NASA Selects Contractor for First Prometheus Mission to Jupiter

yurkin · Sep 24, 2004

but concentrator mass vs. JIMO power sys. mass is the crux of the issue IMO. Sounds fair. Lets look at a nuclear thermal generator versus us a solar thermal generator. Both are running at Jupiter’s orbit, at the same energy levels, on the same cycle and same propulsion. This means that the ships are basically the same except for the boiler. Now on a nuclear generator determining the weight of the power source for the boiler is fairly easy to calculate. According to JPL’s website it could be done for as little as 100kg plutonium. That’s gives it an amazing 1 kW/kg of energy per mass. This is half your estimates for a solar concentrator, which was 2 kW/kg. But you still have to figure in the weight of the internal refractor, fuel cells and gyros and servos to keep it aligned. And most of all you have to figure in the weight of the machine that unfurls the supports and reflector. I really think that no matter what way you try to figure it your going to have a hard time beating the 1 kW/kg potential energy of plutonium.

SteveMick · Sep 24, 2004

Yurkin Your reply leaves me somewhat perplexed as you seem to be "on a different page". 1 AU and if the 1KW/kg concentrator cells are used for electrical power generation, the specific power is 1.1kw/kg at 1 AU. Is this just the weight per pound of the solar cells and concentrator? Nothing else? You are correct, but actually there isn't much more to it mass wise except primarily the power conditioning equip. and radiator(which doubles as concen. support. That's why I previously guessed 2kg/KW at 1 AU. "JIMO by contrast has a projected mass of 50,000lb/100KW or a specific power of 500lb. " Okay let's use half the mass and double JIMO's specific power to say 100kg/KW. At 1AU, solar has 50 times the specific power, and twice at Jupiter. That is an unfair comparison though because much of the propellent JIMO uses is to get its own corpulent self out of orbit. The STR (the preferable mode for Earth escape) does operate at 800 sec. or so compared with the JIMO 3000sec.(it could be higher but would take even longer to escape) and this does mean more propellent per unit mass of spacecraft; however because the thrusts the STR would use occur exclusively at perigee when traveling the fastest, the effective efficiency is about doubled or more as compared to a spiraling trajectory like SMART-1 or JIMO. "efficiency of 33% on the overall sys., 833KW of energy goes into thrust. No it doesn’t mean that. It would mean that 833kW goes from the generator to the engine. You can’t just magically transfer electricity to thrust. They are two different units." I was speaking here of STR mode. There is no generator only a heat exchanger/rocket engine easily scaled up to use the tremendous thermal energy available. In electric mode both solar and JIMO would have the same less than 100% efficient engines. "Ahuh… I’m not sure you know how to calculate thrust. It’s the mass flow rate of the exhaust times its velocity. I don’t see anyt

mrmorris · Sep 24, 2004

"Since a hydrolizer is basically a fuel operating in reverse somewhere around 8-10 pounds per KW." You're assuming that the hydrolizer is exactly the same weight as the fuel cell. Not a bad assumption, so we can go with it, but keep in mind that it *is* an assumption. Ergo -- 26 pounds for 3 kW or ~8.67 pounds/kW (that .67 becomes fairly significant when multiplied by 200). However -- the weight does not include a few things. You have to have a tank with enough hydrogen to provide 'x' hours of power and a second to store the oxygen needed for the fuel cell reaction and a third to store the water resulting from the fuel cell operation (to later be converted back to oxygen and hydrogen). The previous assumption was 15 hours. The article doesn't give any information on usage rates that I see, so I can't even make a guess as to the sizes and masses of the various storage tanks required. Three -- there *may* be 0% losses on the system, but if not -- there must be excess amounts of water or oxygen/hydrogen to compensate. My post contained verifiable figures based on known energy densities of lithium ion batteries. Your post supplies guesses -- followed by statement of opinion put forth in the guise of a statement of fact. I don't argue that fuel cells may well be the better option. Indeed -- if lithium ion batteries were a better/lighter solution, then it's likely the shuttle would be using them rather than fuel cells. However -- you cannot make a numeric statement like "...would still be well less than a third what a nuclear reactor would require" when you've only supplied a tiny fraction of the variables involved in determining that number.

