FromEarth Posted July 7, 2016 Posted July 7, 2016 (edited) I had an idea for a way to make orbital transfers without using fuel and I'd like an expert to pitch in: link removed How feasible would it be? How precisely can we predict the motion of a spacecraft over millions of kilometers? edit: since the link was removed, I'll quote it here: I will explain how it’s possible to go from Low Earth Orbit to the surface of the moon and back to LEO without using a drop of fuel, for unlimited times. What is required for that is: A space station orbiting in low earth orbit (we already have that) A space station orbiting the moon A big coilgun on the surface of the moon and on both space stations A very sturdy spacecraft that can survive high accelerations and decelerations by a coilgun The only fuel required is to take the spacecraft from the surface of the earth to LEO and for only one time. This is how I imagine the process: A spacecraft takes off from the surface of the earth using conventional rockets and meets with the space station in LEO. The spacecraft is accelerated by a coilgun mounted on the space station and sent to the moon. There is an equal and opposite reaction to the space station, which is decelerated. Since the space station is much more massive than the spacecraft, the change in velocity is not enough to de-orbit the space station. The acceleration is carefully timed so that the spacecraft will go inside the coilgun of the space station orbiting the moon. Once the spacecraft goes inside the coilgun of the space station orbiting the moon, it is decelerated to the same speed of the space station in moon’s orbit. The deceleration of the spacecraft causes an increase of velocity for the space station, but not enough to send it out of the Sphere Of Influence of the moon. When the timing is right, the spacecraft is carefully decelerated again so that it will go inside the coilgun on the surface of the moon. The spacecraft is decelerated to a standstill by the coilgun on the surface. The energy is transferred to the moon. Since the moon is much more massive than the spacecraft, the effect on the moon is negligible. To go back to earth, the coilgun on the moon accelerates the spacecraft and sends it inside the coilgun mounted to the space station orbiting the moon, which sends it to the coilgun mounted to the space station orbiting the earth. The spacecraft, both space stations and the moon are now back to their initial orbital energies. Taking a space station from low earth orbit to moon orbit could be done without using a drop of fuel by stealing orbital energy from the moon: A spacecraft is sent from LEO to do a gravity assist around the moon (to steal orbital energy from the moon). The gravity assist is timed so that the spacecraft will rendez-vous with the initial space station again after doing the gravity assist. Once the spacecraft comes back, it is slowed down by the coilgun, transferring energy to the space station, and thus increasing the space station’s orbital energy (in equal amount to that which was stolen from the moon). By doing this over and over, the space station can be raised enough to put it into moon’s orbit. I think two gravity assists could be done in one go: one when going from LEO towards the moon, and the second when going from High Earth Orbit towards LEO. If we can steal energy from the moon and give it to other spacecraft, we could theoretically send a space station from LEO to mars without using fuel. Once we have a space station orbiting mars, by the same concept, we can go to mars and back without using fuel or stealing energy from the moon, unlimited times. The stolen energy could be used to decrease fuel needs to put an object in earth’s orbit as the rocket will only need to lift it to the space station’s altitude and the coil gun would accelerate the object to orbital velocity. Since the ISS will be decommissioned in 4 years, it could be a goodcandidate as an “orbital energy storage” in LEO or around the moon. It would be a waste to deorbit such a massive station. Another idea for fuel-less orbital transfers requires two satellites of identical mass, one equipped with a coilgun, both in the same orbit but going in reverse direction.As the satellites come close, they “attract” eachother using electromagnetic force (or coilgun) generated by solar panels, they pass pass eachother very closely and then they “repel” eachother. Doing this, both satellites accelerate by an equal amount and thus they will meet again at the apoapsis (the same for both) where they do the same thing again, circularizing the orbits and effectively converting the sun’s radiation into orbital energy. I don’t know if I’m the first to have these ideas but if I am, I take credit for them. Edited July 7, 2016 by FromEarth
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Sensei Posted July 7, 2016 Posted July 7, 2016 Coilgun requires electricity to create electromagnets and accelerate object inside. So there would be needed fuel to make all that electricity. If there would be used energy from the Sun, that's it 1367 W/m^2 with 100% efficiency of solar panels (in reality it's ~15% widely used, NASA could have slightly better), to accelerate object with mass 10 tons, to velocity 30 km/s, solar panels area 100 m^2, there would be needed 381 days to get all that energy.. With 15% efficiency, it's 2541 days (nearly 7 years), all day long just on loading accumulators. And it's still with assumption that 100% of energy in accumulators is used to get kinetic energy of object, without any lost..
FromEarth Posted July 7, 2016 Author Posted July 7, 2016 (edited) Coilgun requires electricity to create electromagnets and accelerate object inside. So there would be needed fuel to make all that electricity. If there would be used energy from the Sun, that's it 1367 W/m^2 with 100% efficiency of solar panels (in reality it's ~15% widely used, NASA could have slightly better), to accelerate object with mass 10 tons, to velocity 30 km/s, solar panels area 100 m^2, there would be needed 381 days to get all that energy.. With 15% efficiency, it's 2541 days (nearly 7 years), all day long just on loading accumulators. And it's still with assumption that 100% of energy in accumulators is used to get kinetic energy of object, without any lost.. How about 2 tons and only 4 km/s (the delta-v needed from LEO to LLO)? With 30 km/s you could send it directly into the sun ahahahaha! Btw, the ISS has 2500 square meters of solar panels and makes between 84 and 120 kW. So about 60 kW on average because half of the time the sun is not visible. According to my math, it will take only 74 hours to get the energy needed to send a 2 ton spacecraft to the moon: time = (kinetic energy needed) / (solar power) = ( ( m * v^2 ) / 2 ) / ( P ) = ( ( 2000 kg * (4000 m/s)^2 ) / 2 ) / 60000 W = 266'666 seconds time = 266'666 seconds / (3600 seconds/hour) / (24 hours/day) = 3.1 days Edited July 7, 2016 by FromEarth
pavelcherepan Posted July 11, 2016 Posted July 11, 2016 If we can steal energy from the moon and give it to other spacecraft, we could theoretically send a space station from LEO to mars without using fuel. Once we have a space station orbiting mars, by the same concept, we can go to mars and back without using fuel or stealing energy from the moon, unlimited times. Not true. You can only use gravity assist to increase relative velocity between objects orbiting the same body, i.e. you can use Moon gravity assist to raise your velocity in Earth's FoR potentially to the escape velocity, but once you achieve that you're on the orbit around the Sun moving at about 30 km/s and you won't be able to utilise gravity of the Moon any more. So you still need quite a bit of Delta-v to get to Mars intercept.
Enthalpy Posted July 11, 2016 Posted July 11, 2016 I have no objection based on orbital mechanics against the coilguns on space stations. But practical objections, sure. Departure from Leo to lunar transfer takes about 3.2km/s, braking at the Moon 1km/s. With rockets, both use fuel so mission planners add the expenses, but with the coilgun only the acceleration needs energy. However, 10 tons would demand a heavy space station, because already 50m/s changes the orbit a lot. 10t and 3.2km/s make 51GJ, or 70t lithium batteries and heavier capacitors, with efficient coilguns that don't exist. 60kW from solar panels provide this energy in 10 days. And, well, we don't have the proper electric guns. Gravity assist: I have the biggest doubts that a craft accelerated by the Moon could come back to a station on Leo. There are angle conditions on the craft's orbits before and after the flyby, and if the return angle is symmetric I vaguely feel that no speed has been gained. But I'm not easy with gravity assist. Some other scenarios have been proposed, like Halo and Kolomaro transfers, between Earth and Moon or Mars. I don't know how they work, it could be a tidal effect. For sure, such transfers are very slow, taking many orbits of the Moon or planet.
