicon (not the preferential materials for a blocking agent) into a solar screen. It may be possible, trapping carbon would probably be easier. Another strategy that may have longer term benefits is to use induction to move the orbit of mars and earth further out and away from the sun. The problem with this right now is some parts of the world that are in deepest trouble rely on insolance levels (pan-evaporation rates) as the energy source that drives precipitation inland. If by either means sunlight falls, East Africa could go into a deep prolonged drought killing 10s of millions of people. The best solution is to educate the worlds most vulnerable populations that their behaviors (environmental exploitation) effect them and their neighbors, and their neighbors actions likewise. If we can get a country like china to markedly decrease coal utilization and India to reduce the burning of firewood for cooking and heating, to stop burning of forest in Indochina, Oceania and Latin America as a means of generating cropland then there is some flexibility in the technique used to cool the Earth.

Remediating climate change and Atmospheric carbon reduction are two separate issues. To give an example, to remediate ocean acidification in sensitive fisheries one can take limestone, heat it until it decomposes into calcuim oxide, dropwise add it to distilled water until saturated, mix this with dehydrated salt water and distribute it into the currents that feed the fisheries. That may fix the fisheries but it also adds CO2. There are many similar issues such as widening waterways, national and continental water plans that deal with extreme drought and rainfall events. All of these things require energy and technology that require CO2. Yes you can use Solar panels to move water, but how do you move water close to the coast in a 1:500 year rain event that are now happening every 20 years, given that the infrastructure and power reserves required will not be used but once every 20 years. Carbon Dioxide is very abundant in the ocean, Carbonates are major buffers in sea-water. It is easy enough to take NaCl and split it into Sodium and Chloride. Acidify the sea-water and heat it and CO2 will come out, add the NaOH back to it and you will have sodium rich sea water. The problem is what to do with the CO2 once you have it. No-one really takes carbon capture seriously because no-one as yet is talking about carbon dioxide polymerization. About the cheapest carbon polymer that can be made is formalin, and it requires the addition of two CH bonds, its easily contaminates ground water and is toxic.. In terms of Carboniferous-like tree production, it has to be remembered that tree growth (if you can call them trees) in the early terrestrial ecosystems was very rapid compared to today. That is the problem with lignite coal burning, much of the trees are bark that was contaminated with all kinds of heavy metals. Trees could grow more densely than at present make coal formation much easier. The more rapid alternative is microbial oil synthesis (biofuels) the problem is that the most prosperous that dont require CO2 producing fertilizers undergo nitrogen fixation, the cells of cyanobacterium that fix nitrogen are extremely toxic to most eucaryotes, and so this cannot be done in the oceans and require land use that could otherwise be devoted to other tasks. So before you come up with a strategy for greening the oceans for carbon capture, you first need a way of nutrient priming oil producing microbes that is also carbon neutral on the front side. This is the problem. I should make the point here that its about 10 times easier to conserve fossil fuels than remediate their damage after your burn them. If someone can come up with an efficient means of Hydrogen production (at least half is wasted in production of oxygen, something that is already abundant) then addition of hydrogen to coal emulsification plants or as a cracking agent in coal-tar sands effectively reduces the carbon footprint per btu in half. This is easy, but adding hydrogen to CO2 is significantly harder because of the stability of C=O bond versus C=C bond. A significantly higher amount of energy would be required to crack and hydrate CO bonds to form formic acid which can then be further reduced to methanol. Chain elongation is more difficult but doable. The minimum ground stable carbon polymers are C20 and longer. These steps are on the order of 'beware of the unintended consequences'. The best choice on a planet of 7 billion people is to nudge people to conserve energy ever-harder each year: about deforestation, unneccesary pastoralism, inefficient and carbon-hungry agriculture, power production/transmission inefficiencies and fertility rates. Trying to remove carbon in the current environment will only make the matter worse, the people who are removing the carbon will attempt to exact the costs out of people generating carbon which could lead to war. Get those across-the-board usages downward trending first. With regard to population size one has to remember that most of the worlds population lives within 500 feet of sea-level, so if decadal-intensity cyclones increase in frequency and people are moved out of the highest production areas (such as what has happened in burma) then population will fall anyway. The predictions in the mid-90s that moved the discussion from global warming to climate changes suggest that large areas of dense human habitation were at danger for increased intensity flooding and droughts. If people are squeezed from these areas to lower production areas then there will be wars (such as what is happening along the Burmese and Bangladeshi border) and the population will fall. The US department of defense has already accepted this as something that will happen, and almost all countries have signed the Paris accord so at least in words people deem this is important. In terms of the original post. dryland grasses increase ground water; the grass roots increase the penetration of rainwater deep into the soil and in drought grass die or rely on dormant root systems for seasonal growth. Overgrazing of semi-arid land encourages grass decline and the growth of non-edible trees (such as mexican red cedar in the southwest). This then blocks the absorption of ground water. Native americans knew the value of this and burned grasslands periodically to kill the trees. Cattle brought both mesquite and cedar from Mexico and this has increased the runoff and inhibit the penetration of water into agriculturally benefical aquifers. Africa's deserts may be making a comeback for other reasons. Of course it is smarter to control grazing and replace pastoral-derived intake with agricultural products. Management strategies may fail before they begin because of social instability, some of which has been created by climate change. There are many leading issues with regard to proper land management including composing versus chaff burning, disuse of firewood, recycling human waste and animal waste for methane production, regrading soil to trap water, restoring underground water reserves, etc. Any of these can increase carbon retention. These are all fine but who is going to northern Nigeria and Niger to teach locals how to do these things, war is all the easier way to control population overgrowth. It is, afterall, the outsiders that made matters worse to begin with.

And who exactly is going to build and test the floating venutian ship. The trip to Venus exposes the crew to more radiation than ISS, the earths magnetic field takes about half of the radiation and travelers to venus on average would be exposed to 3 times as much radiation. As a consequence to get to Venus they would have to be able to do so in about 4 earth months. In addition Venus has no magnetic field like earth, it has no moons. There is no protection when astronauts get there. Colonies on mars are doable, just incredibly expensive and suicidal (that is the risk of death each year increases 10 fold that of earth). Reaching the venutian cloud level is simply insane.

you don have to fill the whole cave, just block off sections. Find or make a smaller lava hole. Build a air tight wall on both sides of the lava hole; this will require circumcising the tub until solid rock is reached. The each wall will be connected by a submarine door (or two). Fill the lava hole with moon dust, packing the dust as it fills, as it fills there are empty gas bags distributed at even intervals the bags will be laid out vertically hanging circles and connected to a pressurization system, rigid enough to withstand a couple of atmospheres of pressure. in areas with gaps between circular bags chains would connect the two walls with force sensors on the chains. Once the dust is compacted the lava tube/manhole is capped. As the the most distal chamber is inflated the pressure in the bags is increased so that the balance of pressure on both sides of the wall is the same. This could be done simulataneously at many spots to create a pressure gradient. BTW the material that comprises the lava tube walls is expected to be toxic to humans since the tubes have crumbly material it would need to be dressed. Thus there would need to be a way to coat the walls with an oxygen resistant material. Likewise sealing the walls to the pressure walls. Most of the cements used on earth would not stand up to the vacuum of space, so pressurization with nitrogen during application may be needed.

Magnetic North moves. Also A polar orbit passes over the both magnetic poles. A 100 minute polar orbit passes to within a few kilometers every twenty four hours. The Earth rotates under the orbit. There are 14.4 orbits a day however radiating 28.8 orbits per day at each pole. The Geomagnetic north poles is at 86.5 degrees. A degree at the equator is 60 nautical miles or 110,600 meters. The cosine of 85.6 is 0.0610. 110,600 Cos 85.6 = 6,746 meters. If a satellite cross from equator to NP or vice versa each orbit, and the number of orbits a day is not an even whole number then it creates ~ 29 areas everyday. 360/29 on average differing by 12.413 degrees however its closest proximity on either side are 6.206 degrees which is means the satellited passes within 41 kilometers of the north or south pole every day. Within a week it will pass within 5 kilometer. The magnetic pole of the earth has been moving 35 kilometer each year in a western direction. This means it moves 0.673 km per week. Thus a magnetical polar orbit would in about 2 months, obviate any need to have a magnetic polar orbit. In addition since the satellite would hold magnetic north the longest if the equatorial path was set to 86.5, and south magentic pole is less than 86.5 degrees, then it would be and not 90 degrees, the more intelligent magnetic pole crossing spacecraft would simply have an inclination of 86.5' allowing clustered mulitple north crossings per day and a few close crossings of the south pole.

My guess that at 0.5 c that 150,000 km per second taking 4.5 seconds to reach the opposite side, the suns density is low. It would create a shock on the opposite side ejecting a large amount of plasma into solar orbit and possibly picked up by venus and earth. It might have a cooling effect on venus. Why do you need to collide our sun when you have nice neutron stars you can watch merge.

Yes, yet another bit of scenary for Jeb to fall though and get stuck in.or go poof. There are martians hiding inside the caves waiting to ambush venutian women. Mars needs women!!!!!! It goes without saying that the nation that plants a flag up on these caves will have a advantage in colonization of the moon. At least on problem just disappeared, what to do about space radiation.

Well given that forum members want other forum members to volunteer to be part of the ultimate fix, I would say its now officially in verboten land.

I thought the politics of climate change was verboten.