mrmorris · Sep 24, 2004

"The first and most obvious is to minimize the time in shadow at Jupiter by carefully choosing orbits." Coupla things here, stevemick. While that might be obvious to you -- it's obviously a crock of fertilizer to anyone who gives it more than a few seconds of thought. First of all -- Jupiter is a really big planet. I mean really really big... and it casts a big shadow. If you're orbiting the planet -- there's not much of a way to *not* spend a good bit of time in shadow without doing a polar orbit. Because Jupiter's axial tilt is only ~3 degrees, a probe sent from Earth is going to have to make a massive vector change in order to achieve one. Not an option. However -- let's forget that crud -- because the probe isn't orbiting Jupiter as such. The name of the game is JIMO (Jupiter Icy Moons Orbiter). Since the probe is going to be orbiting the moons, rather than Jupiter itself -- unless you can figure out a way to change the orbits of Callisto, Ganymede and Europa, then JIMO's path around Jupiter once it's orbiting a given moon is pretty well set in stone. Since the orbital inclination of all of these is less than 1 degree -- that pretty well puts them into Jupiter's shadow from time to time. "The 100KW figure is IMO an arbitrary number and not based on supplying particular power needs expressed by scientists prior to that figure being mentioned In other words, scientists didn.t say "I need 100KW, but rather they were told they would have 100KW." You keep making two different arguments here, sm. First -- you say that there's no need for nuclear, because solar could supply the same amount of power as a nuclear reactor and with less mass, cost etc. Then you keep flip-flopping and saying that there's no reason to supply that much power and a solar powered mission could do the same amounts of science with much less. Pick one o

yurkin · Sep 24, 2004

Your reply leaves me somewhat perplexed as you seem to be "on a different page". That’s true I think you need some basics. For one thing STR solar thermal reactor, is different then STP solar thermal propulsion. A solar thermal reactor generates work that could be used to power a generator. The energy from the generator could then be used to power an ion engine or an electric coffee maker. In solar thermal propulsion the energy from the sun is used to heat fuel. The fuel is then vented through the exhaust valve and this produces thrust. So please keep the two straight. JIMO by contrast has a projected mass of 50,000lb/100KW or a specific power of 500 lb/kW Okay let's use half the mass and double JIMO's specific power to say 100kg/KW. At 1AU, solar has 50 times the specific power, and twice at Jupiter. Please use some common sense! The whole craft isn’t going to be power supply. There’s the science package, main bus, engines and fuel. Half the weight is going to fuel. Your comparing the whole weight of the Jimo against the weight of the solar condensor array. and this does mean more propellent per unit mass of spacecraft; however because the thrusts the STR would use occur exclusively at perigee when traveling the fastest, the effective efficiency is about doubled or more as compared to a spiraling trajectory like SMART-1 or JIMO. That’s wrong! Smart is only thrusting around the perigee. If Jimo wanted to do an energy efficient escape it would do the same thing. Even if Jimo was thrusting the whole time it still could get better efficiency then a thermal engine. "efficiency of 33% on the overall sys., 833KW of energy goes into thrust. No it doesn’t mean that. It would mean that 833kW goes from the generator to the engine. Y

pathfinder_01 · Sep 24, 2004

Stevemik, I am no expert but it looks like you are trying to compare two different types of propulsion that have very different characteristics. Thermal Propulsion (i.e. nuclear thermal, solar thermal) and Electrical Propulsion (i.e. plasma, ion ECT.) Thermal Propulsion systems theoretically have greater acceleration than ion propulsion systems but less fuel efficiency (i.e. thrust per pound of expended propellant) and a slower maximum speed. It is sorta like the turtle vs. the hare. Given enough time the ion engine would reach a higher velocity than the thermal one and the ion engine would need to carry less mass in terms of fuel to reach any velocity. There are applications where thermal rockets make more sense and there are applications where ion engines make more sense. For a trip to the moon a solar thermal rocket would be a great way to carry cargo quicker than an ion engine because of the short distance. A solar thermal engine would be likely be cheaper than a chemical rocket but possibly more expensive than an ion one due to the weight in fuel. The JIMO will need to get to Jupiter the brake into orbit around Jupiter and brake into (and break out of) orbit around each of the moons to be studied. The Fuel efficiency of Electric propulsion makes it an attractive technology for this mission. The higher top speed also might make it more attractive. The mission could go directly into orbit to Jupiter instead of having to use mulitple gravitational assists like Galileo. Also a rocket has to go much faster than the escape velocity of earth to reach jupiter. It also looks like you are comparing two different types of power generation. Solar power and Nuclear power. Solar power is not very useful as you get farther away from the sun. There is less and less light available for any sort of operation using light. A solar thermal rocket would require massive solar collectors to brake into Jupiter’s Orbit.

scottb50 · Sep 25, 2004

The system I referenced weighed 13 pounds and produced 3 KW or 4.3 pounds per kilowatt figuring a hydrolizer at the same weight comes to 8.6 pounds per kilowatt of power, or roughly 2,000 pounds for 200KW. The existing ISS solar panels weigh 2400 pounds apiece and a pair produces about 65KW. Because it takes more energy to hydrolize water than it produces and we would want to have a reserve supply, as well as being able to use Hydrogen and Oxygen as chemical propellants we would probably need at least 300KW of generating power or the equivelent of five pairs of ISS solar arrays, again using existing hardware and discounting advancements in solar cell efficiency that have occured in the recent past. So now we are looking at 24,000 pounds for the solar panels and 2,000 pounds for the hydrolizers and fuel cells, or roughly 1/2 what the proposed reactor would weigh. Like you say this doesn't include plumbing and basic structure as well as the water needed to run the system, but, as I have said there are numerous uses for water and Hydrogen and Oxygen, so they would be required with whatever power system you use. <div class="Discussion_UserSignature"> </div>

scottb50 · Sep 25, 2004

You misunderstand. What I propose is using solar power to run hydroliizers that break water into Hydrogen and Oxygen. The Hydrogen and Oxygen are fed into fuel cells that produce electricity and exhaust water that is collected and recycled indefinitely, as long as it is a sealed system. This would work in Space because solar power is available at all times, outside of orbit, and by producing more Hydrogen and Oxygen than needed while in orbit enough would be available when the solar cells are not producing electricity to keep the fuel cells working. Very true it takes more energy to hydrolize water than you can get from reacting them in a solar cell or by combusting them in an engine, but solar power is unlimited, so as long as you have a cushion of supply for times you have no solar power while in orbit, the fuel cells could operate continuously. As for being an infinite energy machine the mechanics of a system would always cause problems, you would have to provide water to the hydrolizers, Hydrogen and Oxygen to the fuel cells, so there would be too may variables to say it would be infinite. <div class="Discussion_UserSignature"> </div>

pathfinder_01 · Sep 25, 2004

Hey Scott, Fuel cells would probably be worse than batteries for a mission like this. The reason why the Shuttle uses fuel cells is because they produce more power in less space with fewer restrictions than solar panels. The space station has to keep it’s enormous panels pointed at the sun, the shuttle does not need to worry about which way it pointed in orbit when it comes to electrical power nor does it need to carry large, heavy solar panels to generate power. Fuel cells also produce more power than batteries alone. For the Shuttle it is a great way to make power since the shuttle is not intended to stay in space for months or years. It will only be in space for two or three weeks at most. You get both electricity and water, both items needed for life support. In fact the shuttle produces more water than the crew can consume and winds up dumping the excess water. To give an idea of the power of the fuel cell vs., a solar panel, the space shuttle generates more power than the Russian MIR space station. For a spacecraft intended to spend months or years in orbit fuel cells make less sense. They need both hydrogen and oxygen to produce power. Once the hydrogen and oxygen runs out then no more power. i.e. The MIR space station would have power indefinitely while the shuttle would eventually run out of fuel for the fuel cells. The cost of sending supplies to a spacecraft is prohibitive so spacecraft intended to spend long periods of time in space use solar power or rarely nuclear power for electricity. If you use solar power then the craft needs some way to store power for times when either the solar panels are oriented away from the sun or when the craft is in the shadows. Usually spacecraft use batteries for this. You could use solar panels to hydrolyze the water back into it’s elements, but then you would need separate tanks for the hydrogen, oxygen, and water not counting all the plumbing the system wou

najab · Sep 25, 2004

>The system I referenced weighed 13 pounds and produced 3 KW or 4.3 pounds per kilowatt figuring a hydrolizer at the same weight comes to 8.6 pounds per kilowatt of power, or roughly 2,000 pounds for 200KW.Good, some numbers, I was waiting for some numbers. The US has ground tested the SAFE-400 reactor which provides 400kW thermal (100kW electrical) and has a mass of - 512 kg (1150lbs). Two of these (for redundancy) are less massive than one ISS solar panels.QED.

scottb50 · Sep 25, 2004

From what I have been able to find out the SAFE-400 reactor would work with a Sterling pump to produce the 100KW of electrical power, so again we are looking at a second unit and plumbing and such, so the weight would be increased somewhat. I would also question the redundacy of just two systems, I know they have pretty much allowed twin engine airliners to fly any overwater route in the last few years, but I would feel a lot better with at least four engines and redundancy. Once in place solar panels are fairly redundant in their own right, a few individual cells fail, or are damaged, the panel will produce power at a degraded rate but still work. Because of the low weight of solar cells failed ones could easily be replaced and carried as spares. I also would expect the weight relative to the ISS system could be cut considerably, using 21st century design rather that mid 20th century technology. I also pointed out a solar based system could be used to provide chemical fuel for transfer purposes that nuclear could not. If nuclear power is used for a Mars transfer vehicle, as an example, you would still need chemical engines to reduce speed enough for an aerocapture, get to and from the surface and break orbit and return to LEO. <div class="Discussion_UserSignature"> </div>

scottb50 · Sep 25, 2004

The reason the Shuttle uses fuel cells is because they can produce power for an extended period of time. Batteries lose power and have to be recharged which is done by a generator in your car or by solar panels at the ISS. The biggest problem is solar panels only provide power for roughly half an orbit, so batteries are needed to store enough energy for the times solar panels are not generating. Another consideration is the stability of the power, a lot of electrical devices, computes for example need a steady supply at a steady voltage, something a solar panel can't provide. <<For a spacecraft intended to spend months or years in orbit fuel cells make less sense.They need both hydrogen and oxygen to produce power. Once the hydrogen and oxygen runs out then no more power. i.e. The MIR space station would have power indefinitely while the shuttle would eventually run out of fuel for the fuel cells. />> My point is you would not run out of Hydrogen and Oxygen, you would recycle the water, rather than dumping the excess and hydrolize it back into Hydrogen and Oxygen. The MIR would have power indefinitly as long as the batteries stayed operable true, all I am saying is replace the batteries with water. <<You could use solar panels to hydrolyze the water back into it’s elements, but then you would need separate tanks for the hydrogen, oxygen, and water not counting all the plumbing the system would need and the effect of shifting that much mass around the spacecraft would have on it’s center of gravity. It would be better to simply go with batteries. Batteries don’t need hydrolizers, tanks or plumbing to be recharged and they are cheaper and a lot simpler than fuel cells. />> Yes you would need storage for Hydrogen and Oxygen as well as water, the amount would depend on how you are using it. If you are talking about a small orbital facility the amounts would be rather modest and if you are talking long ra <div class="Discussion_UserSignature"> </div>

pathfinder_01 · Sep 25, 2004

Scott, I still don’t see the advantage. Fuel cells, like batteries degrade over time. They become less and less efficient. Also the amount of times a battery can be charged depends on the type of battery. The batteries onboard the ISS are built to be recharged 38,000 times before failing. That is how many charges they are excepted to happen over six and a half years. The ones onboard the MER are a different type selected for different reasons. Also nuclear power in the form of RTG could provide decades worth of power with no need of batteries. Also why carry so much water? Other than acting as a radiation shield I can’t see the use for it. Not all rocket engines require hydrogen and oxygen nor are engines that use hydrogen and oxygen the best engines for all applications. If you needed hydrogen and oxygen for the engines it might just be simpler to send them instead of water. The hydrogen and oxygen would be available for use immediately and the spacecraft wouldn’t need to carry around all the equipment needed to convert the water to hydrogen and oxygen. If the water is needed for the crew you could simply recycle oxygen and water in a closed or semi-closed life support system like the way it is done on the ISS. If water is needed for radiation shielding then you could carry just enough water for radiation shielding (or better yet use something else) and use faster, more fuel efficient types of propulsion such as thermal or electrical instead of chemical propulsion both types of propulsion may or may not need water, hydrogen, or oxygen as propelent and there maybe a great advantage to using something else.

SteveMick · Sep 27, 2004

Yurkin Thanks for the response. You write: "Your reply leaves me somewhat perplexed as you seem to be "on a different page". That’s true I think you need some basics. For one thing STR solar thermal reactor, is different then STP solar thermal propulsion. A solar thermal reactor generates work that could be used to power a generator. The energy from the generator could then be used to power an ion engine or an electric coffee maker. In solar thermal propulsion the energy from the sun is used to heat fuel. The fuel is then vented through the exhaust valve and this produces thrust. So please keep the two straight." I have never heard of a solar thermal reactor as such and am of course quite well aquainted with the difference between thermal propulsion and electric power generation. It is not obvious, so I assume that what you are calling a "reactor" is a boiler. With PV why would you want that? Because the PV has to remain at 25 deg. C or so it is impractical to use them as a "topper" and then use their waste heat to run some kind of system as some other alternatives allow. Another possibility is solar MHD which would use the same heat exchanger as the solar thermal rocket. JIMO by contrast has a projected mass of 50,000lb/100KW or a specific power of 500 lb/kW Okay let's use half the mass and double JIMO's specific power to say 100kg/KW. At 1AU, solar has 50 times the specific power, and twice at Jupiter. Please use some common sense! The whole craft isn’t going to be power supply. There’s the science package, main bus, engines and fuel. Half the weight is going to fuel. Your comparing the whole weight of the Jimo against the weight of the solar condensor array." I assume you mean concentrator array. I took the figure of 25,000lb. that you gave for the nuclear electric system in another reply. "and this does mean more propellent per unit mass of spacecraft; however because the thrusts the STR would use

najab · Sep 27, 2004

>Apparently you see an overarching need to operate the science package at full or partly full power in these periods and if so the mass of power storage would of course go up.See, right there, that is why we need space nuclear power. Current day thinking is "Well, if we want to operate the instruments during eclipse then we'd need more batteries. We can't afford the extra mass, so we'll design the science program around the available power."With nuclear power, science isn't driven by the engineering - by developing the next step in the engineering, the science is freed to go where the science wants to go.

SteveMick · Sep 27, 2004

mmorris writes: ""The first and most obvious is to minimize the time in shadow at Jupiter by carefully choosing orbits." Coupla things here, stevemick. While that might be obvious to you -- it's obviously a crock of fertilizer to anyone who gives it more than a few seconds of thought. First of all -- Jupiter is a really big planet. I mean really really big... and it casts a big shadow. If you're orbiting the planet -- there's not much of a way to *not* spend a good bit of time in shadow without doing a polar orbit. Because Jupiter's axial tilt is only ~3 degrees, a probe sent from Earth is going to have to make a massive vector change in order to achieve one. Not an option" Oh really? The Ulysses probe did a gravity assist at Jupiter that called for a polar oriented swingby. Use of a trajectory that approaches the Sun before heading for Jupiter not only saves a lot of propellent but makes plane changes relative to the plane the most planets roughly orbit in thereby allowing an initial orbit at Jupiter that is highly inclined if desired. The tone of your post seemed somewhat condescending and offensive. If I'm reading that correctly I believe I'm owed an apology. I agree that the moons are where they are and there will be times in shadow. The real question is whether this impact can be mitigated and I believe it can be worked around. After all, the proponents of JIMO are content to wait all those years for it to be built and all those years to get to Jupiter, so if the radar observations of the moons by a solar probe are stretched out due to shadow issues, the delay is relatively trivial. ""The 100KW figure is IMO an arbitrary number and not based on supplying particular power needs expressed by scientists prior to that figure being mentioned In other words, scientists didn.t say "I need 100KW, but rather they were told they would have 100KW." You keep making two different arguments here, sm. First -- you say that there's n

bobvanx · Sep 27, 2004

>If science is the goal, solar can do the job... However, if the argument is over whether solar can provide the same power at Jupiter What if the argument is about using what we've got to get what we want? Why can't we use the brains and muscle behind nuclear to set us up a space-based infrastructure? We've got an industry that is so totally committed to developing its product that it has been working against tremendous odds and opposition, and now they get a break, and it feels like you want to take it away. Why?

SteveMick · Sep 28, 2004

Bobvanx: You portray the huge and highly subsidized nuclear industry as if they were a small dedicated band trying against the odds to create a dream. My feeling is that due to various factors, nuclear has not lived up to its promise even though on paper it should or at least could and that giving this huge and dare I say it -failed- industry 400 million to make more paper with the far-off possibility that hardware might actually result while other great ideas are starved is just plain wrong. To be fair, I have asked for solar to be considered in a positive light and I am therefore obligated to consider nuclear in terms of how it COULD be made into something more beneficial. The JIMO project has been a ridiculously easy target since I am still amazed that such incredibly poor specific power numbers have even been written by competent engineers. If a solar alternative was proposed that acheived 250 lb. per KW electric, it would never get past the falling on the floor laughing stage. It is difficult to imagine how a poorer example of nuclear's potential could be dreamed up unless by someone anti-nuke. If a flying mach 3 ramjet could be nearly developed in the sixties which had to have good specific power or it couldn't fly, why would we even consider the JIMO embarassment? To have a reactor making high quality heat and not use that heat for propulsion is baffling to me, but at any rate redesigning it to do so would actually be an advancement. A smaller hotter reactor could weigh a lot less wouldn't it? Admittedly nuclear is not my area, but this dog just won't hunt without a trip to the Vet IMO. Steve

mrmorris · Sep 28, 2004

Oh really? The Ulysses probe did a gravity assist at Jupiter that called for a polar oriented swingby." Ulysses used Jupiter's gravity to swing out of the eliptic plane into a polar orbit around the sun, not a polar orbit around Jupiter. The two aren't interchangeable and have nothing in common. "Use of a trajectory that approaches the Sun before heading for Jupiter not only saves a lot of propellent but makes plane changes relative to the plane the most planets roughly orbit in thereby allowing an initial orbit at Jupiter that is highly inclined if desired. " No. An excursion through the inner solar system has no bearing on the inclination of the final orbit. You have no concept of orbital mechanics. No matter how the velocity to move a probe from earth is generated, be it a direct path using a large chemical booster, or a multi-planet-swingby through Earth/Venus/Mars or an ion drive, the *directional* component of the velocity is going to be approximately 99.986% aligned to the plane of the eliptic of the sun (calculated using the diameter of Jupiter against the *minimum* flight path from Earth- />Jupiter). A true 'polar' orbit of Jupiter would be at almost a 90 degree angle to that vector. There is no way to magically convert the one vector to the other. In any event -- it doesn't matter, as the probe is supposed to orbit the moons, and their path about Jupiter is out of anyone's control. As I indicated in the previous post -- the idea of 'carefully choosing orbits' to minimize the amount of time spent in shadow is ludicrous. The probe will often be in the shade of Jupiter for multiple hours at a time and additionally it will be in the shadow of the moons themselves a good bit of the time. "The tone of your post seemed somewhat condescending and offensive. If I'm reading that correctly I believ

yurkin · Sep 29, 2004

This is simply incorrect. You forget that the inverse square law works both ways. The intensity of sunlight is the same and therefore the temperature at Jupiter and Earth. The intensity of sunlight at Earth and Jupiter are the same? And the temperature is the same as well? I don’t know what to say, I’m not sure I can compete against that level of WOOWOOness.

bobvanx · Sep 29, 2004

I’m not sure I can compete against that level of WOOWOOness. Ayup. It's almost time to get this thread moved over to Free Space.

SteveMick · Sep 30, 2004

Yurkin "This is simply incorrect. You forget that the inverse square law works both ways. The intensity of sunlight is the same and therefore the temperature at Jupiter and Earth. The intensity of sunlight at Earth and Jupiter are the same? And the temperature is the same as well? I don’t know what to say, I’m not sure I can compete against that level of WOOWOOness. Since of the several points I made you have responded only in this one area, am I correct in assuming that you concede the many other points you made which I refuted? If so a general criticism such as woowooness is inappropriate or perhaps you are referring to your own incorrect and even uninformed opinions? You are claiming that a parabolic concentrator mirror will not be able to focus to the same or nearly the same intensity and temperature as distance from the Sun increases. This is simply not true! What I meant was that the smaller apparent size of the Sun at Jupiter is such that the intensity at the focus remains the same and the way I said that was that the inverse square law works both ways. Physics not opinion says you are wrong. Every point that has been raised has been refuted with the sole exception of mmorris's assertion that high power observations must be made in shadow. Since no one knows exactly what observations are to be made since the experiments/instruments haven't been picked for JIMO, this may or may not be a problem. If it is radar shadow or not should make no difference. His objections about entering a polar orbit at Jupiter are not credible as stated. Nearest the Sun (perihelion), or any body you're in orbit around, it takes much less energy to change the plane of your orbit resulting in an orbit that is inclined to the plane of the ecliptic in this case, and which can result in an inclined orbit at Jupiter. This is pretty basic stuff so I suspect that you are being disingenuous and really know better mmorris. The assertion that this

scottb50 · Sep 30, 2004

We don't need more batteries, we need fuel cells and a store of Hydrogen and Oxygen to use during periods solar power is unavailable. Since you would need Hydrogen and Oxygen anyway it would not require that much more mass. A nuclear reactor needs a way to convert the heat output to electricity, it also needs a way to shed heat that isn't needed or usable to produce electricity, look at the cargo bay door radiators on the Shuttle to get some idea of the scale that would be needed. Inevitably the chance a reactor would have to be shut down for service would require batteries for back-up power, and shielding the crew from the reactor itself would take even further mass. <div class="Discussion_UserSignature"> </div>

mrmorris · Sep 30, 2004

"Every point that has been raised has been refuted with the sole exception of mmorris's assertion that high power observations must be made in shadow. Since no one knows exactly what observations are to be made since the experiments/instruments haven't been picked for JIMO, this may or may not be a problem. If it is radar shadow or not should make no difference. " Huh? Radar is one of the highest powered instruments that will likely be onboard. The only reason I can think that you'd make a statement that it wouldn't matter whether it were made in shadow or not is if you're confusing the 'function' of radar with its power requirements. Visible light is not required and radar can indeed function in total darkness -- but only if it has the power to do so. This is one of the primary reasons why it will be of use in imaging the darksides of the three moons in question. All three moons are tidally locked because of their proximity to Jupiter. Since they always present the same face to Jupiter -- their dark side revolves around the surface at the same rate as their orbital period or 3.1, 7.2, and 16.7 days for Europa, Ganymede and Callisto respectively. A probe orbiting these moons is going to have to produce science while periodically being in the shadow of the moons, and Jupiter -- while the 'sunward side' of the moons slowly creeps around. I repeat once again -- a probe that has full power despite the amount of sunlight it's currently receiving makes for a much more productive mission. "His objections about entering a polar orbit at Jupiter are not credible as stated. Nearest the Sun (perihelion), or any body you're in orbit around, it takes much less energy to change the plane of your orbit resulting in an orbit that is inclined to the plane of the ecliptic in this case, and which can result in an inclined orbit at Jupiter. This is pretty basic stuff so I suspect that y

bobvanx · Sep 30, 2004

...a parabolic concentrator mirror will not be able to focus to the same or nearly the same intensity and temperature as distance from the Sun increases. This is simply not true! ... the smaller apparent size of the Sun at Jupiter is such that the intensity at the focus remains the same Sorry, no. Your reflector is a fixed size? Then at Jupiter, it will collect and focus 25 times less solar energy than at Earth. Are you trying to say that we could focus to a smaller point at jupiter, because of the sun's apparent size? This has so little impact that I can't imagine you are serious. and the way I said that was that the inverse square law works both ways. Physics not opinion says you are wrong. Maybe if you drew a diagram to explain how the inverse square law even remotely supports your assertion? Every point that has been raised has been refuted Again, sorry, no. Nearly every point has been ignored. By repeating the same erroneous comments again. That's not a rebuttal nor a refutation. You have to let the new information in, process it, and demonstrate mastery and show how the new info does not apply, or your info is superior or has greater applicability. Repeating nonsense and errors doesn't make them more true, it just makes them more permanent. His objections about entering a polar orbit at Jupiter are not credible as stated. Yeah, fortunately we have a very fine example of a probe that is going to enter a polar orbit eventually, out at Saturn. Cassini has been captured in Saturn's orbital plane, and over the next few years they'll be using Titan gravity assists to swing it into all sorts of orbits, including polar. Unfortunately for your solar/nuclear diatribe, polar orbit around Jupiter doesn't get you into polar orbits around the ice moons. In fact, it makes it e

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