FromEarth Posted July 11, 2016 Author Posted July 11, 2016 (edited) 10t and 3.2km/s make 51GJ, or 70t lithium batteries and heavier capacitors You are right, storing all that energy in batteries is a huge problem. Since batteries can't be discharged fully in less than a second, capacitors are needed. Maybe SMES could solve the problem? https://en.wikipedia.org/wiki/Superconducting_magnetic_energy_storage The energy density is between 4 and 40 kJ/kg. Assuming it's 40 kJ/kg, to store 51 GJ one would need to lift 1'275'000 kg into orbit. If we reduce the requirement to only 2 tons per spacecraft, then it's 10 GJ and 255'000 kg worth of SMES (about half the mass of the ISS). Maybe we could use compressed air for the initial acceleration, increasing the overall energy density in combination with SMES? If the acceleration tube is long enough, we can keep the air from escaping into the vacuum of space. (Compressed air can reach energy densities of 180 kJ/kg according to this page: https://en.wikipedia.org/wiki/Compressed_air_energy_storage#Specific_energy.2C_energy_density_and_efficiency). Gravity assist: I have the biggest doubts that a craft accelerated by the Moon could come back to a station on Leo. There are angle conditions on the craft's orbits before and after the flyby, and if the return angle is symmetric I vaguely feel that no speed has been gained. But I'm not easy with gravity assist. Does anyone know a program that can simulate a gravity assist? I wish I had a program that could allow me to simulate n-body scenarios. Edited July 11, 2016 by FromEarth
FromEarth Posted July 12, 2016 Author Posted July 12, 2016 (edited) I'm playing with Bugale N-Body Simulator and something unexpected happened.... Double lunar gravity assist corn ahahahahaha Edited July 12, 2016 by FromEarth 1
FromEarth Posted July 12, 2016 Author Posted July 12, 2016 (edited) There are angle conditions on the craft's orbits before and after the flyby, and if the return angle is symmetric I vaguely feel that no speed has been gained. After trying for 3 hours, I can confirm that your doubts were 100% right. It's not possible to come back to the starting point after a gravity assist (if any energy is gained). If the apogee is raised, so is the perigee in 100% of the cases. At this point, two stations are needed, one in LEO, the other at an altitude higher than the moon to make the spacecraft decelerate a little and lower the perigee to that of the station in LEO. Edited July 12, 2016 by FromEarth
Enthalpy Posted July 13, 2016 Posted July 13, 2016 I don't know a software for it, but a rather simple (though not instantenously usable) tutorial is available on the Web: "The slingshot effect", R. C. Johnson, University of Durham Energy storage: achieving 4km/s over 400m leaves you 50ms, which is comfortable for aluminium capacitors but not for supercapacitors nor batteries. Have a look at datasheets; aluminium capacitors are comfortable with 350V each. You might try a flywheel with a compulsator, too; I compare the energy density of a flywheel with a pressure vessel there http://www.scienceforums.net/topic/59338-flywheels-store-electricity-cheap-enough/#entry863014 Raising a terrestrial station by lunar slingshot, I have a bad intuition. But you could try to capture a ferry by a lunar (orbital) station and send it back to exploit the Moon's speed without depending on the angles of a slingshot. The ferry would arrive slower than the Moon (as a normal consequence of transfer orbits), the station there get energy from the speed difference, and send the ferry back with an angle less backward than it arrived. I'm not quite sure something is gained, but at least you get some freedom from less coupled speed and angle. Then, the Lunar station will get an elliptic orbit, which can't be summed again and again, but maybe you can annihilate this effect by doing the next operation one-half Lunar orbit later, when the station's apoapsis and periapsis swap their roles. Not very clear to me. Sending the ferry is one difficult task, but do you have a technologically credible script about how to catch it at 3.2km/s?
FromEarth Posted July 13, 2016 Author Posted July 13, 2016 (edited) Sending the ferry is one difficult task, but do you have a technologically credible script about how to catch it at 3.2km/s? I was thinking of using the coilgun but in reverse, charging the capacitors (or flywheel as you suggested which is also a good way to store energy) and decelerating the spacecraft. Energy storage: achieving 4km/s over 400m leaves you 50ms How did you get 50 ms? I'm getting 200 ms: acceleration = velocity / time = 4000 m/s / 0.2 s = 20'000 m/s^2 distance = acceleration * time^2 / 2 = 20'000 m/s^2 * (0.2 s)^2 / 2 = 400 m Maybe we don't need to accelerate the spacecraft to 3.2 km/s in one go if we do it in small increments and with a longer coilgun. If we can split the acceleration in 3 and use a 1 km long coilgun, then the time could be increased to 2 seconds and reduce the acceleration from 2000 g, to only 50 g. Maybe a well trained astronaut could take 50 g for 2 seconds? Maybe it's possible to build an apparatus to keep the eyeballs in place and prevent them from escaping the eye sockets. Then, the Lunar station will get an elliptic orbit, which can't be summed again and again, but maybe you can annihilate this effect by doing the next operation one-half Lunar orbit later, when the station's apoapsis and periapsis swap their roles. Not very clear to me. Good idea. Maybe if we keep the slingshots happening at a constant rate, the orbit of the station will be circularized every 27 days (the orbital period of the moon). EDIT: if anyone is interested in seeing how the "double lunar gravity assist corn" was made, here is a youtube video I just made of the simulation using Bugale N-Body Simulator: The mass of the earth, the mass of the moon, the velocity of the moon, the distance of the moon from earth, the diameter of earth and the diameter of the moon are all simulated. In other words, that orbit is actually possible in real life. Btw, the yellow trail (behind the spacecraft) has a width of 400 km. EDIT 2: I think that, once we have a coilgun in space, all space debris becomes valuable because it can be deorbited and most of the orbital energy added to the station. Edited July 14, 2016 by FromEarth
Enthalpy Posted July 17, 2016 Posted July 17, 2016 How did you get 50 ms? I'm getting 200 ms: My bad. (very very bad). Maybe a well trained astronaut could take 50 g for 2 seconds? No, very strong accelerations around 20g have been survived, but the total speed difference was small, like 20m/s. No hope with 3km/s. I didn't consider astronauts here. I was thinking of using the coilgun but in reverse, charging the capacitors (or flywheel as you suggested which is also a good way to store energy) and decelerating the spacecraft. That's what I supposed from your post, but how does the craft thread into the railgun? I did such things at 70km/h, it was already damn difficult. Maybe we don't need to accelerate the spacecraft to 3.2 km/s in one go if we do it in small increments and with a longer coilgun.If we can split the acceleration in 3 and use a 1 km long coilgun, then the time could be increased to 2 seconds and reduce the acceleration from 2000 g, to only 50 g. That's unclear to me. First, how to split the acceleration? Bringing the craft back for a second run demands an acceleration too. Then, the length of a coilgun depends on the initial speed if it's not zero. Distances vary as time squared at constant acceleration because of that, so accelerating in three episodes doesn't cut the individual distance by three. If you decide that some craft (not the present day ones, and not inhabited) can resist 50g=500m/s2 and you want to achieve 3200m/s, the coilgun must be 10km long.
FromEarth Posted July 21, 2016 Author Posted July 21, 2016 (edited) That's what I supposed from your post, but how does the craft thread into the railgun? I did such things at 70km/h, it was already damn difficult. I think that's the most difficult part, one that I don't know how to solve. That's unclear to me. First, how to split the acceleration? Bringing the craft back for a second run demands an acceleration too. Then, the length of a coilgun depends on the initial speed if it's not zero. Distances vary as time squared at constant acceleration because of that, so accelerating in three episodes doesn't cut the individual distance by three. If you decide that some craft (not the present day ones, and not inhabited) can resist 50g=500m/s2 and you want to achieve 3200m/s, the coilgun must be 10km long. I was thinking of 3 orbits, where the spacecraft is accelerated by 1 km/s for every pass, but I forgot that in the next pass, the spacecraft will be going 1km/s faster relative to the station and thus there will be less time to accelerate by 1 km/s.... Anyway I think the concept of splitting the acceleration in multiple stages could allow us to decrease the acceleration required for the coilgun. Edited July 21, 2016 by FromEarth
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