By now most here are familiar with Keplar's laws of planetary motion. For the most part these are empirical observations, without the mechanical thermodynamics its' relatively difficult to apply theory to. Specific Mechanical Energy (SME) is a subset of all Specific Energy. Specific Energy is the energy per kilogram in its various and sundry forms. SME is the energy within an orbit. To show the relationship a spaceship during a burn converts chemical energy Specific Chemical Energy in fuel entering the reaction chamber into SME in the form of Kinetic Energy. If this along the prograde the specific KE is converted to specific PE such that the SME remains the same until some other form of energy is converted to SME. While Keplarian laws are useful in calculating position over time, understanding SME is very helpful in describing objects that transit in and out of differernt Hill radius, such as a spacecraft traveling between different planets. Critical to understanding SME is the simplified mechanical energy potential Mgh (Mass x agravity x height) where h is the height over which PE is being analyzed. The result in Joules (J) can be deprecated to specific mechanical potential energy by the removal of M and thus the units are J/kg. The equation gh almost never gives a correct answer unless some sort of gravity correction is applied, but for very small heights the answer is generally well within the relative precision of h measurement. Averaging the gravity between the two ends of the potential works well only when plot of gravity versus height is flat, once the scale is such that the plot is visibly non-linear the estimation of velocity will be incorrect. An example of where this might be a problem is during a Suicide burn countdown. The actual equation is the integral of gh which using standard graviational parameter (µ) instead of g and radius replacing height can be converted to SPE = -(µ/ro - u/r1). For a central point mass (CPM) the resulting equation gives and accurate SPE from the surface of the objects hill sphere to 2 Schwarzschild radii . If we makes the basic assumption that SPE between an objects hill sphere (SOI in kSP) radius and infinity is low the calculations can be relatively simplified, in some cases really simple. The reason for this is that in the SPE equation if starting radius (ro) is infinity the SPE is µ/r where r is the radius that one wants to leave orbit from to escape or circularize to on system entry. Upon entering a Hill sphere the SME = SKE + SPE until exit or some other force is applied. Given that a hyperbolic orbit entering a hill sphere at t=0 has a SPE0 At the bottom of the potential all SPE is converted to SKE and speed (Vorbital, Pe) is = √2SPE0 . It really is that simple. In addition the velocity of a circular orbit at Pe is √SPE0 and the dV required to circularize at PE is 0.4142√SPE0 , [ √2 - 1 = 0.4142135]. Therefore for any given circular orbit of radius r to CPM the amount of dV required to escape is 0.4142 √(µ/r). You cannot get simpler than that. In an orbit SPE and SKE are not always interchangable (as in an elliptical orbit) at all magnitudes, this is the fundemental inequity that causes the oberth effect. The limit to which SKE, at any given radius, can be converted to SPE is specified by µ/r. Any SKE acquired at rexceeding u/r resides in SKE. This can result in some rather bizarre physics. For instance if we convert Kerbin into a nuetron star of radius 0.5 km and entering kerbin from its hill sphere, orbital velocity is slightly adjusted to obtain a hyperbolic Pe of 1 km. The equation tells us that at 1 km we have an SKE of at least 3,531,598,400 (v = 84,083 m/s). If exactly at Pe we could add 1 m/s dV (v = 84,084) we can see the effect. In adding 1 dV, SPE (3,531,682,443.3) leaves 84043.327 SKE on system exit. If the space craft leaves into another CPM gravitational influence along the left bodies prograde motion the bodies residual velocity (409 m/s) is added to that bodies prograde motion speed. In this instance, by entering the hill sphere and approaching the dense body closely the addition of 1 m/s of prograde motion resulted in a 84000 fold return of energy and 409 fold return on velocity. While the oberth effect is limiited to objects entering and leaving hill sphere's, the underlying physics are not so limited. An example is burning from Kerbin to Moho. The primary issue here is that an object in Kerbins SOI is traveling through kerbins space-time warp. A key dimension is time, the longer time required to exit, the more time space-time has to act on it. By applying a tremendous amount of dV from a minimal circular orbit the less time the space craft will be influenced. If simply 0.41422 √(µ/r) the resulting craft will exit kerbin with dV slightly different than kerbin. The limit orbit around Kerbin is 70km alt (r =670,000). √(µ/r) = 2295.68 if one applies 2000 dV rapidly along the tangent close to angle to prograde 300 the resulting SKE = 9.226MJ/Kg. The residual SKE on kerbin system exit is 3955987.99 J/kg. This results in a residual velocity of 2812.82. IOW we applied 2000 dV to a stable circular orbit, we overcame the energy required to escape (dv = 950) and majically had 2812.82 'dV' to play with on Kerbin exit. Since the Kerbin orbital velocity is 9284.5 and the exit vector is nearly retrograde the resulting crafts circumKerbol velocity is 6741.81 and allows the resulting orbit to cross the elliptical cylinder that moho resides (noting to intersect the orbit itself requires a plane change in kerbin). So the question is how to practically use SME. So lets take moho as an example because it is one of the most challenging objects to intersect. For the hohmann transfer we want an orbit with a Pe very close to the Moho periapsis (r = 4,210,510,628; 4.2Gm). Kerbin has a orbital radius of 13.599840256 Gm. The semi-major axis is the average of the two, 8905175422. Using vis-viva equation (kerbol µ = 1.17023 x 1018] the velocity at kerbin orbit needs to be √(µ(2/r - 1/a) ) = 6378 m/s. With this number in hand if we need to make a plane change of 7 degrees (assuming that one node is radial to its periapsis and we align 180' but along the same axis through kerbol, then we can calculated antipolar dV required. 6378 * sin 7' = 777.28 (south polar). From Kerbin escape minimum of 9284.5 - 6378 = 2906.3 m/s velocity in the retrograde and 777.28 m/s polar. The speed is thus √(2906.32 + 777.282) = 3008.45 m/s. So now we convert back into Kerbin. SKE on exit is 4,525,385.7 J/Kg. We have to add this to KPE of the orbit we are rising from 5270477 requiring an SKE of 9795833 and requiring a velocity at alt = 70k of 4426.3. Since the orbit requires 2295.7 this means dV added is 2130.3. The savings from making these two burns in low kerbin orbit are 1553.3 dV. However, there is one caveot, because kerbin is rotational axis is perpendicular to the plane of its rotation with a equitorial surface velocity of 174.94 and the moho pre-plane matching orbit about kerbin is not perpendicular to the same axis the launch looses some of that rotational velocity, resulting in a slower acquisition of circular orbit. Exit angle relative to the equitorial plane at angle to prograde 270' can be calculated. Its the arcsin of 777.28/3008.45 = -14.97' OR a bearing of 104.97 relative to kerbins axis of rotation. The amount of dV loss due to non-optimal launch and circularization angle is 174.94 - 174.94 * cosine 14.97 is greater than 5.97 m/s dV. Therefore the amount of savings is less than 1547.3 m/s dV. The remaining question is where to launch. There are two places one can launch from, but in either place the launch either needs to cross the Angle to prograde plane of 90/270 degrees that halves kerbin at 36,000 meters, at which point HSI switches from surface to prograde motion (assume you are using a automated navigation system like MechJeb). Because of Kerbins equitorial surface velocity is not zero, the initial launch angle at when crossing Angle to Prograde of 90' below 36,000m is less than 75' or at AtP of 270 crossing is more than 105 degrees. We can also calculate this but the calculations are inexact because speed reaching 36,000 can vary from 400 m/s to 800 m/s a good guess is to use a handicap of 2 degree (73 or 107 degrees, respective to AtP). Any inaccuracies in the tilt can be corrected at Atp90 or 270 using polar burns in LKO.

And they always carry some extra fuel to make corrective burns. lol. Anyway its the n-body problem, you cannot have a perfect calculation so why not just get a good enough calculation and correct at some later point.

Just making a correction to one of the links provided. For Moho, if you transfer from 70 km kerbin orbit. The best place is the ascending node that is close to Moho apoapsis. 1. Launch at 96 Angle to prograde (6 minutes to midnight ). 2. The angle to use is not 45' This is too steep for plane changing to Moho's plane. instead use 73.5 degrees (heading) from launch and switch to prograde at 36 km altitude. Set circularization altitude for 70.5 km or so. This establishes the maximum heading in orbit of 15 degrees. 3. It takes approximately 2250 dV and the burn should begin between Angle to prograde of 310 to 295 depending on TWR. The apoapsis of the circum kerbol should be about 1.5 days in front of kerbin. A small correction burn in flight may be needed to get an inclination of 7.000. When you cross 270 angle to prograde (high noon) the bearing should be 105'. 4. It takes 2559 dV (moho intersect) and 323 dV (insertion at 500 km altitude) to establish a circular Moho orbit (this is high, I carry three satellites and set up a network on Moho before landing). Another 1000 to land. Total cost to Moho 4200 + 2250 + 2559 + 323 + 1000 (10,032). Kiss ole Jeb goodbye, you need about an additional 60% dV (~16,000) to get him back to kerbin. IF we compare this with to intersect Moho using a strait bearing of 090 and then plane change in kerbol orbit. Spend 2173dV to burn to moho near-perapsis on Hohmann transfer. Since leaving kerbin you have initiall 70 km is around 2300 dv (KE = 2.6Mj) to escape is 3252 (5.3 Mj) 2300 + 2173 = 4473 (10Mj) (10Mj-5.29Mj = 4.71 Mj) leaving 3070 m/s. This means orbital speed around kerbol dropped from 9284 to 6214. The plane change required is 7 degrees at orbits apoapsis is best place to change. 6214 x sin 7' = dv 757 m/s. Consequently a combined burn to intersect Moho and plane change starting in low kerbin orbit (2250 dV) versus a two step burn (2173 + 757 = 2930 dv) saves ~700 dV. There are a few dV lost on the 15' bearing prograde burn because the planets costing about 10 dV.

Uh, huh as you get closer to the sun the intensity of magnetic storms increases. The affect of no gravity during the flight are comparable. Before Elon Musk can die on Mars he first has to survive the trip to Mars (or Venus). So a rotating ship as mentioned by kerbiloid with 1 meter thick composite with a density of 2.0 is 2000kg per cubic meter. For a rotating ship to work the human body would on average need to experience 1 g of force, this means either 2 to 0 g. But the effect of less gravity also includes the eyes. Consequently the head needs at least half of a g-force. The average human of 6 feet tall would need to be in a cylinder. And the device needs to be balanced, iow it would hold two individuals. Such a device 2/3 of a meter by 1/2 a meter (minimally) or about 7/12 of a meter in inside diameter and an exterior diameter of 19/12ths of a meter would have a crosssectional area of 1.968 square meters at the base (v=1.968 m3) and 1.7002 along the column. In order for the g force to be 4.9a at top of head and average 9.8a over then entire length of the body (4.9 + X)/2 = 9.8 then at the feet G should be 14.7 a. The column should be 2.74 meters in length( v = 4.65 m3) The mass of the body, alone would be 2 x (4.65 + 1.968) = 13.25 m3 x 2000 kg/m3 = 26,504 kgs. See other thread about the amount of ground rocket to escape velocity. Basically you need An atlas IV rocket to send 3000kg to escape velocity. Double that to reach mars and retro into orbit. So to get 26,504 from a launch pad to Mars you would need the equivilent of 17 atlas 4 rockets. This does not include non-g areas of ship, its food and water supplies or the mars or venutian lander. So lets just say a launch rocket about 5 times the size of a SaturnV rocket. Yeah, no, thats not a solution.

eccentricity is a geometric parameter that defines the relative shape of an ellipse. If you have a point mass in which the system center and point mass are virtually indistinguishable then any 3 coordinate 3-vector velocity produces a single eccentricity. (In a 2-body system three dimensions can be reduced to 2 dimensions). Noting that for an point mass moving around another point mass in space, the relative states all points and velocity reference frames are valid, we can give the central point mass (CPM) a coordinate of 0,0,0 and a velocity of 0,0,0 and that at any moment we can give an object a two dimensional position of X = 0 and Y (r) = distance from CPM, a radial velocity of dr/dt, a CPM relative speed and dX/dt which is the (Speed2 - (dr/dt)2)0.5. For any motion of movement the triangle of the distance traveled as one side and the CPM is exactly the same. Roughly this is 0.5 * Y * dX/dt* t. From this you can calculate a (semi-major axis). Also you can derive moment of Radians/sec. dr2/dt = (radians/sec)2r- Mu/r^2. r1 = r0 +dr/dt + dr2/dt. Since total energy has to remain the same no matter where the object is and thermodynamic energy is determined the integral of r to infrMu, then dX2/dt can be calculated from SQRT(KE0-dPE) = speed and new dX/dt is derived as above. IOW non-inertial objects orbit with no regard to eccentrity, theoretically if the warping of spacetime about an object did not exhibit perfect radial symmetry, object orbits would not be elliptical and eccentricity could not define position. This is in-fact reality, as objects move around earth the density and elevation of the earth shifts slightly, consequently the orbits are not elliptical. If you were to look at this from a more macro - quantum perspective you would say that the eccentricity of an orbit falls within a confidence interval in which all positions relative to the average positions at whatever sigma you like. This might have some pertinence for objects orbiting a black hole. Another example, the two neutron stars that merge both spinning rapidly themselves with each very powerful magnetic fields. Objects in that environments are not expected to exhibit true keplarian motion even when 3-body solutions are given. To simplify, eccentricity is an approximation of a vector quantity just like centripetal force and gravity it is a faux parameters. Eccentricity is used because many orbital calculations begin with: given: a central point mass assume it is infinitely greater than an orbit object. When you are given these two the assumption is that space-time exhibits perfect radial symmetry, as a consequence mu is perfectly dependent or radius.

Oh boy. I would say before we do all this arguing why dont we create a thread 'which do you think is worse on the human body, space radiation or lack of gravity? Maybe we can get people to actually think about the nuances of space travel.

I have a actual scale RL10b engine you can put on the back of it. lol. Its just so easy to make fuel tanks and engines if you make a decent part send me the file and I'll use it. The hard part is the command modules because of the hatch transforms and the interiors. Actually I don't really ever look at interiors. For me its all about the functional stuff. First Off, most all of KSP parts can be rescaled. if you have 1 1.5 meter diameter rocket, just take a 0.625 radius rocket and copy the cfg and -rename the cfg e.g. factorOnesection ----> factorOneSection5over4 -open the cfg -change the parts name. -change the rescaleFactor to 1.2 -change the title -change the mass . . . . .new mass = old mass x 1.728 -save the cfg go to space station view. Open cheat window (Alt -F12) choose database. Reload database. Go to VAB. Add the new size of part. done. Forgot to add if they are a fuel tank increase fuel by 72.8% If it is an coupler or stack separator you might want to increase the ejection force by 44%.

8000 * SQRT(2) - 8000 = 3313 m/s. This is the approximate dV required to escape earths orbit from LEO. It differs with LEO chosen. What this means is that a 4550kg payload can be carried from LEO to escape velocity with dV of 3475. Large payloads might not reach that height, smaller payloads can go farther. IOW this is a rocket spec. The RL10B-2 can run for a rated 220 seconds at 110,000 N/sec of thrust for 24,200,000. It has an ISP of 475 (4700 m/sec exhaust velocity). 23.404 kg of fuel per sec. This is 5148 kg of fuel. The engine weighs 277 kg and the fuel tanks are probably 12% of the fuel weight so its fixed mass is 6042 kg. With no payload is can achieve dV of 8700 m/s second. It can carry a PL of 3814 kg and achieve the same velocity spec as Arian 5 rocket. If this is the desire you would need a rocket that can carry a LEO payload of 9856 kg (~10 tonnes) into LEO. Note that in order for the RL10B-2 to achieve a higher payload all that needs to be done is re-rate the engines to a longer burn time and increase the proportion of fuel. The problem with at is a exit trajectory requires a fast burn at a single theta. A burn of 220 seconds covers a theta of 16 degrees. If one extends the burn time then one also has a longer burn window and some of the fuel is used in raising the perigee of the orbit. The following launch vehicles can launch such a payload. (*** retired, * in developement) Take your pick. BFR (Space X)* -15x ENergia 5b* - 8x -15x Saturn 5*** - 14x LongMarch 9* - 14x SLS Block 2 w/EUS* - 13x SLS Block 1b* - 10.5 Energia** - 10x N1** - 9.5x SLS1* - 8.5x Falcon heavy* - 6.4x New Glenn *- 4.5x Buran *** - 3.0x Delta4 heavy - 2.9x Lng March 5 - 2.5x Space Shuttle*** - 2.4x Angara A5 - 2.4x Proton M - 2.4x Falcon 9 - 2.3x , less if recycled first stage. Titan 4*** - 2.2x Arian 5 ECA - 2.1x Atlas V - 1.4 to 2.0 and everyones most recent entry into the world of national or commercial space flight. Given that one can increase the fuel, re-rate the engine and increase the payload its certainly possible to achieve higher rating. The spec they did not provide is final stage thrust. On a delta 4 heavy one could have a much larger payload but confined to 'kicking' the space craft at perigee several times to achieve maximum efficient exit.

Ambitious isn't he. I think he should set his sites on not dying during the 9.5 month transfer.

Well, Elon wants to die on Mars. I wouldn't say that it is a practical ambition but it does take away the need for landing deltaV. lol. Lets summarize the problem, the exploitable utility of Mars is about 1/100th of the most impractical location on surface of the Earth. Its better than being in a volcanic crater but worse than the south pole. The resupply missions to the South Pole being a tiny fraction of those to mars. For the cost of one colony on mars we could have a tunnel through the antarctic to the south pole allowing for an almost balmy walk there. As Tyson points out Mars is not a practical Earth back-up either. There is no reasoning there for a colony, this might change in 200 million years. It might be more practical to scoot mars out of the system and find a way to elevate Earths orbit than to terraform Mars. This leaves not a colony but a science station or science base. So you are not sending Elon Musk and a flying tomb, but high-end engineers and scientist. So lets compare Mars to South Pole Mars South Pole Vacuous 0.8 Earth Atm Oxygen percentage <1% Oxygen Percentage 29% Scarce Equitorial Water Water abundant with a heat source No Living Organic Resources Resources 2000 miles away Dim light for Horticulture No light six months a year Very Dusty Lots of blowing snow. Requires EVA suits Requires Arctic clothing Abundant CO2 for plants Nominal CO2 for plants growth On 'Terra' firma Dig 2 miles So CO2 is better on Mars for growing plants, almost certainly a greenhouse would be subterranian on Mars (either a tunnel or a greenhouse covered in radiation blocking substrate) so they are equal, both require LED lit greenhouse. Though on Earth it does not need to be pressurized. The drawback of South pole is that it is on an amorphous solid subject to change. Ask yourself the basic question. . . . who would conceive and raise offspring on Mars or on a Venusian cloud floater? Before you ask that question think about the practicality of raising offspring at the south pole. Hey Johnny go outside and play, . . . not. Diaper duty takes on a whole new meaning in a sealed container. Romantic, say go live with Inuvet citizens through a cold canadian winter. Then you are talking about colonization potentials. I should also point out that Mars has a moon that about to disintegrate and bombard the surface, survival of inhabitants would be unlikely. So as a backup to earth its not the place. I cannnot Nix a manned science mission. Right now its a matter of practicality in terms of trip length and health issues. Mars colonization/exploration/science base is more or less a suicide mission.

Moho is a special case because of the dV requirement to bend and because its orbital period about the star is fast relative to kerbin. Two optimizing parameters to look at. 1. Where are the ascending and descending nodes? like you say +/- bearing burns can reduce the amount of DV need out of kerbins orbit. 2. That the theta of kerbin should intersect as close as possible to Moho's apoapsis in it orbit. This way when you transit you are going to be as close as possible to its pE when you intersect. Even though you will have more energy trying to intercept a moho Apo, your differential velocity relative to Moho will be less. This will almost never correspond to a window that allows a transfer to Moho. Fortunately Moho provides a solution. One the planes match look for the 'bugs' that are one above the ships orbit and directly below on moho's orbit. This indicates a retrorgrade dV at Moho's pE while send those bugs counter clockwise to intercept. If correction in inclination have to be made during the Hohmann transition remember that the amount of dV = Orbital velocity * sin ('change). If the degree change is 7 and you do the plane change at 18,000 m/s that means 2193 dV need to be added. When the craft exits kerbin it is traveling at 6000 or so m/s the amount of dV required is one third.

Delta-v breaks into two groups. First to delta-v to create a hohmann transition and second the delta-v to circularize the orbit. First let me say to things about the question. The simple assumption is that you are moving from one nearly circular orbit to another. I almost never do this and I generally find ways not to do this. There is a simple reason for this. Lets take the Kerbin Duna example. Once a duna launch window is optimized we are going to wait until a certain kerbin 'Angle to prograde' (as in MechJeb) and theoretically you want to burn a certain amount of DV from as low a kerbin orbit as possible. I strongly recommend not doing this below 70k because if your acceleration is high enough and the target is DeltaE high enough, you are going to have a heat problem. 70.00001 km suffices to avoid that. That ideal transfer from kerbin is never going to happen, there are acceleration limits and the exact shape of the transition is hard to estimate. So typically for example from Kerbin you might be on angle to prograde of 150' and start at 2 minute burn at 0.3 TWR. As your burning your spacecraft is lifting and you get less of the oberth like gravity well effect. So different rockets will produce slightly different dV. The idea here is this. Keeping kerbin in mind. From any given circular orbit you need to add x V to leave kerbin (meaning you escape but linger close to kerbin). removing m, E/m = 1/2v^2. Lets say at 70k the orbital velocity is 2300. So here is a little trick for any given circular orbit, the velocity required to escape is SQRT(2)* circular velocity. 3252 therefore gets you out of the Kerbin system. At circular orbit your KE is 2.65E6 J/kg (or 2.65 MJ/kg) and you need a KE of 5.29 MJ/Kg to leave the system from that orbit. SO lets say at Kerbin orbit (Kerbol) radius you need 1000 dV to reach duna radius. That translates into 500,000 j/kg of KE. So lets add .5 MJ/kg to escape energy of 5.29 MJ and that equals 5.79 MJ. That tidbit tells us how much energy we need a kerbin 70K orbit. SQRT(5.79 MJ/kg * 2) = 3402 m/sec. You started with 2300 and you need 1102 or slightly more than your orbit breaking speed. Had you created an orbit 80M meters (essentially requiring 950 dv for the transition and a few dV to make orbit. YOu would have need to add a trace less than 1000 dV at the angle to prograde of 90' to reach Duna. How come it is cheaper to burn from low kerbin than high kerbin. How does a high kerbin orbit loose energy. High kerbin orbit did not loose energy, what you do in a low kerbin orbital burn is you increase Apoapsis without increasing Periapsis. In a high orbital burn you have put (thermodynamic) energy into the periapsis altitude and increased apoapsis. The way I like to look at this once you are in a circular orbit your centipedal force and gravitational forces (two very interesting faux forces) cancel. The time you spend fighting gravity to reach a higher orbit. If you throw all the energy you need in at once, you spend less time fighting kerbins gravity and when you leave the system you get that energy back to use to fight Kerbols gravity. If you raise your orbit in bits and then transfer you have spent unnecessary bits of time fighting gravity which you could have simply escaped from quickly. This is the weakness of ION drive, although have very high ISP, because of the very low acceleration of true Ion drives, they waste almost half there ISP in interplanetary transfers fighting their way out of the gravitational well before they transfer. Transfers are also a problem with NTR because the low thrust to weight of these rockets means they spend a considerable amount of energy raising their departure planets gravity well. Where both excel is moving between bodies in which time spent near pE is long. For example an Ion drive burning close to the sun might generate good system leaving velocities (getting to closer to the sun and not poofing is the problem, lol) OK so now how do we find out the cost of reaching Duna. The amount of energy in an orbit is a function of mu of the central body, semimajor axis of the orbit, and the keplarian laws of motion. Its not easy to calculate except at periapsis and apoapsis. We assume that kerbin is the periapsis and Duna is apoapsis. The Major axis is 13.3Gm + (19 to 21)Gm + diameter of kerbol + radii of kerbin and duna. Resulting semi-major is about 17Gm. From there you can calculate based upon keplarian laws of planetary motion what your velocity should be at 13.3Gm and subtract kerbins orbital velocity. For planets with relatively close orbits that budget is not so high. So thats that bit. Unless you have started your transfer on an ascending or descending node and calculated that into your kerbin burn. you need to make roughly 80 dV of plane change burn. The next bit is harder and a place were many space-flight end in ruin. In the case of an a prograde burn from low orbit you know where you are and you simply need to burn until you intercept the next orbit, your burn is not idea, but your going to lose a few dV as your vertical velocity increases during the burn. On entry you need to do several things. 1. Determine the closest point you can retrograde burn. This may differ based on mission (atmospheric reentry versus complex landing and docking) you need to determine the orbital velocity of that altitude. (r = altitude + surface radius). This is omega^2 * r = mu/r^2 also written as v^2/r = mu/r^2. We can rewrite the second as v = SQRT(mu/r). For Duna at 50 km altitude (r = 370,000) this translates into 920 m/sec orbital speed. You put into the system 1000 dV to get from kerbin to duna, more or less you have to take it out, its not as important as you think because on Duna entry 0.5 MJ but 2 x .451 MJ added from duna escape to duna 50km alt of which we keep 0.451 in the orbit. As a result we get to our burn point with 1674 velocity but we need to burn to 920 and this leaves 750 dV approximately . (Mu = Mass of central body times gravitational constant). So the problem is where to burn. This is were good engines com in handy. If you are reversing 750 m/s and your accel max is 2g (20 a) then you need 37.5 seconds or lead the periapsis of 52 km with about 18 sec burn prior. Its not going to be perfect because 18 second prior you will be above 50 km. Most efficient deep space engines do not have TWR with fuel load of 2g, so this where sacrifices are made. You burn for to short and you will have to correct apoapsis and reburn at periapsis. you burn too early and your Pe ends up in the atmosphere. So, My math is not perfect, you should get actual mu from Kerbal wiki, dV spent on plane change will vary. Also on kerbin or duna entry you may have to deal with interfering bodies (like that pest Ike). You need about 2500 dV to reach Duna, your are not going to nail it (at least if you use stock game) and you should have a flexible system that allows re-purposing fuel, such as for landing or return. Unless you are using a flight computer, both the velocity and altitude of Planets changes. Its not simply a matter of knowing were a planet will be, but also how fast it is going when you get there. A word of advice, worry less about dV requires a guess is often good. Worry more about getting good dV. I have found weaknesses in the game in terms of structural parts, I have added my own parts like piping and pipe connectors that allow polyfunctional craft with better efficiency. If a sz0 part will do the job on a sz1 lander, use the size zero. I dont use mono on landers, a sz0 reaction wheel suffices. Less wheel, less battery, fewer panels. Duna is about drag, for instance, dV to land is almost zero if you design the shape of the craft with drag on bottom and aero on top.

If the neutron star that forms is heavier then it simply forms a black hole, a small one that would generally go unnoted. Note that two merging neutron stars of 10km diameter will only produce a result of 12.6km in diameter, two small for even the most sensitive telescopes to see. You can see the effects of neutron stars because of the intense magnetic fields that spin around them, these can be noisy or quiet. Thus a nuetron star can disappear and appear to be a black hole. Black holes are may not be visible but when they absorb matter they release UV and Xrays and cause a glow of excited plasma at their 'poles'. Once you have a merger, there is a certain inevitability that it will form a black hole. The problem is that such a small black hole would begin to disintegrate unless more material is added over time due to hawkings radiation, which converts a higher percentage of mass the smaller the black hole is.

Gravitational waves are created before the two neutron stars collide. As they approach the two objects circle each other with higher and higher omega. This increases the tone of the energy pulse carried on space-time. Since space-time is massless the pulse travels at the speed of light, just like all massless waveforms. The tone increases and the two neutron stars collide. Neutron stars are the remnant of supernova, such stars create clouds of gas that surround them at distances too far from the star to coalesce at the systems center, however as two neutron stars pass close they draft debris from the outer regions other Neutron stars system. Neutron stars also have very intense magnetic fields. Neutron stars are typically 10 km in diameter and as they approach they are multiples of this distance apart, the gravitational waves in space-time we are measuring are from a distance of 100,000,000 of light years. a light year is 10E13 km. IOW the intensity of the perceived acceleration in 10E40 times stronger between the two worlds relative to what we can detect. The closer they get the faster things close by get pulsed. Now imagine your little KSP craft entering the atmosphere from say 1700 m/s versus 4000 m/s, that brighter glow is cause be collisions of particles being acceleration E = hv, higher energy collisions result in brighter light. Electrons are being stripped from their atoms from increasingly energetic shells and when the recombine they create more energetic light. Prior to us detecting the gravitational waves the approaching stars are causing low energy radiation from less energetic collisions. As they get closer the orbiting stars cause some of this radiation to be blue shifted (toward us) and red shift(away from us). As they get really close the shift increases more. Consequently the higher energy pre-GRB hv observed is expected prior to the formal collision. Conceptionally however I should make the point that when two stars cross each others systems boundaries, a collision of sorts has already happened, there is already material colliding and producing radiation. If you imagine relative interstellar speeds in the 10k m/s to 100k m/s range you are looking at a fairly bright glow of relatively low intensity. Such collisions will produce plasma and can be subject to the accelerations of the magnetic field. The problem is that unless your are already looking at that system, you probably will never set your instruments in a mode to detect the radiation emitted. Its only because we can see the pulsing gravitational waves that we see things are going to happen. If you then take a close look, of course you will see something. GRB are a known commodity they were detecting using nuclear bomb surveylance satellites. However, we don't often see hv emitted before GRB until more or less recently. More sensitive gravitatonal wave detection will give better spectrographic information of the entire process of a merger, not just the last few seconds. I should also correct one media fallacy. It is generally accepted that light is not the only massless field that travels at C. In this regard the discovery is non-exceptional. What is more valuable is the finding that the energies and material present are suitable for heavier than iron element formation. The increasing frequency of light and finally x-rays and gamma suggests that these mergers not only have sufficient energy but a means of blowing material away from the merging stars. This is important because many neutron star mergers will create black holes from which no matter escapes. When two black holes merge only the lightest elements in the surrounding gases escape due to x-ray pressure.