FromEarth

I think you could touch things in a blackhole.

I think that's the most difficult part, one that I don't know how to solve. 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.

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. 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. 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.

The only way to see a clock run backwards is to go faster than light.

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.

I'm playing with Bugale N-Body Simulator and something unexpected happened.... Double lunar gravity assist corn ahahahahaha

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). 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.

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

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: